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

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

Step 1

Factor each term.

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 $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

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

Rewrite $5$ as $- 1$ plus $6$

$\frac{2 x^{2} + (- 1 + 6) x - 3}{x + 3} = 7$

Apply the distributive property.

$\frac{2 x^{2} - 1 x + 6 x - 3}{x + 3} = 7$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(2 x^{2} - 1 x) + 6 x - 3}{x + 3} = 7$

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

$\frac{x (2 x - 1) + 3 (2 x - 1)}{x + 3} = 7$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$\frac{(2 x - 1) (x + 3)}{x + 3} = 7$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{(2 x - 1) (x + 3)}{x + 3}$ by cancelling the common factors.

Cancel the common factor.

$\frac{(2 x - 1) (x + 3)}{x + 3} = 7$

Rewrite the expression.

$\frac{2 x - 1}{1} = 7$

Divide $2 x - 1$ by $1$ .

$2 x - 1 = 7$

Step 2

Solve the equation.

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

Add $1$ to both sides of the equation.

$2 x = 7 + 1$

Add $7$ and $1$ .

$2 x = 8$

Divide each term in $2 x = 8$ by $2$ and simplify.

Divide each term in $2 x = 8$ by $2$ .

$\frac{2 x}{2} = \frac{8}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{8}{2}$

Divide $x$ by $1$ .

$x = \frac{8}{2}$

Simplify the right side.

Divide $8$ by $2$ .

$x = 4$

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

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