2.2 Linear expressions
From Förberedande kurs i matematik 1
| 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
- Solve the equation \displaystyle x+3=7.
Subtract \displaystyle 3 from both sides- \displaystyle x+3-3=7-3.
- \displaystyle x=7-3=4.
- Solve the equation \displaystyle 3x=6.
Divide both sides by \displaystyle 3- \displaystyle \frac{3x}{3} = \frac{6}{3}\,.
- \displaystyle x=\frac{6}{3} = 2.
- 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.
- \displaystyle x = \frac{4}{2} = 2.
- \displaystyle 2x=5-1.
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]](/wikis/2008/bridgecourse1-ImperialCollege/img_auth.php/metapost/5/c/f/5cf29d3d71396023b27b2e26cd868e7c.png)
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
- 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. - 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.
|
|
| |
| Line y = 2x - 1 | Line y = 2 - x/2 |
![[Image]](/wikis/2008/bridgecourse1-ImperialCollege/img_auth.php/metapost/c/3/1/c31d10da27b90635a6743357c3fed013.png)
![[Image]](/wikis/2008/bridgecourse1-ImperialCollege/img_auth.php/metapost/4/3/1/4315cd505e70fc08c578e2fc9d28153f.png)
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}.
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]](/wikis/2008/bridgecourse1-ImperialCollege/img_auth.php/metapost/a/4/0/a40744821c6c8cd59c30c09292cf4926.png)
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
- The lines \displaystyle y=3x-1 and \displaystyle y=3x+5 are parallel.
- 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
- 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. - 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
- 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]](/wikis/2008/bridgecourse1-ImperialCollege/img_auth.php/metapost/b/e/f/beff31399f9e094624a3955795665c60.png)
- 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]](/wikis/2008/bridgecourse1-ImperialCollege/img_auth.php/metapost/f/6/9/f698c636d69d550c46687c51eac40238.png)
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]](/wikis/2008/bridgecourse1-ImperialCollege/img_auth.php/metapost/e/6/1/e61506118ea86b81827e64ee907a585a.png)
Points that satisfy both inequalities form a band-like region where both coloured areas overlap.
![[Image]](/wikis/2008/bridgecourse1-ImperialCollege/img_auth.php/metapost/6/4/9/649cbdd936a52663de8af03e3b98807b.png)
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]](/wikis/2008/bridgecourse1-ImperialCollege/img_auth.php/metapost/4/f/e/4fe11327941ab6fd1e943157b94a5cc0.png)
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.
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
