Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
6. Systems of Linear Inequalities
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Exercise 42 Page 405

Practice makes perfect
a In order to find the value of x, we first have to find the slope of the line 2x+y=3. This will allow us to find the slope of a line perpendicular to the given one. The given equation is in the standard form, so we will rearrange it so that it is written in slope-intercept form first.
2x+y=3
y=3-2x
y=-2x+3
In this equation the coefficient of the x variable gives the slope of the line, -2. Perpendicular lines have negative reciprocals for their slopes. In this case, this will be 12. Since we can calculate the slope using the two given points on the line, ( -1, 6) and ( x, 2), we can use the Slope Formula to solve for the missing x value.
m=y_1-y_2/x_1-x_2
1/2=6- 2/-1- x
â–Ľ
Solve for x
1/2=4/-1-x
1=8/-1-x
-1-x=8
- x=9
x=-9
b Now, having the slope and a point on the line, we can write a new equation using the point-slope form. We have found that the slope is 12, or 0.5, and we can use the point (-1,6), so m= 0.5 and ( x_1, y_1)=( -1, 6).
y-y_1=m(x-x_1)
y- 6= 0.5(x-( -1))
â–Ľ
Write in slope-intercept form
y-6=0.5(x+1)
y-6=0.5x+0.5
y=0.5x+6.5
The equation of the line perpendicular to the given line 2x+y=3 and through the given points is y=0.5x+6.5.
c To find a point that is a solution to both equations, we need to find their point of intersection. One way to do this is by equating the lines and solving for x.
y_1=y_2
-2x+3=0.5x+6.5
â–Ľ
Solve for x
-2x-0.5x+3=6.5
(-2-0.5)x+3=6.5
- 2.5 x + 3 = 6.5
-2.5 x=3.5
x=3.5/-2.5
x=-1.4
The x-coordinate of the common point is - 1.4. Now we can substitute this value of x into one of the equations to find the corresponding y-coordinate.
y=-2x+3
y=-2 ( -1.4)+3
y=2.8+3
y=5.8
The y-coordinate of the common point is 5.8. Therefore, the ordered pair is (1.4,5.8).