Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 3.2
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Exercise 87 Page 184

Practice makes perfect
a Examining the system of equations, we notice that both equations are solved for y. Therefore, we should use the Substitution Method to solve the system.
y=5x-2 & (I) y=2x+10 & (II)
y=5x-2 5x-2=2x+10
â–Ľ
(II): Solve for x
y=5x-2 5x=2x+12
y=5x-2 3x=12
y=5x-2 x=4
Having isolated x, we substitute this into (I) to solve for y.
y=5x-2 x=4
y=5( 4)-2 x=4
â–Ľ
(II): Evaluate right-hand side
y=20-2 x=4
y=18 x=4
Having solved for x and y, we can check our solution by substituting these values into both equations. If the left-hand side and right-hand side are equal, the solution is correct.
y=5x-2 & (I) y=2x+10 & (II)

(I), (II): x= 4, y= 18

18? =5( 4)-2 18? =2( 4)+10

(I), (II): Multiply

18? =20-2 18? =8+10

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

18=18 âś“ 18=18 âś“
The solution is correct.
b Since the first equation is solved for x, we should use the Substitution Method to solve the system.
x=-2y-1 & (I) 2x+y=-20 & (II)
x=-2y-1 2( -2y-1)+y=-20
â–Ľ
(II): Solve for x
x=-2y-1 - 4y-2+y=-20
x=-2y-1 - 3y-2=-20
x=-2y-1 - 3y=-18
x=-2y-1 y=6
Having isolated y, we substitute this into (I) to solve for x.
x=-2y-1 y=6
x=-2( 6)-1 y=6
x=-12-1 y=6
x=-13 y=6
Having solved for y, we can check our solution by substituting these values into both equations. If the left-hand side and right-hand side are equal, the solution is correct.
x=-2y-1 & (I) 2x+y=-20 & (II)

(I), (II): x= -13, y= 6

-13? =-2( 6)-1 2( -13)+ 6? =-20

(I), (II): Multiply

-13? =-12-1 - 26+6? =-20

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

-13=-13 âś“ - 20=-20 âś“
The solution is correct.