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# (1/8)(56x-24)=(8/7)(21x-7)+7

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(1/8)(56x-24)=(8/7)(21x-7)+7

Guest Sep 10, 2017
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I will solve for x in the equation $$\frac{1}{8}(56x-24)=\frac{8}{7}(21x-7)+7$$.

 $$\frac{1}{8}(56x-24)=\frac{8}{7}(21x-7)+7$$ Distribute the 1/8 to both terms inside of the parentheses. $$7x-3=\frac{8}{7}(21x-7)+7$$ Multiply both sides by 7 to get rid of all the fractions in this equation. $$49x-21=8(21x-7)+49$$ Inside the parentheses, let's factor our a GCF of 7 from both terms of the parentheses. $$8(21x-7)=(8*7)(3x-1)=56(3x-1)$$ Now, plug that back into the equation. $$49x-21=56(3x-1)+49$$ Divide all sides of the equation by its GCF, 7. This should ease computation since the numbers will be easier to work with. $$7x-3=8(3x-1)+7$$ Distribute the 8 to both terms in the parentheses. $$7x-3=24x-8+7$$ Simplify the left hand side. $$7x-3=24x-1$$ Subtract 7x on both sides. $$-3=17x-1$$ Add 1 to both sides. $$-2=17x$$ Divide by 17 on both sides. $$x=-\frac{2}{17}$$
TheXSquaredFactor  Sep 10, 2017

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