Solve by Factoring 8/(x+3)=1/x+1

$\frac{8}{x + 3} = \frac{1}{x} + 1$

Step 1

Move all the expressions to the left side of the equation.

Subtract $\frac{1}{x}$ from both sides of the equation.

$\frac{8}{x + 3} - \frac{1}{x} = 1$

Subtract $1$ from both sides of the equation.

$\frac{8}{x + 3} - \frac{1}{x} - 1 = 0$

Step 2

Simplify $\frac{8}{x + 3} - \frac{1}{x} - 1$ .

To write $\frac{8}{x + 3}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$\frac{8}{x + 3} \cdot \frac{x}{x} - \frac{1}{x} - 1 = 0$

To write $- \frac{1}{x}$ as a fraction with a common denominator, multiply by $\frac{x + 3}{x + 3}$ .

$\frac{8}{x + 3} \cdot \frac{x}{x} - \frac{1}{x} \cdot \frac{x + 3}{x + 3} - 1 = 0$

Write each expression with a common denominator of $(x + 3) x$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{8}{x + 3}$ by $\frac{x}{x}$ .

$\frac{8 x}{(x + 3) x} - \frac{1}{x} \cdot \frac{x + 3}{x + 3} - 1 = 0$

Multiply $\frac{1}{x}$ by $\frac{x + 3}{x + 3}$ .

$\frac{8 x}{(x + 3) x} - \frac{x + 3}{x (x + 3)} - 1 = 0$

Reorder the factors of $(x + 3) x$ .

$\frac{8 x}{x (x + 3)} - \frac{x + 3}{x (x + 3)} - 1 = 0$

Combine the numerators over the common denominator.

$\frac{8 x - (x + 3)}{x (x + 3)} - 1 = 0$

Simplify the numerator.

Apply the distributive property.

$\frac{8 x - x - 1 \cdot 3}{x (x + 3)} - 1 = 0$

Multiply $- 1$ by $3$ .

$\frac{8 x - x - 3}{x (x + 3)} - 1 = 0$

Subtract $x$ from $8 x$ .

$\frac{7 x - 3}{x (x + 3)} - 1 = 0$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{x (x + 3)}{x (x + 3)}$ .

$\frac{7 x - 3}{x (x + 3)} - 1 \cdot \frac{x (x + 3)}{x (x + 3)} = 0$

Combine $- 1$ and $\frac{x (x + 3)}{x (x + 3)}$ .

$\frac{7 x - 3}{x (x + 3)} + \frac{- (x (x + 3))}{x (x + 3)} = 0$

Combine the numerators over the common denominator.

$\frac{7 x - 3 - (x (x + 3))}{x (x + 3)} = 0$

Simplify the numerator.

Apply the distributive property.

$\frac{7 x - 3 - x \cdot x - x \cdot 3}{x (x + 3)} = 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$\frac{7 x - 3 - (x \cdot x) - x \cdot 3}{x (x + 3)} = 0$

Multiply $x$ by $x$ .

$\frac{7 x - 3 - x^{2} - x \cdot 3}{x (x + 3)} = 0$

Multiply $3$ by $- 1$ .

$\frac{7 x - 3 - x^{2} - 3 x}{x (x + 3)} = 0$

Subtract $3 x$ from $7 x$ .

$\frac{4 x - 3 - x^{2}}{x (x + 3)} = 0$

Reorder terms.

$\frac{- x^{2} + 4 x - 3}{x (x + 3)} = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 3 = 3$ and whose sum is $b = 4$ .

Factor $4$ out of $4 x$ .

$\frac{- x^{2} + 4 (x) - 3}{x (x + 3)} = 0$

Rewrite $4$ as $1$ plus $3$

$\frac{- x^{2} + (1 + 3) x - 3}{x (x + 3)} = 0$

Apply the distributive property.

$\frac{- x^{2} + 1 x + 3 x - 3}{x (x + 3)} = 0$

Multiply $x$ by $1$ .

$\frac{- x^{2} + x + 3 x - 3}{x (x + 3)} = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(- x^{2} + x) + 3 x - 3}{x (x + 3)} = 0$

Factor out the greatest common factor (GCF) from each group.

$\frac{x (- x + 1) - 3 (- x + 1)}{x (x + 3)} = 0$

Factor the polynomial by factoring out the greatest common factor, $- x + 1$ .

$\frac{(- x + 1) (x - 3)}{x (x + 3)} = 0$

Factor $- 1$ out of $- x$ .

$\frac{(- (x) + 1) (x - 3)}{x (x + 3)} = 0$

Rewrite $1$ as $- 1 (- 1)$ .

$\frac{(- (x) - 1 (- 1)) (x - 3)}{x (x + 3)} = 0$

Factor $- 1$ out of $- (x) - 1 (- 1)$ .

$\frac{- (x - 1) (x - 3)}{x (x + 3)} = 0$

Rewrite $- (x - 1)$ as $- 1 (x - 1)$ .

$\frac{- 1 (x - 1) (x - 3)}{x (x + 3)} = 0$

Move the negative in front of the fraction.

$- \frac{(x - 1) (x - 3)}{x (x + 3)} = 0$

Step 3

Set the numerator equal to zero.

$(x - 1) (x - 3) = 0$

Step 4

Solve the equation for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 1 = 0$

$x - 3 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $(x - 1) (x - 3) = 0$ true.

$x = 1, 3$

Solve by Factoring 8/(x+3)=1/x+1

Related Questions

Support LCMGCF.com by disabling your adblocker.