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3 years ago

Solve the system using substitution.

Mathematics

1 answer:

GaryK [48]3 years ago

4 0

Answer: x = 7, y = -3, z = 1

Step-by-step explanation:

EQ 1: 3x + 4y - z = 8 solve for z

+z -8 +z - 8

3x + 4y - 8 = z

EQ 3: 5x + y - 2z = 30 substitute 3x + 4y - 8 for z and solve for x

5x + y - 2(3x + 4y - 8) = 30

5x + y - 6x - 8y + 16 = 30

-x - 7y + 16 = 30

-16 -16

-x - 7y = 14

+7y +7y

-x = 7y + 14

x = -7y - 14

EQ 2: x - 5y + 2z = 24 substitute 3x+4y - 8 for z and -7y - 14 for x

(-7y - 14) - 5y + 2(3x + 4y - 8) = 24

-7y - 14 - 5y + 6(-7y - 14) + 8y - 16 = 24

-7y - 14 - 5y - 42y - 84 + 8y - 16 = 24

-46y - 114 = 24

+114 +114

-46y = 138

÷-46 ÷-46

y = -3

x = -7y - 14

= -7(-3) - 14

= 21 - 14

= 7

z = 3x + 4y - 8

= 3(7) + 4(-3) - 8

= 21 - 12 - 8

= 1

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