1
$\begingroup$

I don't know how to get this result: $\frac{38}{17}$

This is the equation, can anyone explain why?

$(\frac{x}{4}) - (x - \frac{5}{6}) = (1 + \frac{2(x-5)}{3})$

I did this : LCM $= 12$.

Then: $$3(x) - 2(x-5) = 4(1 + 2(x-5)$$

$$3x - 2x + 10 = 4(1 + 2x - 10)$$

$$3x - 2x + 10 = 4 + 8x - 40$$

$$-7x = -46$$

$$\frac{(-46)}{-7}$$

$\endgroup$
  • $\begingroup$ What steps did you take? Show us what you did, and please read the MathJax tutorial. $\endgroup$ – Sean Roberson Jun 3 '17 at 15:14
  • 4
    $\begingroup$ How would WE know how YOU got your result? $\endgroup$ – user223391 Jun 3 '17 at 15:15
  • $\begingroup$ Here's a MathJax Tutorial $\endgroup$ – Sahiba Arora Jun 3 '17 at 15:15
  • $\begingroup$ I can't go to this result, i made in calculator, my results is always different, i did LCM to multiply, but i just can't go to the final result, i don't know why. $\endgroup$ – Ryan Zagon Jun 3 '17 at 15:16
  • 1
    $\begingroup$ $12 \times (x - \frac{5}{6})$ is not equal to $2(x - 5)$. $\endgroup$ – NickD Jun 3 '17 at 15:24
2
$\begingroup$

$$ \frac{x}{4} - \left(x - \frac{5}{6}\right) = 1 + \frac{2(x-5)}{3} $$

First add the $x/4$ and $-x$ on the LHS and distribute the $2/3$ on the RHS,

$$ \frac{-3}{4}x + \frac{5}{6} = 1 + \frac{2}{3}x - \frac{10}{3} $$

Move all the constants to the LHS and $x$'s to the RHS,

$$ \frac{5}{6} - \frac{6}{6} + \frac{20}{6} = \frac{8}{12}x + \frac{9}{12}x$$

Simplify,

$$ \frac{19}{6} = \frac{17}{12} x $$

Multiply both sides by $\frac{12}{17}$,

$$ \frac{38}{17} = x $$

$\endgroup$
2
$\begingroup$

Hint:

Multiply both sides by 12 to remove all denominators.

Some details:

After multiplication, one gets $$3x-(12x-10)=12+8(x-5)\iff40-12+10=12x-3x+8x.$$

$\endgroup$
0
$\begingroup$

we have $$\frac{x}{4}-x+\frac{5}{6}=1+\frac{2x-10}{3}$$ multipliying by $12$ $$3x-12x+10=12+8x-40$$ $$-9x+10=-28+8x$$ $$-17x=-38$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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