# Question 2b76d

Feb 14, 2017

In this case, the set of like terms are $\frac{3}{5} x$ and $\frac{7}{8} x$. To combine them, let's subtract $\frac{3}{5} x$ from both sides. On the left side, the positive and negative $\frac{3}{5} x$ cancel each other out or become equal to $0$.
$\frac{3}{5} x - \left(\frac{3}{5} x + 33 = \frac{7}{8} x\right) - \frac{3}{5} x$

So this is how our solution looks like.
$33 = \frac{7}{8} x - \frac{3}{5} x$

Let's focus on those two fractions. We'll be subtracting dissimilar fractions. First, we find the LCD (Least Common Denominator) of $8$ and $5$, which is $40$.
$\frac{7}{8} x - \frac{3}{5} x = \frac{-}{40}$

Next, we divide $40$ by $8$ and $5$. Then, the quotient of $40$ and $8$ is $5$ and will be multiplied to $7$. The quotient of $40$ and $5$ is $8$ and will be multiplied by $3$. The solution looks like this:
$\frac{7}{8} x - \frac{3}{5} x = \frac{35 - 24}{40}$

Subtract 35 and 24 to get the following:
$33 = \frac{11}{40} x$

Now let's isolate the variable $x$. We could apply cross-multiplication by multiplying $33$ by $40$, $11$ by $1$ (which is underneath $33$ this whole time). Solution becomes like this:
$\frac{33}{1} = \frac{11}{40} x$ ==> $1320 = 11 x$

DIvide both sides by $11$ to get $x = 120$.
$\frac{1320}{11} = \frac{11}{11} x$ ==> $120 = x$

Feb 14, 2017

$x = 120$

#### Explanation:

$\frac{3}{\text{5" .x + 33 = 7/"8}} . x$

Now, subtracting $\frac{3}{\text{5}} . x$ from the both sides

cancel(3/"5".x) + 33 -cancel( 3/"5".x) = 7/"8".x - 3/"5".x

$33 = x . \left(\frac{7}{\text{8" - 3/"5}}\right)$

$33 = x . \left(\text{7(5) - 3(8)"/"40}\right)$

$33 = x . \left(\text{35 - 24"/"40}\right)$

$33 = x . \frac{11}{\text{40}}$

Multiplying the both sides by 40
33 × 40 = x. 11/cancel("40" )× cancel(40)

33 × 40 = x. 11

Now, dividing the both sides by 11

$\cancel{\text{(33)"^3 × 40/cancel"(11)" = x. cancel"(11)"/cancel"(11)}}$

3 × 40 = x#

$120 = x$