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3x – 7y = 5 9y = 5x + 5
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A. (–3, 2) B. (8, 5) C. (–10, –5) D. (0, 0)
3x - 7y = 5 y = 5/9 x + 5/9 3x - 7(5/9 x + 5/9) = 5 3x - 35/9 x - 35/9 = 5 3x - 35/9x = 35/9 + 5 3x - 35/9x = 80/9 -8/9 x = 80/9 multiply both sides by the reciprocal x = (80/9)(9/-8) x = -10
c
3(-10) -7y = 5 -30 -7y = 5 -7y = 35 y = -5
Yep
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Do you know how to do these systems?
The method I used is called substitution, there is also solving through matrices (you will learn this in algebra 2) and elimination.
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