# How do you solve (7x) /(2x+5) + 1 = (10x - 3) /( 3x)?

May 10, 2018

$x = \frac{29 \pm \sqrt{421}}{14}$

#### Explanation:

The first thing you do is incorporate the 1 on the left side into the fraction. In order to do that you substitute it by a fraction that is the left denominator divided by itself, like this:

$\frac{7 x}{2 x + 5} + \frac{2 x + 5}{2 x + 5} = \frac{10 x - 3}{3 x}$

Now we can add the fractions on the left side. Since they have the same denominator, we only add the numerators.

$\frac{7 x + 2 x + 5}{2 x + 5} = \frac{10 x - 3}{3 x}$

$\frac{9 x + 5}{2 x + 5} = \frac{10 x - 3}{3 x}$

Now we multiply both sides of the equation to remove the $3 x$ from the bottom of the right side.

$\frac{3 x \cdot \left(9 x + 5\right)}{2 x + 5} = \frac{10 x - 3}{\cancel{3 x}} \cdot \cancel{3 x}$

Multiply the top on the left and we get:

$\frac{3 x \cdot 9 x + 3 x \cdot 5}{2 x + 5} = 10 x - 3$

$\frac{27 {x}^{2} + 15 x}{2 x + 5} = \left(10 x - 3\right)$

Do the same thing with the left denominator. Multiply both sides by it in order to remove it.

$\cancel{2 x + 5} \frac{27 {x}^{2} + 15 x}{\cancel{2 x + 5}} = \left(10 x - 3\right) \cdot \left(2 x + 5\right)$

Now we multiply out the right side:

$27 {x}^{2} + 15 x = \left(10 x - 3\right) \left(2 x + 5\right)$

$27 {x}^{2} + 15 x = 20 {x}^{2} + 50 x - 6 x - 15$

$27 {x}^{2} + 15 x = 20 {x}^{2} + 44 x - 15$

Finally we can move everything from the right side of the equation to the left:

$27 {x}^{2} + 15 x - 20 {x}^{2} - 44 x + 15 = 0$

$7 {x}^{2} - 29 x + 15 = 0$

Now we have a single equation if the form of $a {x}^{2} + b x + c = 0$

Which means we can use the quadratic formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ to solve for $x$

When we plug in the numbers, we get:

$x = \frac{29 \pm \sqrt{{\left(- 29\right)}^{2} - 4 \cdot 7 \cdot 15}}{2 \cdot 7}$

$x = \frac{29 \pm \sqrt{841 - 420}}{14}$

$x = \frac{29 \pm \sqrt{421}}{14}$