Solve for x (x^2+x-12)/(x-3)=15

$\frac{x^{2} + x - 12}{x - 3} = 15$

Step 1

Factor each term.

Factor $x^{2} + x - 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

$\frac{(x - 3) (x + 4)}{x - 3} = 15$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{(x - 3) (x + 4)}{x - 3}$ by cancelling the common factors.

Cancel the common factor.

$\frac{(x - 3) (x + 4)}{x - 3} = 15$

Rewrite the expression.

$\frac{x + 4}{1} = 15$

Divide $x + 4$ by $1$ .

$x + 4 = 15$

Step 2

Move all terms not containing $x$ to the right side of the equation.

Subtract $4$ from both sides of the equation.

$x = 15 - 4$

Subtract $4$ from $15$ .

$x = 11$

Solve for x (x^2+x-12)/(x-3)=15

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