Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 3 down vote accepted

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

share|cite|improve this answer
why was this downvoted? It is perfectly correct answer! – MonK Jul 1 '14 at 7:54
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 $$x = \frac{3}{8\cdot 8\cdot 3} = \frac1{8\cdot 8}=\frac1{64}$$

share|cite|improve this answer

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

I your case;


As a solution to

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

Substituting for $x$ gives;

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

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



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

For ways to find the solution see the other answers.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.