+2440 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. |
| \(x=-\frac{2}{17}\) |
