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Solve using elimination. 2y - 5x = -1 x = 2y + 5
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@phi
\(2y - 5x = -1\) \(x = 2y + 5\) I rearranged the second equation to get to \(2y - x = -5\) I now have \(2y - 5x = -1\) \(-\) \(2y - x = -5\) \(-4x = 4\) \(x = -1\)
yes, that looks good.
\(-1 = 2y + 5\) \(-1 - 5 = 2y + 5 -5\) \(-6 = 2y\) \(-3 = y\) Solution \((-1, -3)\) Would I be correct?
yes
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Okay. And when I solve it using substitution it would be \(2y - 5x = -1\) \(x = 2y + 5\) \(2y - 5(2y + 5) = -1\) \(2y - 10y - 25 = -1\) \(-8y - 25 = -1\) \(-8y = 24\) \(y = -3\) \(x = 2(-3) + 5\) \(x = -6 + 5\) \(x = -1\) Solution \((-1, -3)\)
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