# How do you simplify (5x)/(3x^2+19x-14)-1/(9x^2-12x+4)?

Jun 3, 2017

$\frac{15 {x}^{2} - 11 x - 7}{9 {x}^{3} + 51 {x}^{2} - 80 x + 28}$ OR $\left(15 {x}^{2} - 11 x - 7\right) \frac{\setminus}{\left(x + 7\right) {\left(3 x - 2\right)}^{2}}$
Both are equal in value but the one on the right may be considered more simplified the one on the left.

#### Explanation:

In an equation like this there are several steps. First I'll show you the equation to get to the answer, and then walk you through it.

((5x(9x^2−12x+4))-(1(3x^2+19x-14)))/((3x^2(9x^2-12x+4))+(19x(9x^2-12x+4))+(-14(9x^2-12x+4))

So by doing that equation you will be able to simplify the equation, but how do you get to that equation in the first place?

First the top half, you cross multiply.

Multiple the numerator on the left with the denominator on the right and subtract* it from what you get after multiplying the numerator on the right from the denominator on the left.

*(If the fractions from the original equation were being added together instead of subtracted that would be changed to 'add' instead of subtract)

Next, the bottom half you multiply the denominators.

Each term* of the denominator on the left gets multiplied against the entire denominator on the right with the products of each getting added together.

*($3 {x}^{2}$would be one term $19 x$ another and $- 14$ another)

After you have gotten the above equation you then need to solve each half (top and bottom) like you would any other algebra equation. Once done your equation should look like this: $\frac{45 {x}^{3} - 63 {x}^{2} + x + 14}{27 {x}^{4} + 135 {x}^{3} - 342 {x}^{2} + 244 x - 56}$

Next you need to factor either half:
$\frac{\left(45 {x}^{3} - 63 {x}^{2} + x + 14\right)}{\left(27 {x}^{4} + 135 {x}^{3} - 342 {x}^{2} + 244 x - 56\right)}$
$= \frac{\left(3 x - 2\right) \left(15 {x}^{2} - 11 x - 7\right)}{\left(x + 7\right) {\left(3 x - 2\right)}^{3}}$

Remove common terms:
$\left(15 {x}^{2} - 11 x - 7\right) \frac{\setminus}{\left(x + 7\right) {\left(3 x - 2\right)}^{2}}$

(You can then open the bottom half back up if needed to:
$\frac{15 {x}^{2} - 11 x - 7}{9 {x}^{3} + 51 {x}^{2} - 80 x + 28}$ but the other form should be fine.)

.

$\left(15 {x}^{2} - 11 x - 7\right) \frac{\setminus}{\left(x + 7\right) {\left(3 x - 2\right)}^{2}}$

Jun 3, 2017

 (15x^2 - 11x -7)/((3x-2)^2(x+7)

#### Explanation:

(5x)/(3x^2+19x-14) - 1/(9x^2-12x+4

$= \frac{5 x}{3 {x}^{2} + 21 x - 2 x - 14} - \frac{1}{3 x - 2} ^ 2$

$= \frac{5 x}{3 x \left(x + 7\right) - 2 \left(x + 7\right)} - \frac{1}{3 x - 2} ^ 2$

$= \frac{5 x}{\left(x + 7\right) \left(3 x - 2\right)} - \frac{1}{3 x - 2} ^ 2$

= (5x(3x-2) - (x+7))/((3x-2)^2(x+7)

= (15x^2 - 10x -x-7)/((3x-2)^2(x+7)

= (15x^2 - 11x -7)/((3x-2)^2(x+7) [Ans]

Jun 3, 2017

$\frac{15 {x}^{2} - 11 x - 7}{\left(x + 7\right) {\left(3 x - 2\right)}^{2}}$

#### Explanation:

$\text{before we can subtract the fractions we require them}$
$\text{to have a "color(blue)"common denominator}$

$\text{factorising the denominators gives}$

$\frac{5 x}{\left(3 x - 2\right) \left(x + 7\right)} - \frac{1}{3 x - 2} ^ 2$

$\text{to obtain a common denominator}$

$\text{multiply the numerator/denominator of the fraction on}$
$\text{the left by } \left(3 x - 2\right)$

$\text{multiply the numerator/denominator of the fraction on}$
$\text{the right by } \left(x + 7\right)$

$\frac{5 x}{\left(3 x - 2\right) \left(x + 7\right)} \times \frac{3 x - 2}{3 x - 2} - \frac{1}{3 x - 2} ^ 2 \times \frac{x + 7}{x + 7}$

$= \frac{5 x \left(3 x - 2\right)}{{\left(3 x - 2\right)}^{2} \left(x + 7\right)} - \frac{x + 7}{{\left(3 x - 2\right)}^{2} \left(x + 7\right)}$

$\text{the fractions now have a common denominator so }$
$\text{we can subtract the numerators, leaving the denominator}$

$\Rightarrow \frac{15 {x}^{2} - 10 x - x - 7}{{\left(3 x - 2\right)}^{2} \left(x + 7\right)}$

$= \frac{15 {x}^{2} - 11 x - 7}{{\left(3 x - 2\right)}^{2} \left(x + 7\right)} \rightarrow \left(x \ne \frac{2}{3} , x \ne - 7\right)$