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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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