2.2 Linear expressions

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{{Selected tab|[[2.2 Linear expressions|Theory]]}}
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{{Not selected tab|[[2.2 Exercises|Exercises]]}}
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{{Info|
{{Info|
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'''Innehåll:'''
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'''Contents:'''
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*Förstagradsekvationer
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*First degree equations
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*Räta linjens ekvation
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* Equation of a straight line
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*Geometriska problem
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*Geometrical problems
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*Områden som definieras av olikheter
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*Regions that are defined using inequalities
}}
}}
{{Info|
{{Info|
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'''Lärandemål:'''
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'''Learning outcomes:'''
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Efter detta avsnitt ska du ha lärt dig att:
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After this section you will have learned how to:
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*Lösa algebraiska ekvationer som efter förenkling leder till förstagradsekvationer.
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*Solve algebraic equations, which after simplification results in first degree equations.
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*Omvandla mellan formerna ''y'' = ''kx'' + ''m'' och ''ax'' + ''by'' + ''c'' = 0.
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*Convert between the forms ''y'' = ''kx'' + ''m'' and ''ax'' + ''by'' + ''c'' = 0.
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*Skissera räta linjer utgående från ekvationen.
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*Sketch straight lines from their equation.
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*Lösa geometriska problem som innehåller räta linjer.
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* Solve geometric problems that contain straight lines.
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*Skissera områden som ges av linjära olikheter och bestämma arean av dessa.
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*Sketch regions defined by linear inequalities and determine the area of these regions.
}}
}}
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== Förstagradsekvationer ==
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== First degree equations ==
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För att lösa förstagradsekvationer (även kallade linjära ekvationer) utför vi räkneoperationer på båda leden samtidigt, som successivt förenklar ekvationen och till slut gör att vi får <math>x</math> ensamt i ena ledet.
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To solve first degree equations (also known as linear equations) we perform calculations on both sides simultaneously. This gradually simplifies the equation and ultimately leads to <math>x</math> being alone on one side of the equation.
<div class="exempel">
<div class="exempel">
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'''Exempel 1'''
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''' Example 1'''
<ol type="a">
<ol type="a">
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<li>Lös ekvationen <math>x+3=7</math>.<br/><br/>
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<li>Solve the equation <math>x+3=7</math>.<br/><br/>
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Subtrahera <math>3</math> från båda led
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Subtract <math>3</math> from both sides
:<math>x+3-3=7-3</math>.
:<math>x+3-3=7-3</math>.
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Vänsterledet förenklas då till <math>x</math> och vi får att
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The left-hand side then simplifies to <math>x</math> and we get
:<math>x=7-3=4</math>.</li>
:<math>x=7-3=4</math>.</li>
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<li>Lös ekvationen <math>3x=6</math>. <br/><br/>
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<li>Solve the equation <math>3x=6</math>. <br/><br/>
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Dividera båda led med <math>3</math>
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Divide both sides by <math>3</math>
:<math>\frac{3x}{3} = \frac{6}{3}\,</math>.
:<math>\frac{3x}{3} = \frac{6}{3}\,</math>.
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Efter att ha förkortat bort <math>3</math> i vänsterledet har vi att
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After having cancelled <math>3</math> on the left-hand side we have
:<math> x=\frac{6}{3} = 2</math>.</li>
:<math> x=\frac{6}{3} = 2</math>.</li>
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<li> Lös ekvationen <math>2x+1=5</math><br/><br/>
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<li> Solve the equation <math>2x+1=5\,\mbox{.}</math><br/><br/>
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Först subtraherar vi båda led med <math>1</math> för att få <math>2x</math> ensamt i vänsterledet
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First we subtract <math>1</math> from both sides to get <math>2x</math> on its own on the left-hand side
:<math>2x=5-1</math>.<br/>
:<math>2x=5-1</math>.<br/>
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Sedan dividerar vi båda led med <math>2</math> och får svaret
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We then divide both sides by <math>2</math> to get the answer
:<math>x = \frac{4}{2} = 2</math>.</li>
:<math>x = \frac{4}{2} = 2</math>.</li>
</ol>
</ol>
</div>
</div>
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En förstagradsekvation kan skrivas på normalformen <math>ax=b</math>. Lösningen är då helt enkelt <math>x=b/a</math> (man måste anta att <math>a\not=0</math>).
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A first degree equation can be written in the normal form <math>ax=b</math>. The solution is then simply <math>x=b/a</math> (we must assume that <math>a\not=0</math>).
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De eventuella svårigheter som kan uppstå när man läser en förstagradsekvation gäller alltså inte själva lösningsformeln utan snarare de förenklingar som kan behövas för att komma till normalformen. Här nedan visas några exempel som har det gemensamt att en ekvation förenklas till linjär normalform och därmed får en unik lösning.
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The possible difficulties that may occur when you solve a first degree equation are thus not in themselves the form of the solution, but rather the simplifications that may be needed to achieve the normal form. Below are a couple examples in which the equation can be simplified to a linear normal form, thus having a unique solution.
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<div class="exempel">
<div class="exempel">
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'''Exempel 2'''
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'''Example 2'''
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Lös ekvationen <math>\,2x-3=5x+7</math>.
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Solve the equation<math>\,2x-3=5x+7</math>.
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Eftersom <math>x</math> förekommer både i vänster- och högerledet subtraherar vi <math>2x</math> från båda led
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Since <math>x</math> occurs on both the left and right hand sides we subtract <math>2x</math> from both sides
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{{Fristående formel||<math>2x-3-2x=5x+7-2x</math>}}
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{{Displayed math||<math>2x-3-2x=5x+7-2x</math>}}
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och får <math>x</math> samlat i högerledet
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and now <math>x</math> only appears on the right-hand side
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{{Fristående formel||<math>-3 = 3x+7 \; \mbox{.}</math>}}
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{{Displayed math||<math>-3 = 3x+7 \; \mbox{.}</math>}}
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Nu subtraherar vi 7 från båda led
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We now subtract 7 from both sides
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{{Fristående formel||<math>-3 -7 = 3x +7-7</math>}}
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{{Displayed math||<math>-3 -7 = 3x +7-7</math>}}
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och får <math>3x</math> ensamt kvar i högerledet
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and get <math>3x</math> on its own on the right-hand side
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{{Fristående formel||<math>-10=3x\,\mbox{.}</math>}}
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{{Displayed math||<math>-10=3x\,\mbox{.}</math>}}
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Det sista steget är att dividera båda led med <math>3</math>
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The last step is to divide both sides by <math>3</math>
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{{Fristående formel||<math>\frac{-10}{3} = \frac{3x}{3}</math>}}
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{{Displayed math||<math>\frac{-10}{3} = \frac{3x}{3}</math>}}
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och detta ger att
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giving
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{{Fristående formel||<math>x=-\frac{10}{3}\,\mbox{.}</math>}}
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{{Displayed math||<math>x=-\frac{10}{3}\,\mbox{.}</math>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 3'''
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'''Example 3'''
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Lös ut <math>x</math> från ekvationen <math>ax+7=3x-b</math>.
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Solve for <math>x</math> the equation <math>ax+7=3x-b</math>.
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Genom att subtrahera båda led med <math>3x</math>
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By subtracting <math>3x</math> from both sides
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{{Fristående formel||<math>ax+7-3x=3x-b-3x</math>}}
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{{Displayed math||<math>ax+7-3x=3x-b-3x</math>}}
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{{Fristående formel||<math>ax+7-3x=-b</math>}}
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{{Displayed math||<math>ax+7-3x=\phantom{3x}{}-b\phantom{{}-3x}</math>}}
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och sedan med <math>7</math>
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and then subtracting <math>7</math>
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{{Fristående formel||<math>ax+7-3x -7=-b-7</math>}}
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{{Displayed math||<math>ax+7-3x -7=-b-7</math>}}
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{{Fristående formel||<math>ax-3x=-b-7</math>}}
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{{Displayed math||<math>ax\phantom{{}+7}{}-3x\phantom{{}-7}{}=-b-7</math>}}
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har vi samlat alla termer som innehåller <math>x</math> i vänsterledet och övriga termer i högerledet.
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we have gathered together all the terms that contain <math>x</math> on the left-hand side and all other terms on the right-hand side. Since the terms on the left-hand side have <math>x</math> as a common factor <math>x</math>, they can be factored out
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Eftersom termerna i vänsterledet har <math>x</math> som en gemensam faktor kan <math>x</math> brytas ut
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{{Displayed math||<math>(a-3)x = -b-7\; \mbox{.}</math>}}
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{{Fristående formel||<math>(a-3)x = -b-7\; \mbox{.}</math>}}
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Divide both sides with <math>a-3</math> giving
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Dividera båda led med <math>a-3</math>
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{{Displayed math||<math>x= \frac{-b-7}{a-3}\; \mbox{.}</math>}}
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{{Fristående formel||<math>x= \frac{-b-7}{a-3}\; \mbox{.}</math>}}
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</div>
</div>
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Det är inte alltid uppenbart att man har att göra med en förstagradsekvation. I följande två exempel förvandlas den ursprungliga ekvationen genom förenklingar till en förstagradsekvation.
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It is not always obvious that we are dealing with a first degree equation. In the following two examples simplifications turn the original equation into a first degree equation.
<div class="exempel">
<div class="exempel">
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'''Exempel 4'''
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''' Example 4'''
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Lös ekvationen <math>\ (x-3)^2+3x^2=(2x+7)^2</math>.
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Solve the equation<math>\ (x-3)^2+3x^2=(2x+7)^2</math>.
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Utveckla kvadratuttrycken i båda leden
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Expand the quadratic expressions on both sides
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{{Fristående formel||<math>x^2-6x+9+3x^2=4x^2+28x+49</math>}}
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{{Displayed math||<math>x^2-6x+9+3x^2=4x^2+28x+49\,\mbox{,}</math>}}
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{{Fristående formel||<math>4x^2-6x+9=4x^2+28x+49</math>}}
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{{Displayed math||<math>4x^2-6x+9=4x^2+28x+49\,\mbox{.}</math>}}
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Subtrahera <math>4x^2</math> från båda led
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Subtract <math>4x^2</math> from both sides
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{{Fristående formel||<math>-6x +9 = 28x +49\; \mbox{.}</math>}}
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{{Displayed math||<math>-6x +9 = 28x +49\; \mbox{.}</math>}}
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Addera <math>6x</math> till båda led
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Add <math>6x</math> to both sides
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{{Fristående formel||<math>9 = 34x +49\; \mbox{.}</math>}}
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{{Displayed math||<math>9 = 34x +49\; \mbox{.}</math>}}
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Subtrahera <math>49</math> från båda led
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Subtract <math>49</math> from both sides
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{{Fristående formel||<math>-40=34x\; \mbox{.}</math>}}
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{{Displayed math||<math>-40=34x\; \mbox{.}</math>}}
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Dividera båda led med <math>34</math>
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Divide both sides by <math>34</math>
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{{Fristående formel||<math>x = \frac{-40}{34}= - \frac{20}{17}\; \mbox{.}</math>}}
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{{Displayed math||<math>x = \frac{-40}{34}= - \frac{20}{17}\; \mbox{.}</math>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 5'''
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''' Example 5'''
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Lös ekvationen <math>\ \frac{x+2}{x^2+x} = \frac{3}{2+3x}</math>.
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Solve the equation <math>\ \frac{x+2}{x^2+x} = \frac{3}{2+3x}</math>.
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Flytta över båda termerna i ena ledet
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Collect both terms to one side
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{{Fristående formel||<math>\frac{x+2}{x^2+x}-\frac{3}{2+3x}= 0\; \mbox{.}</math>}}
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{{Displayed math||<math>\frac{x+2}{x^2+x}-\frac{3}{2+3x}= 0\; \mbox{.}</math>}}
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Förläng termerna så att de får samma nämnare
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Convert the terms so that they have the same denominator
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{{Fristående formel||<math>\frac{(x+2)(2+3x)}{(x^2+x)(2+3x)}-\frac{3(x^2+x)}{(2+3x)(x^2+x)}= 0</math>}}
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{{Displayed math||<math>\frac{(x+2)(2+3x)}{(x^2+x)(2+3x)}-\frac{3(x^2+x)}{(2+3x)(x^2+x)}= 0</math>}}
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och förenkla täljaren
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and simplify the numerator
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{{Fristående formel||<math>\frac{(x+2)(2+3x)-3(x^2+x)}{(x^2+x)(2+3x)} = 0,</math>}}
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{{Displayed math||<math>\frac{(x+2)(2+3x)-3(x^2+x)}{(x^2+x)(2+3x)} = 0,</math>}}
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{{Fristående formel||<math>\frac{3x^2+8x+4-(3x^2+3x)}{(x^2+x)(2+3x)} = 0,</math>}}
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{{Displayed math||<math>\frac{3x^2+8x+4-(3x^2+3x)}{(x^2+x)(2+3x)} = 0,</math>}}
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{{Fristående formel||<math>\frac{5x +4}{(x^2+x)(2+3x)} = 0\,\mbox{.}</math>}}
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{{Displayed math||<math>\frac{5x +4}{(x^2+x)(2+3x)} = 0\,\mbox{.}</math>}}
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Denna ekvation är uppfylld bara när täljaren är lika med noll (samtidigt som nämnaren inte är lika med noll),
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This equation is only satisfied when the numerator is equal to zero (whilst the denominator is not equal to zero);
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{{Fristående formel||<math>5x+4=0</math>}}
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{{Displayed math||<math>5x+4=0</math>}}
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vilket ger att <math>\,x = -\frac{4}{5}</math>.
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which gives that <math>\,x = -\frac{4}{5}</math>.
</div>
</div>
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== Räta linjer ==
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== Straight lines ==
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Funktioner av typen
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Functions such as
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{{Fristående formel||<math>y = 2x+1</math>}}
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{{Displayed math||<math>y = 2x+1</math>}}
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{{Fristående formel||<math>y = -x+3</math>}}
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{{Displayed math||<math>y = -x+3</math>}}
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{{Fristående formel||<math>y = \frac{1}{2} x -5 </math>}}
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{{Displayed math||<math>y = \frac{1}{2} x -5 </math>}}
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är exempel på linjära funktioner och de kan allmänt skrivas i formen
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are examples of linear functions. They have the general form
<div class="regel">
<div class="regel">
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{{Fristående formel||<math>y = kx+m</math>}}
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{{Displayed math||<math>y = kx+m</math>}}
</div>
</div>
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där <math>k</math> och <math>m</math> är konstanter.
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where <math>k</math> and <math>m</math> are constants.
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Grafen till en linjär funktion är alltid en rät linje och konstanten <math>k</math> anger linjens lutning mot <math>x</math>-axeln och <math>m</math> anger <math>y</math>-koordinaten för den punkt där linjen skär <math>y</math>-axeln.
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The graph of a linear function is always a straight line. The constant <math>k</math> indicates the slope of the line with respect to the <math>x</math>-axis and <math>m</math> gives the <math>y</math>-coordinate of the point where the line intersects the <math>y</math>-axis.
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<center>{{:2.2 - Figur - Linjen y = kx + m}}</center>
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<center>{{:2.2 - Figure - The line y = kx + m}}</center>
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<center><small>Linjen ''y'' = ''kx'' + ''m'' har lutning ''k'' och skär ''y''-axeln i (0,''m'')</small></center>
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<center><small>The line ''y'' = ''kx'' + ''m'' has slope ''k'' and cuts the ''y''-axis at (0,''m'')</small></center>
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Konstanten <math>k</math> kallas för linjens riktningskoefficient och innebär att en enhetsförändring i positiv <math>x</math>-led på linjen ger <math>k</math> enheters förändring i positiv <math>y</math>-led. Det gäller därmed att om
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The constant <math>k</math> is called the slope and states that given a unit change in the positive <math>x</math>-direction, along the line there is a <math>k</math> unit change in the positive <math>y</math>-direction. Thus if
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*<math>k>0\,</math> så lutar linjen uppåt
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*<math>k>0\,</math> the line slopes upwards,
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*<math>k<0\,</math> så lutar linjen nedåt
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*<math>k<0\,</math> the line slopes downwards.
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För en horisontell linje (parallell med <math>x</math>-axeln) är <math>k=0</math> medan en vertikal linje (parallell med <math>y</math>-axeln) inte har något <math>k</math>-värde (en sådan linje kan inte skrivas i formen <math>y=kx+m</math>).
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For a horizontal line (parallel to the <math>x</math>-axis) <math>k=0</math>, whereas a vertical line (parallel to the <math>y</math>-axis) does not have a <math>k</math> value ( a vertical line cannot be written in the form <math>y=kx+m</math>).
<div class="exempel">
<div class="exempel">
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'''Exempel 6'''
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''' Example 6'''
<ol type="a">
<ol type="a">
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<li> Skissera linjen <math>y=2x-1</math>. <br/><br/>
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<li> Sketch the line <math>y=2x-1</math>. <br/><br/>
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Jämför vi linjens ekvation med <math>y=kx+m</math> så ser vi att <math>k=2</math> och <math>m=-1</math>. Detta betyder att linjens riktningskoefficient är <math>2</math> och att den skär <math>y</math>-axeln i punkten <math>(0,-1)</math>. Se figuren till vänster nedan.</li>
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Comparing with the standard equation <math>y=kx+m</math>, we see that <math>k=2</math> and <math>m=-1</math>. This gives the line's slope as <math>2</math> and that it cuts the <math>y</math>-axis at <math>(0,-1)</math>. See the figure below-left.</li>
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<li>Skissera linjen <math>y=2-\tfrac{1}{2}x</math>.<br/><br/>
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<li>Sketch the line <math>y=2-\tfrac{1}{2}x</math>.<br/><br/>
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Linjens ekvation kan skrivas som <math>y= -\tfrac{1}{2}x + 2</math> och då ser vi att dess riktningskoefficient är <math>k= -\tfrac{1}{2}</math> och att <math>m=2</math>. Se figuren nedan till höger.
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The equation of the line can be written as <math>y= -\tfrac{1}{2}x + 2</math>. From this we see that has the slope <math>k= -\tfrac{1}{2}</math>and that <math>m=2</math>. See the figure below to the right.
</ol>
</ol>
{| align="center" padding="20px"
{| align="center" padding="20px"
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|align="center"|{{:2.2 - Figur - Linjen y = 2x - 1}}
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|align="center"|{{:2.2 - Figure - The line y = 2x - 1}}
|width="20px"|
|width="20px"|
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|align="center"|{{:2.2 - Figur - Linjen y = 2 - x/2}}
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|align="center"|{{:2.2 - Figure - The line y = 2 - x/2}}
|-
|-
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|align="center"|<small>Linjen ''y'' = 2''x'' - 1</small>
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|align="center"|<small>Line ''y'' = 2''x'' - 1</small>
||
||
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|align="center"|<small>Linjen ''y'' = 2 - ''x''/2</small>
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|align="center"|<small>Line ''y'' = 2 - ''x''/2</small>
|}
|}
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 7'''
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''' Example 7'''
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Vilken riktningskoefficient har den räta linje som går genom punkterna <math>(2,1)</math> och <math>(5,3)</math>?
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What is the slope of the straight line that passes through the points <math>(2,1)</math> and <math>(5,3)</math>?
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Ritar vi upp punkterna och linjen i ett koordinatsystem så ser vi att <math>5-2=3</math> steg i <math>x</math>-led motsvaras av <math>3-1=2</math> steg i <math>y</math>-led på linjen. Det betyder att <math>1</math> steg i <math>x</math>-led måste motsvaras av <math>k=\frac{3-1}{5-2}= \frac{2}{3}</math> steg i <math>y</math>-led. Alltså är linjens riktningskoefficient <math>k= \frac{2}{3}</math>.
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If we plot the points and draw the line in a coordinate system, we see that <math>5-2=3</math> steps in the <math>x</math>-direction corresponds to <math>3-1=2</math> steps in the <math>y</math>-direction along the line. This means that <math>1</math> step in the <math>x</math>-direction corresponds to <math>k=\frac{3-1}{5-2}= \frac{2}{3}</math> steps in the <math>y</math>-direction. So the line's slope is <math>k= \frac{2}{3}</math>.
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<center>{{:2.2 - Figur - Linje genom punkterna (2,1) och (5,3)}}</center>
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<center>python</center>
</div>
</div>
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Två räta linjer som är parallella har uppenbarligen samma riktningskoefficient. Det går också att se (t.ex. i figuren nedan) att två linjer som är vinkelräta har riktningskoefficienter <math>k_1</math> respektive <math>k_2</math> som uppfyller <math>k_2 = -\frac{1}{k_1}</math>, vilket också kan skrivas som <math>k_1 k_2 = -1</math>.
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Two straight lines that are parallel clearly have the same slope. It is also possible to see (such as in the figure below) that for two lines with slopes <math>k_1</math> and <math>k_2</math> that are perpendicular to one another, then <math>k_2 = -\frac{1}{k_1}</math>, which also can be written as <math>k_1 k_2 = -1</math>.
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<center>{{:2.2 - Figur - Riktningskoefficient för vinkelräta linjer}}</center>
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<center>{{:2.2 - Figure - The slope of perpendicular lines}}</center>
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Den räta linjen i figuren till vänster har riktningskoefficient <math>k</math>, dvs. <math>1</math> steg i <math>x</math>-led motsvaras av <math>k</math> steg i <math>y</math>-led. Om linjen vrids <math>90^\circ</math> motsols får vi linjen i figuren till höger, och den linjen har riktningskoefficient <math>-\frac{1}{k}</math> eftersom nu motsvaras <math>-k</math> steg i <math>x</math>-led av <math>1</math> steg i <math>y</math>-led.
+
The straight line in the figure on the left has slope <math>k</math>, that is <math>1</math> step in the <math>x</math>-direction corresponds to <math>k</math> steps in the <math>y</math>-direction. If the line is rotated <math>90^\circ</math> clockwise we get the line in the figure to the right. That line has slope <math>-\frac{1}{k}</math> because now <math>-k</math> steps in the <math>x</math>-direction corresponds to <math>1</math> step in the <math>y</math>-direction
<div class="exempel">
<div class="exempel">
-
'''Exempel 8'''
+
''' Example 8'''
<ol type="a">
<ol type="a">
-
<li>Linjerna <math>y=3x-1</math> och <math>y=3x+5</math> är parallella.
+
<li> The lines <math>y=3x-1</math> and <math>y=3x+5</math> are parallel.
-
<li>Linjerna <math>y=x+1</math> och <math>y=2-x</math> är vinkelräta.
+
<li> The lines <math>y=x+1</math> and <math>y=2-x</math> are perpendicular.
</ol>
</ol>
</div>
</div>
-
Alla räta linjer (även den vertikala linjen) kan skrivas i den allmänna formen
+
All straight lines (including vertical lines) can be put into the general form
<div class="regel">
<div class="regel">
-
{{Fristående formel||<math>ax+by=c</math>}}
+
{{Displayed math||<math>ax+by=c</math>}}
</div>
</div>
-
där <math>a</math>, <math>b</math> och <math>c</math> är konstanter.
+
where <math>a</math>, <math>b</math> and <math>c</math> are constants.
<div class="exempel">
<div class="exempel">
-
'''Exempel 9'''
+
''' Example 9'''
<ol type="a">
<ol type="a">
-
<li>Skriv linjen <math>y=5x+7</math> i formen <math>ax+by=c</math>.<br/><br/>
+
<li>Put the line <math>y=5x+7</math> into the form <math>ax+by=c</math>.<br/><br/>
-
Flytta över <math>x</math>-termen till vänsterledet <math>-5x+y=7</math>.</li>
+
Move the <math>x</math>-term to the left-hand side: <math>-5x+y=7</math>.</li>
-
<li>Skriv linjen <math>2x+3y=-1</math> i formen <math>y=kx+m</math>.<br/><br/>
+
<li> Put the line <math>2x+3y=-1</math> into the form <math>y=kx+m</math>.<br/><br/>
-
Flytta över <math>x</math>-termen i högerledet <math>3y=-2x-1</math> och dela båda led med <math>3</math>
+
Move the <math>x</math>-term to the right-hand side <math>3y=-2x-1 </math> and divide both sides by <math>3</math>
-
{{Fristående formel||<math>y=-\frac{2}{3}x - \frac{1}{3}\,\mbox{.}</math>}}
+
{{Displayed math||<math>y=-\frac{2}{3}x - \frac{1}{3}\,\mbox{.}</math>}}
</ol>
</ol>
</div>
</div>
-
[http://www.cut-the-knot.org/Curriculum/Calculus/StraightLine.shtml '''Här'''] kan du se hur linjens ekvation kan skrivas utifrån att man vet koordinaterna för två punkter på linjen.
+
[http://www.cut-the-knot.org/Curriculum/Calculus/StraightLine.shtml '''Here'''] you can see how an equation for a line can be obtained if we know the coordinates of two points on the line.
-
[http://www.theducation.se/hemsida//gymnasium_komvux/webbaserade_laromedel_och_webbstod/matematik_3000/experimentera_med_den_rata_linjen/index.asp '''Här'''] kan du ändra på k och m och se hur detta påverkar linjens egenskaper.
+
[http://www.theducation.se/hemsida//gymnasium_komvux/webbaserade_laromedel_och_webbstod/matematik_3000/experimentera_med_den_rata_linjen/index.asp '''Here'''] you can vary k and m and see how this affects the line's characteristics.
-
== Områden i koordinatsystem ==
+
==Regions in a coordinate system ==
-
Genom att tolka olikheter geometriskt kan de användas för att beskriva områden i planet.
+
By geometrically interpreting inequalities one can describe regions in the plane.
<div class="exempel">
<div class="exempel">
-
'''Exempel 10'''
+
''' Example 10'''
<ol type="a">
<ol type="a">
-
<li>Skissera området i <math>x,y</math>-planet som uppfyller <math>y\ge2</math>. <br/><br/>
+
<li>Sketch the region in the <math>x,y</math>-plane that satisfies <math>y\ge2</math>. <br/><br/>
-
Området ges av alla punkter <math>(x,y)</math> vars <math>y</math>-koordinat är <math>2</math> eller större, dvs. alla punkter på eller ovanför linjen <math>y=2</math>.<br/>
+
The region is given by all the points <math>(x,y)</math> for which the <math>y</math>-coordinate is equal or greater than <math>2</math> that is all points on or above the line <math>y=2</math>.<br/>
-
<center>{{:2.2 - Figur - Området y ≥ 2}}</center></li>
+
<center>{{:2.2 - Figure - The region y ≥ 2}}</center></li>
-
<li>Skissera området i <math>x,y</math>-planet som uppfyller <math>y < x</math>. <br/><br/>
+
<li>Sketch the region in the <math>x,y</math>-plane that satisfies <math>y < x</math>. <br/><br/>
-
En punkt <math>(x,y)</math> som uppfyller olikheten <math>y < x</math> har en <math>x</math>-koordinat som är större än dess <math>y</math>-koordinat. Området består alltså av alla punkter till höger om linjen <math>y=x</math>.<br/>
+
A point <math>(x,y)</math> that satisfies the inequality <math>y < x</math> must have an <math>x</math>-coordinate that is larger than its <math>y</math>-coordinate. Thus the area consists of all the points to the right of the line <math>y=x</math>.<br/>
-
<center>{{:2.2 - Figur - Området y mindre än x}}</center>
+
<center>{{:2.2 - Figure - The region y less than x}}</center>
-
Att linjen <math>y=x</math> är streckad betyder att punkterna på linjen inte tillhör det färgade området.
+
The fact that the line <math>y=x</math> is dashed means that the points on the line do not belong to the coloured area.
</ol>
</ol>
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 11'''
+
''' Example 11'''
-
Skissera området i <math>x,y</math>-planet som uppfyller <math>2 \le 3x+2y\le 4</math>.
+
Sketch the region in the <math>x,y</math>-plane that satisfies <math>2 \le 3x+2y\le 4</math>.
-
Den dubbla olikheten kan delas upp i två olikheter
+
The double inequality can be divided into two inequalities
-
{{Fristående formel||<math>3x+2y \ge 2 \quad</math> och <math>\quad 3x+2y\le4 \;\mbox{.}</math>}}
+
{{Displayed math||<math>3x+2y \ge 2 \quad</math> and <math>\quad 3x+2y\le4 \;\mbox{.}</math>}}
-
Flyttar vi över <math>x</math>-termerna till högerledet och delar båda led med <math>2</math> får vi
+
We move the <math>x</math>-terms into the right-hand side and divide both sides by <math>2</math> giving
-
{{Fristående formel||<math>y \ge 1-\frac{3}{2}x \quad</math> och <math>\quad y\le 2-\frac{3}{2}x \;\mbox{.}</math>}}
+
{{Displayed math||<math>y \ge 1-\frac{3}{2}x \quad</math> and <math>\quad y\le 2-\frac{3}{2}x \;\mbox{.}</math>}}
-
De punkter som uppfyller den första olikheten ligger på och ovanför linjen <math>y \ge 1-\tfrac{3}{2}x</math> medan de punkter som uppfyller den andra olikheten ligger på eller under linjen <math>y\le 2-\tfrac{3}{2}x</math>.
+
The points that satisfy the first inequality are on and above the line <math>y = 1-\tfrac{3}{2}x</math> while the points that satisfy the other inequality are on or below the line <math>y = 2-\tfrac{3}{2}x</math>.
-
<center>{{:2.2 - Figur - Områdena 3x + 2y ≥ 2 och 3x + 2y ≤ 4}}</center>
+
<center>{{:2.2 - Figure - The regions 3x + 2y ≥ 2 and 3x + 2y ≤ 4}}</center>
-
<center><small>Figuren till vänster visar området <math>3x+2y\ge 2</math> och figuren till höger området <math>3x+2y\le 4</math>.</small></center>
+
<center><small>The figure on the left shows the region <math>3x+2y\ge 2</math> and figure to the right shows the region <math>3x+2y\le 4</math>.</small></center>
-
Punkter som uppfyller båda olikheterna tillhör det bandformade område som de färgade områdena ovan har gemensamt.
+
Points that satisfy both inequalities form a band-like region where both coloured areas overlap.
-
<center>{{:2.2 - Figur - Området 2 ≤ 3x + 2y ≤ 4}}</center>
+
<center>{{:2.2 - Figure - The region 2 ≤ 3x + 2y ≤ 4}}</center>
-
<center><small>Figuren visar området <math>2\le 3x+2y\le 4</math>.</small></center>
+
<center><small>The figure shows the region <math>2\le 3x+2y\le 4</math>.</small></center>
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 12'''
+
''' Example 12'''
-
Om vi ritar upp linjerna <math>y=x</math>, <math>y=-x</math> och <math>y=2</math> så begränsar dessa linjer en triangel, i koordinatsystemet.
+
If we draw the lines <math>y=x</math>, <math>y=-x</math> and <math>y=2</math> then these lines bound a triangle in a coordinate system.
-
<center>{{:2.2 - Figur - Triangel begränsad av y = x, y = 2 och y = -x}}</center>
+
<center>{{:2.2 - Figure - The triangle bounded by y = x, y = 2 and y = -x}}</center>
-
Vi upptäcker att för att en punkt skall ligga i denna triangel så måste vi sätta en del krav på den.
+
We find that for a point to lie in this triangle, it has to satisfy certain conditions.
-
Vi ser att dess <math>y</math>-koordinat måste vara mindre än <math>2</math>. Samtidigt ser vi att triangeln nedåt begränas av <math> y=0</math>.
+
We see that its <math>y</math>-coordinate must be less than <math>2</math>. At the same time, we see that the triangle is bounded by <math> y=0</math> below. Thus the <math>y</math> coordinates must be in the range <math> 0\le y\le2</math>.
-
<math>y</math>-koordinaten måste således ligga i intervallet <math> 0\le y\le2</math>.
+
For the <math>x</math>-coordinate the situation is a little more complicated.
 +
We see that the <math>x</math>-coordinate must satisfy the fact that all points lie above the lines <math>y=-x</math> and <math>y=x</math>. We see that this is satisfied if <math>-y\le x\le y</math>. Since we already have restricted the <math>y</math>-coordinates we find that <math>x</math> cannot be larger than <math>2</math>
 +
or less than <math>-2</math>.
-
För <math>x</math>-koordinaten blir det lite mer komplicerat. Vi ser att <math>x</math>-koordinaten måste ligga ovanför linjerna <math>y=-x</math> och <math>y=x</math>. Vi ser att detta är uppfyllt då <math>-y\le x\le y</math>.
+
This gives the base of the triangle as <math>4</math> units of length and a height of <math>2</math> units of length.
-
 
+
-
Eftersom vi redan har begränsningar för <math>y</math>-koordinaten så ser vi att <math>x</math> inte kan vara större än <math>2</math>
+
-
och mindre än <math>-2</math> automatiskt.
+
-
 
+
-
Vi ser att basen i triangeln blir <math>4</math> längdenheter och höjden <math>2</math> längdenheter.
+
-
Arean av denna triangel blir alltså <math> 4\cdot 2/2=4</math> areaenheter.
+
The area of this triangle is therefore <math> 4\cdot 2/2=4</math> square units.
</div>
</div>
-
[[2.2 Övningar|Övningar]]
+
[[2.2 Exercises|Exercises]]
-
<div class="inforuta">
+
<div class="inforuta" style="width:580px;">
-
'''Råd för inläsning'''
+
'''Study advice'''
-
'''Grund- och slutprov'''
+
'''Basic and final tests'''
-
Efter att du har läst texten och arbetat med övningarna ska du göra grund- och slutprovet för att bli godkänd på detta avsnitt. Du hittar länken till proven i din student lounge.
+
After you have read the text and worked through the exercises you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.
-
'''Tänk på att...'''
+
'''Keep in mind that... '''
-
Rita egna figurer när du löser geometriska problem och att vara noggrann när du ritar! En bra figur kan vara halva lösningen, men en dålig figur kan lura en.
+
You should draw your own diagrams when you solve geometrical problems and draw them carefully and accurately! A good diagram can mean you are halfway to a solution, however a poor diagram may well fool you.
-
'''Lästips'''
+
<!--
-
 
+
'''Reviews'''
-
för dig som vill fördjupa dig ytterligare eller behöver en längre förklaring vill vi tipsa om:
+
-
[http://matmin.kevius.com/linje.html Läs mer om räta linjens ekvation i Bruno Kevius matematiska ordlista]
+
For those of you who want to deepen your studies or need a more detailed explanation, consider the following references:
 +
[http://matmin.kevius.com/linje.html Learn more about linear equations in Bruno Kevius mathematical glossary (Swedish).]
 +
-->
-
'''Länktips'''
+
'''Useful web sites'''
-
[http://www.cut-the-knot.org/Curriculum/Calculus/StraightLine.shtml Experimentera med Räta linjens ekvation]
+
[http://www.cut-the-knot.org/Curriculum/Calculus/StraightLine.shtml Experiment with Equations of a Straight Line]
-
[http://www.cut-the-knot.org/Curriculum/Geometry/ArchimedesTriangle.shtml Experimentera med Archimedes triangel & andragradskurvor ]
+
[http://www.cut-the-knot.org/Curriculum/Geometry/ArchimedesTriangle.shtml Experiment with Archimedes Triangle & Squaring of Parabola.]
</div>
</div>

Current revision

       Theory          Exercises      

Contents:

  • First degree equations
  • Equation of a straight line
  • Geometrical problems
  • Regions that are defined using inequalities

Learning outcomes:

After this section you will have learned how to:

  • Solve algebraic equations, which after simplification results in first degree equations.
  • Convert between the forms y = kx + m and ax + by + c = 0.
  • Sketch straight lines from their equation.
  • Solve geometric problems that contain straight lines.
  • Sketch regions defined by linear inequalities and determine the area of these regions.

First degree equations

To solve first degree equations (also known as linear equations) we perform calculations on both sides simultaneously. This gradually simplifies the equation and ultimately leads to \displaystyle x being alone on one side of the equation.

Example 1

  1. Solve the equation \displaystyle x+3=7.

    Subtract \displaystyle 3 from both sides
    \displaystyle x+3-3=7-3.
    The left-hand side then simplifies to \displaystyle x and we get
    \displaystyle x=7-3=4.
  2. Solve the equation \displaystyle 3x=6.

    Divide both sides by \displaystyle 3
    \displaystyle \frac{3x}{3} = \frac{6}{3}\,.
    After having cancelled \displaystyle 3 on the left-hand side we have
    \displaystyle x=\frac{6}{3} = 2.
  3. Solve the equation \displaystyle 2x+1=5\,\mbox{.}

    First we subtract \displaystyle 1 from both sides to get \displaystyle 2x on its own on the left-hand side
    \displaystyle 2x=5-1.
    We then divide both sides by \displaystyle 2 to get the answer
    \displaystyle x = \frac{4}{2} = 2.

A first degree equation can be written in the normal form \displaystyle ax=b. The solution is then simply \displaystyle x=b/a (we must assume that \displaystyle a\not=0). The possible difficulties that may occur when you solve a first degree equation are thus not in themselves the form of the solution, but rather the simplifications that may be needed to achieve the normal form. Below are a couple examples in which the equation can be simplified to a linear normal form, thus having a unique solution.

Example 2

Solve the equation\displaystyle \,2x-3=5x+7.


Since \displaystyle x occurs on both the left and right hand sides we subtract \displaystyle 2x from both sides

\displaystyle 2x-3-2x=5x+7-2x

and now \displaystyle x only appears on the right-hand side

\displaystyle -3 = 3x+7 \; \mbox{.}

We now subtract 7 from both sides

\displaystyle -3 -7 = 3x +7-7

and get \displaystyle 3x on its own on the right-hand side

\displaystyle -10=3x\,\mbox{.}

The last step is to divide both sides by \displaystyle 3

\displaystyle \frac{-10}{3} = \frac{3x}{3}

giving

\displaystyle x=-\frac{10}{3}\,\mbox{.}

Example 3

Solve for \displaystyle x the equation \displaystyle ax+7=3x-b.


By subtracting \displaystyle 3x from both sides

\displaystyle ax+7-3x=3x-b-3x
\displaystyle ax+7-3x=\phantom{3x}{}-b\phantom{{}-3x}

and then subtracting \displaystyle 7

\displaystyle ax+7-3x -7=-b-7
\displaystyle ax\phantom{{}+7}{}-3x\phantom{{}-7}{}=-b-7

we have gathered together all the terms that contain \displaystyle x on the left-hand side and all other terms on the right-hand side. Since the terms on the left-hand side have \displaystyle x as a common factor \displaystyle x, they can be factored out

\displaystyle (a-3)x = -b-7\; \mbox{.}

Divide both sides with \displaystyle a-3 giving

\displaystyle x= \frac{-b-7}{a-3}\; \mbox{.}

It is not always obvious that we are dealing with a first degree equation. In the following two examples simplifications turn the original equation into a first degree equation.

Example 4

Solve the equation\displaystyle \ (x-3)^2+3x^2=(2x+7)^2.


Expand the quadratic expressions on both sides

\displaystyle x^2-6x+9+3x^2=4x^2+28x+49\,\mbox{,}
\displaystyle 4x^2-6x+9=4x^2+28x+49\,\mbox{.}

Subtract \displaystyle 4x^2 from both sides

\displaystyle -6x +9 = 28x +49\; \mbox{.}

Add \displaystyle 6x to both sides

\displaystyle 9 = 34x +49\; \mbox{.}

Subtract \displaystyle 49 from both sides

\displaystyle -40=34x\; \mbox{.}

Divide both sides by \displaystyle 34

\displaystyle x = \frac{-40}{34}= - \frac{20}{17}\; \mbox{.}

Example 5

Solve the equation \displaystyle \ \frac{x+2}{x^2+x} = \frac{3}{2+3x}.


Collect both terms to one side

\displaystyle \frac{x+2}{x^2+x}-\frac{3}{2+3x}= 0\; \mbox{.}

Convert the terms so that they have the same denominator

\displaystyle \frac{(x+2)(2+3x)}{(x^2+x)(2+3x)}-\frac{3(x^2+x)}{(2+3x)(x^2+x)}= 0

and simplify the numerator

\displaystyle \frac{(x+2)(2+3x)-3(x^2+x)}{(x^2+x)(2+3x)} = 0,
\displaystyle \frac{3x^2+8x+4-(3x^2+3x)}{(x^2+x)(2+3x)} = 0,
\displaystyle \frac{5x +4}{(x^2+x)(2+3x)} = 0\,\mbox{.}

This equation is only satisfied when the numerator is equal to zero (whilst the denominator is not equal to zero);

\displaystyle 5x+4=0

which gives that \displaystyle \,x = -\frac{4}{5}.


Straight lines

Functions such as

\displaystyle y = 2x+1
\displaystyle y = -x+3
\displaystyle y = \frac{1}{2} x -5

are examples of linear functions. They have the general form

\displaystyle y = kx+m

where \displaystyle k and \displaystyle m are constants.

The graph of a linear function is always a straight line. The constant \displaystyle k indicates the slope of the line with respect to the \displaystyle x-axis and \displaystyle m gives the \displaystyle y-coordinate of the point where the line intersects the \displaystyle y-axis.

[Image]

The line y = kx + m has slope k and cuts the y-axis at (0,m)

The constant \displaystyle k is called the slope and states that given a unit change in the positive \displaystyle x-direction, along the line there is a \displaystyle k unit change in the positive \displaystyle y-direction. Thus if

  • \displaystyle k>0\, the line slopes upwards,
  • \displaystyle k<0\, the line slopes downwards.

For a horizontal line (parallel to the \displaystyle x-axis) \displaystyle k=0, whereas a vertical line (parallel to the \displaystyle y-axis) does not have a \displaystyle k value ( a vertical line cannot be written in the form \displaystyle y=kx+m).

Example 6

  1. Sketch the line \displaystyle y=2x-1.

    Comparing with the standard equation \displaystyle y=kx+m, we see that \displaystyle k=2 and \displaystyle m=-1. This gives the line's slope as \displaystyle 2 and that it cuts the \displaystyle y-axis at \displaystyle (0,-1). See the figure below-left.
  2. Sketch the line \displaystyle y=2-\tfrac{1}{2}x.

    The equation of the line can be written as \displaystyle y= -\tfrac{1}{2}x + 2. From this we see that has the slope \displaystyle k= -\tfrac{1}{2}and that \displaystyle m=2. See the figure below to the right.

[Image]

[Image]

Line y = 2x - 1 Line y = 2 - x/2

Example 7

What is the slope of the straight line that passes through the points \displaystyle (2,1) and \displaystyle (5,3)?


If we plot the points and draw the line in a coordinate system, we see that \displaystyle 5-2=3 steps in the \displaystyle x-direction corresponds to \displaystyle 3-1=2 steps in the \displaystyle y-direction along the line. This means that \displaystyle 1 step in the \displaystyle x-direction corresponds to \displaystyle k=\frac{3-1}{5-2}= \frac{2}{3} steps in the \displaystyle y-direction. So the line's slope is \displaystyle k= \frac{2}{3}.

python

Two straight lines that are parallel clearly have the same slope. It is also possible to see (such as in the figure below) that for two lines with slopes \displaystyle k_1 and \displaystyle k_2 that are perpendicular to one another, then \displaystyle k_2 = -\frac{1}{k_1}, which also can be written as \displaystyle k_1 k_2 = -1.

[Image]

The straight line in the figure on the left has slope \displaystyle k, that is \displaystyle 1 step in the \displaystyle x-direction corresponds to \displaystyle k steps in the \displaystyle y-direction. If the line is rotated \displaystyle 90^\circ clockwise we get the line in the figure to the right. That line has slope \displaystyle -\frac{1}{k} because now \displaystyle -k steps in the \displaystyle x-direction corresponds to \displaystyle 1 step in the \displaystyle y-direction

Example 8

  1. The lines \displaystyle y=3x-1 and \displaystyle y=3x+5 are parallel.
  2. The lines \displaystyle y=x+1 and \displaystyle y=2-x are perpendicular.

All straight lines (including vertical lines) can be put into the general form

\displaystyle ax+by=c

where \displaystyle a, \displaystyle b and \displaystyle c are constants.

Example 9

  1. Put the line \displaystyle y=5x+7 into the form \displaystyle ax+by=c.

    Move the \displaystyle x-term to the left-hand side: \displaystyle -5x+y=7.
  2. Put the line \displaystyle 2x+3y=-1 into the form \displaystyle y=kx+m.

    Move the \displaystyle x-term to the right-hand side \displaystyle 3y=-2x-1 and divide both sides by \displaystyle 3
    \displaystyle y=-\frac{2}{3}x - \frac{1}{3}\,\mbox{.}

Here you can see how an equation for a line can be obtained if we know the coordinates of two points on the line.

Here you can vary k and m and see how this affects the line's characteristics.


Regions in a coordinate system

By geometrically interpreting inequalities one can describe regions in the plane.

Example 10

  1. Sketch the region in the \displaystyle x,y-plane that satisfies \displaystyle y\ge2.

    The region is given by all the points \displaystyle (x,y) for which the \displaystyle y-coordinate is equal or greater than \displaystyle 2 that is all points on or above the line \displaystyle y=2.

    [Image]

  2. Sketch the region in the \displaystyle x,y-plane that satisfies \displaystyle y < x.

    A point \displaystyle (x,y) that satisfies the inequality \displaystyle y < x must have an \displaystyle x-coordinate that is larger than its \displaystyle y-coordinate. Thus the area consists of all the points to the right of the line \displaystyle y=x.

    [Image]

    The fact that the line \displaystyle y=x is dashed means that the points on the line do not belong to the coloured area.

Example 11

Sketch the region in the \displaystyle x,y-plane that satisfies \displaystyle 2 \le 3x+2y\le 4.


The double inequality can be divided into two inequalities

\displaystyle 3x+2y \ge 2 \quad and \displaystyle \quad 3x+2y\le4 \;\mbox{.}

We move the \displaystyle x-terms into the right-hand side and divide both sides by \displaystyle 2 giving

\displaystyle y \ge 1-\frac{3}{2}x \quad and \displaystyle \quad y\le 2-\frac{3}{2}x \;\mbox{.}

The points that satisfy the first inequality are on and above the line \displaystyle y = 1-\tfrac{3}{2}x while the points that satisfy the other inequality are on or below the line \displaystyle y = 2-\tfrac{3}{2}x.

[Image]

The figure on the left shows the region \displaystyle 3x+2y\ge 2 and figure to the right shows the region \displaystyle 3x+2y\le 4.


Points that satisfy both inequalities form a band-like region where both coloured areas overlap.

[Image]

The figure shows the region \displaystyle 2\le 3x+2y\le 4.

Example 12

If we draw the lines \displaystyle y=x, \displaystyle y=-x and \displaystyle y=2 then these lines bound a triangle in a coordinate system.

[Image]

We find that for a point to lie in this triangle, it has to satisfy certain conditions.

We see that its \displaystyle y-coordinate must be less than \displaystyle 2. At the same time, we see that the triangle is bounded by \displaystyle y=0 below. Thus the \displaystyle y coordinates must be in the range \displaystyle 0\le y\le2.

For the \displaystyle x-coordinate the situation is a little more complicated. We see that the \displaystyle x-coordinate must satisfy the fact that all points lie above the lines \displaystyle y=-x and \displaystyle y=x. We see that this is satisfied if \displaystyle -y\le x\le y. Since we already have restricted the \displaystyle y-coordinates we find that \displaystyle x cannot be larger than \displaystyle 2 or less than \displaystyle -2.

This gives the base of the triangle as \displaystyle 4 units of length and a height of \displaystyle 2 units of length.

The area of this triangle is therefore \displaystyle 4\cdot 2/2=4 square units.


Exercises

Study advice

Basic and final tests

After you have read the text and worked through the exercises you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.


Keep in mind that...

You should draw your own diagrams when you solve geometrical problems and draw them carefully and accurately! A good diagram can mean you are halfway to a solution, however a poor diagram may well fool you.


Useful web sites

Experiment with Equations of a Straight Line

Experiment with Archimedes Triangle & Squaring of Parabola.

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