McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
6. Graphing Inequalities in Two Variables
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Exercise 45 Page 321

Practice makes perfect
a Let's solve each inequality for y!

A. 7(y+6) ≤ 21x+14

To solve the inequality, we have to isolate y on one side. We can do it just as if we were solving an equation!
7(y+6) ≤ 21x+14
7y+42 ≤ 21x+14
7y ≤ 21x-28
y ≤ 21x-28/7
y ≤ 21/7x-28/7
y ≤ 3x-4
The solution of this inequality is y ≤ 3x-4.

B. -3y ≤ 3x-12

Again, let's isolate y. In this case, we have to remember that when dividing an inequality by a negative number we change its sign.
-3y ≤ 3x-12
y ≥ 3x-12/-3
y ≥ 3x/-3-12/-3
y ≥ - x-(-4)
y ≥ - x+4
The solution of this inequality is y ≥ - x+4.
b To graph the inequalities we will graph the boundary lines first. Then we will test a point to check which region to shade.

Boundary lines

To graph the boundary lines, we will treat the inequalities as equations. Therefore, we want to graph the following functions. y= 3x-4 and y = - x+4 Let's determine the slopes and the y-intercepts of these equations!
Feature y=3x-4 y=- x+4
Slope 3 -1
y-intercept -4 4

Now we can plot the boundary lines!

Testing a Point and Graphing

Let's test the point (0,0) to see which regions we must shade. We will start with inequality A.
y ≤ 3x-4
â–Ľ
Test (0,0)
0 ? ≤ 3( 0) -4
0 ? ≤ 0-4
0 ≰ -4
Now, let's test the same point for inequality B.
y ≥ - x+4
â–Ľ
Test (0,0)
0 ? ≥ - ( 0) +4
0 ? ≥ 0 +4
0 ≱ 4
We can tell that the point (0,0) does not satisfy our inequalities. Therefore, in each case we will shade the region that does not contain the point (0,0).
c The overlapping region is where both inequalities are satisfied. Therefore, it represents the solutions that make both inequalities A and B true.