Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
2. Section 4.2
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Exercise 60 Page 168

Practice makes perfect
a Because the first equation is already solved for x, we will use the Substitution Method to find the point of intersection.
x=-2y-3 & (I) 4y-x=9 & (II)
x=-2y-3 4y-( -2y-3)=9
â–Ľ
(II): Solve for y
x=-2y-3 4y+2y+3=9
x=-2y-3 6y+3=9
x=-2y-3 6y=6
x=-2y-3 y=1
Having found the value of y, we can substitute this in the first equation to determine x.
x=-2y-3 & (I) y=1 & (II)
x=-2( 1)-3 y=1
x=-2-3 y=1
x=-5 y=1
To check our solution, we have to substitute x=-5 and y=1 in both equations.
x=-2y-3 & (I) 4y-x=9 & (II)

(I), (II): x= -5, y= 1

-5? =-2( 1)-3 4( 1)-( -5)? =9

(I), (II): a * 1=a

-5 ? = -2-3 4+5 ? = 9

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

-5=-5 9=9
The solution is correct.
b In this case the coefficients of x in each equation are opposite numbers. Therefore, we should use the Elimination Method to solve the system.
x+5y=8 & (I) - x+2y=-1 & (II)
x+5y=8 - x+2y+( x+5y)=-1+ 8
â–Ľ
(II): Solve for y
x+5y=8 - x+2y+x+5y=-1+8
x+5y=8 7y=7
x+5y=8 y=1
Having solved the second equation for y, we can substitute this into the first equation and solve for x.
x+5y=8 y=1
x+5( 1)=8 y=1
x+5=8 y=1
x=3 y=1
To check if the solution is correct, we will substitute x=3 and x=1 in both equations and make sure the left-hand side and right-hand side are equal in both of them.
x+5y=8 & (I) - x+2y=-1 & (II)

(I), (II): x= 3, y= 1

3+5( 1)? =8 - 3+2( 1)? =-1

(I), (II): a * 1=a

3+5 ? = 8 -3+2 ? = -1

(I), (II): Add terms

8=8 -1=-1
The solution is correct.
c Since the second equation is in slope-intercept form, let's rewrite the first equation on this form as well.
4x-2y=5 & (I) y=2x+10 & (II)
-2y=-4x+5 y=2x+10
2y=4x-5 y=2x+10
y=2x-2.5 y=2x+10
Having rewritten the equations, we see that they have the same slope but different y-intercepts. This makes them parallel lines, which means they do not intersect. Thus, this system of equations does not have any solution.