Houghton Mifflin Harcourt Algebra 1, 2015
HM
Houghton Mifflin Harcourt Algebra 1, 2015 View details
2. Solving Linear Systems by Substitution
Continue to next subchapter

Exercise 1 Page 400

Practice makes perfect
a Since the first equation already has an isolated y-variable, we can substitute the value it is equal to into the second equation. Then we can solve for x.

y=12 & (I) 2x-y=4 & (II)
y=12 2x- 12=4
y=12 2x=16
y=12 x=8

The solution is the point (8, 12).

b The first equation does not have a slope which means it's a horizontal line. Additionally, since the equation is y=12 we know it's a horizontal line passing through (0,12).
To determine the second graph, we need two points. We can determine the first point by identifying the y-intercept of the equation. This requires us to rewrite the equation in slope-intercept form. 2x-y=4 ⇔ y=2x - 4 The y-intercept is (0, - 4). To find the second point, we can substitute an arbitrary value for x or y and solve for the corresponding x or y-coordinate. For example, we can find the x-intercept by substituting y=0 into the equation and solving for x.

2x-y=4
2x- 0=4
â–¼
Solve for x
2x=4
x=2

Now that we know a second point on the line, we can graph the line. The solution of the system is the point at which the lines intersect.

We can see that the solution is, once again, (8, 12). Comparing our answers from using both methods, we can see that they are the same.