# Is my equation result correct?

I have the following equation:

$20x + \dfrac{3/4}{2} = 8x \left(\dfrac{3}{2} + 4\right)$

For the above equation I have achieved this result: $x = \dfrac{-41}{-4}$

And if it is not, can please tell me why? by going through it? Thanks

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$20x + (\frac{3}{4} / 2) = 8x (\frac{3}{2} + 4)$ (This is my interpretation of your question.)

$20x+\frac{3}{8} = 8x(\frac{3}{2}+4)$ (As $3/4/2 = 3/8$)

$160x+3=8x(12+32)$ (This is multiplying each side by 8 to remove fractions where $\frac{3}{2}*8 = 12$.)

$160x+3 = 352x$ ($44 * 8 = 352$)

$192x = 3$ (Collecting like terms on each side with x on the left side and positive.)

$x = \frac{1}{64}$

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Because I made a miscalculation the first time I wrote this out. – JB King Jul 1 '14 at 7:57
And apparently the downvote was reversed. – Mark Fantini Jul 1 '14 at 7:57

First, expand the parentheses on the right and simplify the double fraction on the left. You get

$$20 x + \frac38 = 12 x + 32 x = 44 x$$

Carry the $x$ values to the left, giving $$\frac 38 = 44x - 20x = 24 x$$

Dividing by $24$ then gives $$\frac{3}{8\cdot 24} = x$$

Which you can simplify to $$x = \frac{3}{8\cdot 8\cdot 3} = \frac1{8\cdot 8}=\frac1{64}$$

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You can test a proposed solution by substituting the value that the solution arrives at;

$x=\frac{-41}{-4}=\frac{41}{4}$

As a solution to

$20x + \dfrac{3/4}{2} = 8x \left(\dfrac{3}{2} + 4\right)$

Substituting for $x$ gives;

$20\dfrac{41}{4} + \dfrac{3}{8} = 8\dfrac{41}{4} \left(\dfrac{3}{2} + 4\right)$

$5*41 + \dfrac{3}{8} = 2*41 \dfrac{11}{2}$

$205+\dfrac{3}{8}=41*11$

$205.375=451$

This is false so your proposed solution can not be correct.

For ways to find the solution see the other answers.

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