# How do you solve 7/(x-4) = 1 + 9/(x+4)?

Apr 8, 2018

$x = - 8$ or $x = 10$

#### Explanation:

first, a common denominator for all the fractions needs to be found.

this is the lowest common multiple of $1$, $x - 4$, and $x + 4$.

since $x + 4$ and $x - 4$ have no common factors, their lowest common multiple is the two expressions multiplied together.

this gives $\left(x + 4\right) \left(x - 4\right)$.

$\left(x + 4\right) \left(x - 4\right)$ can be used as the denominator for both fractions:

$\frac{7}{x - 4} = \frac{7 \left(x + 4\right)}{\left(x - 4\right) \left(x + 4\right)}$

$\frac{9}{x + 4} = \frac{9 \left(x - 4\right)}{\left(x - 4\right) \left(x + 4\right)}$

$1 = \frac{\left(x - 4\right) \left(x + 4\right)}{\left(x - 4\right) \left(x + 4\right)}$

$\frac{7}{x - 4} = 1 + \frac{9}{x + 4}$

$\frac{7 \left(x + 4\right)}{\left(x - 4\right) \left(x + 4\right)} = \frac{\left(x - 4\right) \left(x + 4\right)}{\left(x - 4\right) \left(x + 4\right)} + \frac{9 \left(x - 4\right)}{\left(x - 4\right) \left(x + 4\right)}$

the denominator can then be taken away by multiplying everything by $\left(x - 4\right) \left(x + 4\right) :$

$7 \left(x + 4\right) = \left(x - 4\right) \left(x + 4\right) + 9 \left(x - 4\right)$

then the brackets can be expanded:

$7 \left(x + 4\right) = 7 x + 28$

$\left(x - 4\right) \left(x + 4\right) = {x}^{2} - 16$

(difference of two squares identity: $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$)

$9 \left(x - 4\right) = 9 x - 36$

$\left(7 x + 28\right) = \left({x}^{2} - 16\right) + \left(9 x - 36\right)$

$7 x + 28 = {x}^{2} - 16 + 9 x - 36$

collect like terms:

$7 x + 28 = {x}^{2} + 9 x - 16 - 36$

$7 x + 28 = {x}^{2} + 9 x - 52$

${x}^{2} + 9 x - 52 = 7 x + 28$

then subtract $7 x :$

${x}^{2} + 2 x - 52 = 28$

subtract $28$ so that $0$ is on the right-hand side:

${x}^{2} - 2 x - 80 = 0$

this forms a quadratic eqaution that can be factorised.

to do this, find two numbers that add to make $- 2$

and multiply to make $- 80$.

$8 + \left(- 10\right) = 8 - 10 = - 2$
$8 \cdot - 10 = - 80$

${x}^{2} - 2 x - 80 = \left(x + 8\right) \left(x - 10\right)$

$\left(x + 8\right) \left(x - 10\right) = 0$

for the number on the right-hand side to be $0$, at least one of the numbers in brackets has to be $0$.

either $x + 8 = 0$ or $x - 10 = 0$

if $x + 8 = 0$, then $x = - 8$.

if $x - 10 = 0$, then $x = 10$.

this means that the two possible values for $x$ are $- 8$ and $10$.

Apr 10, 2018

$x$ equals $8 \mathmr{and} - 10$

#### Explanation:

7/(x−4)=1/(1)+9/(x+4)     Solve for $x$

1) Give the addends on the right side a common denominator

7/(x−4)=(x+4)/(x+4)+9/(x+4)

2) Add the like fractions on the right by adding the numerators and keeping the common denominator

7/(x−4)=(x+4+9)/(x+4)

3) Combine like terms in the numerator on the right

7/(x−4)=(x+13)/(x+4)

4) Clear the first fraction by multiplying both sides by $x - 4$ and letting the denominator on the left cancel

$7 = \frac{\left(x - 4\right) \left(x + 13\right)}{x + 4}$

5) Clear the second fraction by multiplying both sides by $x + 4$ and letting the right side denominator cancel

7(x + 4)=(x−4)(x+13)

6) Clear the parentheses by distributing the $7$ and by multiplying the binomials

$7 x + 28 = {x}^{2} + 9 x - 52$

7) Subtract $7 x$ from both sides to collect all the $x$ terms together

$28 = {x}^{2} + 2 x - 52$

8) Set the equation to $0$ by subtracting $28$ from both sides

${x}^{2} + 2 x - 80 = 0$

9) Factor

$\left(x + 10\right) \left(x - 8\right) = 0$

10) Set the factors equal to $0$ and solve for $x$

$x + 10 = 0$
$x = - 10$

$x - 8 = 0$
$x = 8$

$\left(8 , - 10\right)$

$\textcolor{w h i t e}{m m m m m m m m}$―――――――――

Check

Sub in $8$ in the original equation to be sure it still is an equality
7/(x−4)=1/(1)+9/(x+4)

7/(8−4) should equal $\frac{1}{1} + \frac{9}{8 + 4}$

$\frac{7}{4}$ should still equal $\frac{12}{12} + \frac{9}{12}$

$\frac{7}{4}$ should still equal $\frac{21}{12}$

Reduce the fraction to lowest terms

$\frac{7}{4}$ does equal $\frac{7}{4}$

$C h e c k$