1
$\begingroup$

I have this equation: $x+7-(\frac{5x}8 + 10) = 3 $

I've used step-by-step calculators online but I simply don't understand it. Here is how I've tried to solve the problem:

$$x+7-\left(\frac{5x}8+10\right) = x + 7 - \frac{5x}8 - 10 = 3$$

$$x + 7 - \frac{5x}8 - 10 + 10 = 3 + 10$$

$$x + 7 - 7 - \frac{5x}8 = 13 - 7$$

$$x - \frac{5x}8 = 6$$

$$x - 8\times\frac{5x}8 = 6\times8$$

$$x - 5x = 48$$

$$\frac{-4x}{-4} = \frac{48}4$$

$$x = -12$$

Now obviously, it's wrong. The right answer is $16$, but I don't know how to get to that answer. Therefore, I'm extremely thankful if someone truly can show what I need to do, and why I need to do it, because I'm completely lost right now. Thanks.

$\endgroup$
2
$\begingroup$

At the line :

$$x - 8\times \frac{5x}8 = 6\times 8$$ you should instead write $$ 8\times(x-\frac{5x}8) = 6 \times 8.$$

Indeed, you're multiplicate the entire left side and the entire right side of your equation.

Note that you can also calculate $x-\frac{5x}8 = \frac{8x}8 - \frac{5x}8 = \frac{3x}8$.

$\endgroup$
2
$\begingroup$

$x-8\cdot\frac{5x}{8}=6\cdot8$ is wrong, it should be $8x-8\cdot\frac{5x}{8}=6\cdot8$. Then you get $3x=48$, or $x=16$.

$\endgroup$
2
$\begingroup$

4th line from the end you forgot to multiply the $x$ by 8 as well.

$\endgroup$
2
$\begingroup$

You just need to rearrange the equation as follows:

$x+7-(\frac{5x}{8}+10) = 3$

$\frac{3x}{8} -3 = 3$

$\frac{3x}{8} = 6$

$3x = 48$

$x = 16$

If I've gone a bit too quickly please let me know and I'll add a few intermediate steps.

$\endgroup$
2
$\begingroup$

$x+7-(\frac{5x}{8}+10)=3$

$\Rightarrow x+7-(\frac{5x+80}{8})=3$

$\Rightarrow x+7+\frac{-5x-80}{8}=3$

$\Rightarrow \frac{8x+7*8-5x-80}{8}=3$

$\Rightarrow \frac{8x+7*8-5x-80}{8}=3$

$\Rightarrow 8x+56-5x-80=3*8$

$\Rightarrow 8x-5x=24+80-56$

$\Rightarrow 3x=48$

$\Rightarrow x=16$

$\endgroup$

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.