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Solve the following system of equations by substitution. 3y = 12 2x + y = -16 A. (-4, -10) B. (-10, 4) C. (4, 10) D. (3, 3)
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@Uri would you mind to help me??
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3y = 12 -- divide both sides by 3 y = 4 now sub 4 in for y in the other equation to find x 2x + y = -16 2x + 4 = -16 2x = -16 - 4 2x = - 20 x = -10 now sub -10 in for x 2(-10) + y = -16 -20 + y = -16 y = -16 + 20 y = 4 solution is : (-10,4)
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