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In the below system, solve for y in the first equation. x + 3y = 6 2x – y = 10 one thirdx + 2 negative one thirdx + 6 –x + 2 negative one thirdx + 2
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make x+3y=6 into x=6-3y use substitution, plug in 6-3y into the 2nd equation anywhere there is an x After you solve that 2nd equation for y, plug in the answer for y into the place in the 1st equation where there is a y and you will then solve for x.
would the answer be negative one thirdx + 2 ? @IndianChick500
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