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Solve with substitution, show your work.
3x+2y+z=7
5x+5y+4z=3
3x+2y+3z=1

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3x + 2y + z = 7 ⇒ z = 7 - 3x - 2y

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5x + 5y + 4z = 3

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

= 5x + 5y + 28 - 12x - 8y = 3

= -7x -3y + 28 = 3

= -7x - 3y = -25

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3x + 2y + 3z = 1

= 3x + 2y + 3(7 - 3x - 2y) = 1

= 3x + 2y + 21 - 9x - 6y = 1

= -6x - 4y + 21 = 1

= -6x - 4y = -20

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-7x - 3y = -25 ⇒ 4(-7x - 3y = -25) ⇒ -28x - 12y = -100

-6x - 4y = -20 ⇒ -3(-6x - 4y = -20) ⇒ 18x + 12y = 60

-10x = -40

x = 4

-7x - 3y = -25

-7(4) - 3y = -25

-28 - 3y = -25

-3y = 3

y = -1

z = 7 - 3x - 2y

= 7 - 3(4) - 2(-1)

= 7 - 12 + 2

= -3

Answer: x = 4, y = -1, z = -3

User Shabari Nath K
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