Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
3. Section 2.3
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Exercise 105 Page 129

Practice makes perfect
a Let's first solve (1). Since the x-terms have opposite coefficients, we will use the Elimination Method to solve the system of equations.
y=- x+8 & (I) y=x-2 & (II)
y=- x+8 y+ y=x-2+( - x+8)
(II): Solve for y
y=- x+8 y+y=x-2-x+8
y=- x+8 2y= 6
y=- x+8 y=3
Having solved for y, we can find x by substituting this value for y in the first equation.
y=- x+8 y=3
3=- x+8 y=3
(I): Solve for x
x+3=8 y=3
x=5 y=3
The solution to (1) is (5,3). Let's also solve (2). Again, since both equations have y on the left-hand side and with opposite coefficients, we should use the Elimination Method.
2x-y=10 & (I) y=- 4x+2 & (II)
2x-y=10 y+( 2x-y)=- 4x+2+ 10
(II): Solve for x
2x-y=10 y+2x-y=- 4x+2+10
2x-y=10 2x=- 4x+12
2x-y=10 6x=12
y=- x+8 x=2
We have found that x=2. By substituting this value for x in the second equation, we can find y.
2x-y=10 x=2
2( 2)-y=10 x=2
(I): Solve for y
4-y=10 x=2
- y=6 x=2
y=- 6 x=2
The solution to (2) is (2,- 6).
b The equation of a line is most commonly written in slope-intercept form.
c y= mx+ b [0.2em] [-1em] &Slope: m &y-intercept: b Since both graphs cut the y-axis at lattice points, we can identify the y-intercepts by examining the diagram.

Now we can start writing the equations. |c|c|c|c| Line &Equation & m & b q&y= mx+ 8 & m & 8 p&y= mx+ 3 & m & 3 We also have to find the lines slope. If we divide the vertical distance by the horizontal distance between two identifiable points on each graph, we can determine the slopes.

Now we can complete the equations. |c|c|c|c| Line &Equation & m & b [0.2em] [-0.8em] q&y= - 1/2x+ 8 & - 1/2 & 8 [0.8em] [-0.8em] p&y= 2x+ 3 & 2 & 3 [0.3em]

c Two lines are perpendicular when the product of their slopes equals -1.
m_1 m_2=- 1 This seems to be the case for our lines. However, let's investigate this proposition to make sure.
m_1 m_2=- 1
2( - 1/2)? =- 1
Evaluate left-hand side
- 2* 1/2? =- 1
- 2/2? =- 1
- 1=- 1 ✓
They are perpendicular.
d Let's combine the equations into a system of equations.
y=- 12x+3 & (I) y=2x+8 & (II) Since both equations are solved for y, we should use the Substitution Method to solve the system.
y=- 12x+3 & (I) y=2x+8 & (II)
2x+8=- 12x+3 y=2x+8
(I): Solve for x
2x+5=- 12x y=2x+8
- 4x-10=x y=2x+8
- 10=5x y=2x+8
- 2=x y=2x+8
x=- 2 y=2x+8
Having solved for x, we can find the value of y by substituting this value for x in the second equation.
x=-2 y=2x+8
x=-2 y=2( -2)+8
(II): Evaluate right-hand side
x=-2 y=- 4+8
x=-2 y=4
The solution is (-2,4). To determine if it matches the solution in the graph, we have to find the x- and y-coordinate of the point where the lines intersect.

We are correct.