Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
1. Section 9.1
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Exercise 10 Page 485

Practice makes perfect
a We are asked to solve the given system twice. By graphing the given equations, we can determine the solution to the system. This will be the point at which the lines intersect. To do this we will identify the slope m and y-intercept b from their slope-intercept form.
Given Equation Slope-Intercept Form Slope m y-intercept b
y=- 2x+6 y=- 2x+ 6 - 2 (0, 6)
y=1/2x-9 y=1/2x+( -9) 1/2 (0, -9)

Graphing the System

To graph these equations, we will start by plotting their y-intercepts. Then, we will use the slope to determine another point that satisfies each equation and connect the points with a line.

The point of intersection at (6,- 6) is the system's solution.

Solving Algebraically

Since y is solved for in both equations, we should use the Substitution Method to solve the system.
y=- 2x+6 & (I) y=1/2x-9 & (II)
y=- 2x+6 - 2x+6=1/2x-9
â–Ľ
(II): Solve for x
y=- 2x+6 - 2x+15=1/2x
y=- 2x+6 - 4x+30=x
y=- 2x+6 30=5x
y=- 2x+6 6=x
y=- 2x+6 x=6
Having solved for x in the second equation, we can substitute this value in the first equation and solve for y.
y=- 2x+6 x=6
y=- 2( 6)+6 x=6
y=- 12+6 x=6
y=- 6 x=6
Algebraically, we get the solution (6,- 6).

Verifying the Solution

To verify the solution we substitute it into both equations and check if the left-hand side and right-hand side are equal. If they are, the solution is correct.
y=- 2x+6 & (I) y=1/2x-9 & (II)

(I), (II): x= 6, y= - 6

- 6? =- 2( 6)+6 - 6? =1/2( 6)-9

(I), (II): Multiply

- 6? =- 12+6 - 6? =3-9

(I), (II): Add and subtract terms

- 6=- 6 - 6=- 6
Since the solution's coordinates satisfied both equations, we know that the solution is correct.
b Let's check the equation's graphs.
It looks like the lines are perpendicular to each other. If this is true, the product of their slopes equals - 1. In other words, the slopes will be negative reciprocals. m_1* m_2=- 1 Examining the equations, we see that one has a slope of - 2 and the other has a slope of - 12. By substituting these into the formula above, we can determine if the graphs are perpendicular.
m_1* m_2=- 1
( - 2)( 1/2)? =- 1
â–Ľ
Simplify
2(1/2)? =1
2/2? =1
1=1
Yes, the lines are perpendicular.