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__NOTOC__ {| border="0" cellspacing="0" cellpadding="0" height="30" width="100%" | style="border-bottom:1px solid #797979" width="5px" | &nbsp; {{Vald flik|[[2.2 Linjära uttryck|Theory]]}} {{Ej vald flik|[[2.2 Övningar|Exercises]]}} | style="border-bottom:1px solid #797979" width="100%"| &nbsp; |} {{Info| '''Contents:''' *First degree equations * Equation of a straight line *Geometrical problems *Regions that are defined using inequalities }} {{Info| '''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'' &#061; ''kx'' + ''m'' and ''ax'' + ''by'' + ''c'' &#061; 0. *Sketch straight lines from their equation. * Solve geometric problems which 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 calculation on both sides simultaneously, which gradually simplifies the equation and ultimately leads to <math>x</math> being alone on one side of the equation. <div class="exempel"> ''' Example 1''' <ol type="a"> <li>Solve the equation <math>x+3=7</math>.<br/><br/> Subtract <math>3</math> from both sides :<math>x+3-3=7-3</math>. The left-hand side then simplifies to <math>x</math> , and we get :<math>x=7-3=4</math>.</li> <li>Solve the equation <math>3x=6</math>. <br/><br/> Divide both sides by <math>3</math> :<math>\frac{3x}{3} = \frac{6}{3}\,</math>. After having cancelled <math>3</math> on the left-hand side, we have :<math> x=\frac{6}{3} = 2</math>.</li> <li> Solve the equation <math>2x+1=5\,\mbox{.}</math><br/><br/> 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/> Then we divide both sides by <math>2</math> and get the answer :<math>x = \frac{4}{2} = 2</math>.</li> </ol> </div> 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>). 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 that have in common that an equation can be simplified to linear normal form and thus has a unique solution. <div class="exempel"> '''Example 2''' Solve the equation<math>\,2x-3=5x+7</math>. Since <math>x</math> occurs on both the left- and right-hand sides we subtract <math>2x</math> from both sides {{Fristående formel||<math>2x-3-2x=5x+7-2x</math>}} and now <math>x</math> only appears on the right-hand side {{Fristående formel||<math>-3 = 3x+7 \; \mbox{.}</math>}} Now we subtract 7 from both sides {{Fristående formel||<math>-3 -7 = 3x +7-7</math>}} and get <math>3x</math> on its own on the right-hand side {{Fristående formel||<math>-10=3x\,\mbox{.}</math>}} The last step is to divide both sides by <math>3</math> {{Fristående formel||<math>\frac{-10}{3} = \frac{3x}{3}</math>}} and this gives that {{Fristående formel||<math>x=-\frac{10}{3}\,\mbox{.}</math>}} </div> <div class="exempel"> '''Example 3''' Solve for <math>x</math> in the equation <math>ax+7=3x-b</math>. By subtracting <math>3x</math> from both sides {{Fristående formel||<math>ax+7-3x=3x-b-3x</math>}} {{Fristående formel||<math>ax+7-3x=\phantom{3x}{}-b\phantom{{}-3x}</math>}} and then subtract <math>7</math> {{Fristående formel||<math>ax+7-3x -7=-b-7</math>}} {{Fristående formel||<math>ax\phantom{{}+7}{}-3x\phantom{{}-7}{}=-b-7</math>}} 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> can be factored out {{Fristående formel||<math>(a-3)x = -b-7\; \mbox{.}</math>}} Divide both sides with <math>a-3</math> giving {{Fristående formel||<math>x= \frac{-b-7}{a-3}\; \mbox{.}</math>}} </div> It is not always obvious that you 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"> ''' Example 4''' Solve the equation<math>\ (x-3)^2+3x^2=(2x+7)^2</math>. Expand the quadratic expressions on both sides {{Fristående formel||<math>x^2-6x+9+3x^2=4x^2+28x+49\,\mbox{,}</math>}} {{Fristående formel||<math>4x^2-6x+9=4x^2+28x+49\,\mbox{.}</math>}} Subtract <math>4x^2</math> from both sides {{Fristående formel||<math>-6x +9 = 28x +49\; \mbox{.}</math>}} Add <math>6x</math> to both sides {{Fristående formel||<math>9 = 34x +49\; \mbox{.}</math>}} Subtract <math>49</math> from both sides {{Fristående formel||<math>-40=34x\; \mbox{.}</math>}} Divide both sides by <math>34</math> {{Fristående formel||<math>x = \frac{-40}{34}= - \frac{20}{17}\; \mbox{.}</math>}} </div> <div class="exempel"> ''' Example 5''' Solve the equation <math>\ \frac{x+2}{x^2+x} = \frac{3}{2+3x}</math>. Collect both terms to one side {{Fristående formel||<math>\frac{x+2}{x^2+x}-\frac{3}{2+3x}= 0\; \mbox{.}</math>}} Convert the terms so that they have the same denominator {{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>}} and simplify the numerator {{Fristående formel||<math>\frac{(x+2)(2+3x)-3(x^2+x)}{(x^2+x)(2+3x)} = 0,</math>}} {{Fristående formel||<math>\frac{3x^2+8x+4-(3x^2+3x)}{(x^2+x)(2+3x)} = 0,</math>}} {{Fristående formel||<math>\frac{5x +4}{(x^2+x)(2+3x)} = 0\,\mbox{.}</math>}} This equation only is satisfied when the numerator is equal to zero (whilst the denominator is not equal to zero); {{Fristående formel||<math>5x+4=0</math>}} which gives that <math>\,x = -\frac{4}{5}</math>. </div> == Straight lines == Functions such as {{Fristående formel||<math>y = 2x+1</math>}} {{Fristående formel||<math>y = -x+3</math>}} {{Fristående formel||<math>y = \frac{1}{2} x -5 </math>}} are examples of linear functions, and they generally can be put into the form <div class="regel"> {{Fristående formel||<math>y = kx+m</math>}} </div> where <math>k</math> and <math>m</math> are constants. The graph of a linear function is always a straight line and 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. <center>{{:2.2 - Figur - Linjen y = kx + m}}</center> <center><small>The line ''y'' = ''kx'' + ''m'' has slope ''k'' and cuts the ''y''-axis at (0,''m'')</small></center> The constant <math>k</math> is called the slope and means that a unit change in the positive <math>x</math>-direction along the line gives <math>k</math> units change in the positive <math>y</math>-direction. Thus if *<math>k>0\,</math> the line slopes upwards, *<math>k<0\,</math> the line slopes downwards. 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"> ''' Example 6''' <ol type="a"> <li> Sketch the line <math>y=2x-1</math>. <br/><br/> Comparing with the standard equation <math>y=kx+m</math> we see that <math>k=2</math> and <math>m=-1</math>. This means that the line's slope is <math>2</math> and that it cuts the <math>y</math>-axis at <math>(0,-1)</math>. See the figure below to the left.</li> <li>Sketch the line <math>y=2-\tfrac{1}{2}x</math>.<br/><br/> The equation of the line can be written as <math>y= -\tfrac{1}{2}x + 2</math> , and then we see that its slope is <math>k= -\tfrac{1}{2}</math>and that <math>m=2</math>. See the figure below to the right. </ol> {| align="center" padding="20px" |align="center"|{{:2.2 - Figur - Linjen y = 2x - 1}} |width="20px"| |align="center"|{{:2.2 - Figur - Linjen y = 2 - x/2}} |- |align="center"|<small>Line ''y'' = 2''x'' - 1</small> || |align="center"|<small>Line ''y'' = 2 - ''x''/2</small> |} </div> <div class="exempel"> ''' Example 7''' What is the slope of the straight line that passes through the points <math>(2,1)</math> and <math>(5,3)</math>? 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>. <center>{{:2.2 - Figur - Linje genom punkterna (2,1) och (5,3)}}</center> </div> 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 having slopes <math>k_1</math> and <math>k_2</math> and also are perpendicular then <math>k_2 = -\frac{1}{k_1}</math>, which also can be written as <math>k_1 k_2 = -1</math>. <center>{{:2.2 - Figur - Riktningskoefficient för vinkelräta linjer}}</center> 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, and 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"> ''' Example 8''' <ol type="a"> <li> The lines <math>y=3x-1</math> and <math>y=3x+5</math> are parallel. <li> The lines <math>y=x+1</math> and <math>y=2-x</math> are perpendicular. </ol> </div> All straight lines (including vertical lines) can be put into the general form <div class="regel"> {{Fristående formel||<math>ax+by=c</math>}} </div> where <math>a</math>, <math>b</math> and <math>c</math> are constants. <div class="exempel"> ''' Example 9''' <ol type="a"> <li>Put the line <math>y=5x+7</math> into the form <math>ax+by=c</math>.<br/><br/> Move the <math>x</math>-term to the left-hand side: <math>-5x+y=7</math>.</li> <li> Put the line <math>2x+3y=-1</math> into the form <math>y=kx+m</math>.<br/><br/> 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>}} </ol> </div> [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 '''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. <div class="exempel"> ''' Example 10''' <ol type="a"> <li>Sketch the region in the <math>x,y</math>-plane that satisfies <math>y\ge2</math>. <br/><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> <li>Sketch the region in the <math>x,y</math>-plane that satisfies <math>y < x</math>. <br/><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> 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> </div> <div class="exempel"> ''' Example 11''' Sketch the region in the <math>x,y</math>-plane that satisfies <math>2 \le 3x+2y\le 4</math>. The double inequality can be divided into two inequalities {{Fristående formel||<math>3x+2y \ge 2 \quad</math> and <math>\quad 3x+2y\le4 \;\mbox{.}</math>}} 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> and <math>\quad y\le 2-\frac{3}{2}x \;\mbox{.}</math>}} The points that satisfy the first inequality are on and above the line <math>y \ge 1-\tfrac{3}{2}x</math> while the points that satisfy the other inequality are on or below the line <math>y\le 2-\tfrac{3}{2}x</math>. <center>{{:2.2 - Figur - Områdena 3x + 2y ≥ 2 och 3x + 2y ≤ 4}}</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> 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><small>The figure shows the region <math>2\le 3x+2y\le 4</math>.</small></center> </div> <div class="exempel"> ''' Example 12''' 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> We find that for a point to lie in this triangle, it has to satisfy certain conditions. 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>. 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 see that <math>x</math> cannot be larger than <math>2</math> or less than <math>-2</math>. Thus the base of the triangle is <math>4</math> units of length and the height <math>2</math> units of length. The area of this triangle is therefore <math> 4\cdot 2/2=4</math> units of area. </div> [[2.2 Övningar|Exercises]] <div class="inforuta" style="width:580px;"> '''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 ... ''' Draw your own diagrams when you solve geometrical problems and draw carefully and accurately! A good diagram can mean you are halfway to a solution, but a poor diagram may well fool you. <!-- '''Reviews''' For those of you who want to deepen your studies or need more detailed explanations consider the following references: [http://matmin.kevius.com/linje.html Learn more about linear equations in Bruno Kevius mathematical glossary (Swedish).] --> '''Useful web sites''' [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 Experiment with Archimedes Triangle & Squaring of Parabola.] </div>

• 2.2 - Figure - The line y = 2 - x/2
• 2.2 - Figure - The line y = 2x - 1
• 2.2 - Figure - The line y = kx + m
• 2.2 - Figure - The region 2 ≤ 3x + 2y ≤ 4
• 2.2 - Figure - The region y less than x
• 2.2 - Figure - The region y ≥ 2
• 2.2 - Figure - The regions 3x + 2y ≥ 2 and 3x + 2y ≤ 4
• 2.2 - Figure - The slope of perpendicular lines
• 2.2 - Figure - The triangle bounded by y = x, y = 2 and y = -x
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