# 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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3/4 /2 .. do you mean $\dfrac{3/4}{2}$ ? – Debashish Jul 1 '14 at 7:42
ow yes, I edit it now – Mostafa Talebi Jul 1 '14 at 7:43
@Debashish Don't add the homework tag if the OP hasn't explicitly said so. His other question was about an equation he created, so it's very likely this is not homework. Also, this is not a systems of equations question. Please be more careful in your edits. – Mark Fantini Jul 1 '14 at 7:56
@Fantini .. thanks for correcting me ! I will be careful next time – Debashish Jul 1 '14 at 7:57
No this is not the homework. And about tags, isn't this an equation? And if so, isn't this linear-equation? Though StackExchange has explicitly said not to use systems-of-equation without a specific sub-tag, I didn't find any linear-equation to do so. – Mostafa Talebi Jul 1 '14 at 8:33

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