Solve 5x-5/2x+5=2x-11/x-5

$$\frac { 5 x - 5 } { 2 x + 5 } = \frac { 2 x - 11 } { x - 5 }$$

Solve for x

$$x=8$$

$$x=10$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{5}{2},5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(2x+5\right), the least common multiple of 2x+5,x-5.

$$\left(x-5\right)\left(5x-5\right)=\left(2x+5\right)\left(2x-11\right)$$

Use the distributive property to multiply x-5 by 5x-5 and combine like terms.

$$5x^{2}-30x+25=\left(2x+5\right)\left(2x-11\right)$$

Use the distributive property to multiply 2x+5 by 2x-11 and combine like terms.

$$5x^{2}-30x+25=4x^{2}-12x-55$$

Subtract 4x^{2} from both sides.

$$5x^{2}-30x+25-4x^{2}=-12x-55$$

Combine 5x^{2} and -4x^{2} to get x^{2}.

$$x^{2}-30x+25=-12x-55$$

Add 12x to both sides.

$$x^{2}-30x+25+12x=-55$$

Combine -30x and 12x to get -18x.

$$x^{2}-18x+25=-55$$

Add 55 to both sides.

$$x^{2}-18x+25+55=0$$

Add 25 and 55 to get 80.

$$x^{2}-18x+80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -18 for b, and 80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 80}}{2}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 80}}{2}$$

Multiply -4 times 80.

$$x=\frac{-\left(-18\right)±\sqrt{324-320}}{2}$$

Add 324 to -320.

$$x=\frac{-\left(-18\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$x=\frac{-\left(-18\right)±2}{2}$$

The opposite of -18 is 18.

$$x=\frac{18±2}{2}$$

Now solve the equation x=\frac{18±2}{2} when ± is plus. Add 18 to 2.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{18±2}{2} when ± is minus. Subtract 2 from 18.

$$x=\frac{16}{2}$$

Divide 16 by 2.

$$x=8$$

The equation is now solved.

$$x=10 x=8$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{5}{2},5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(2x+5\right), the least common multiple of 2x+5,x-5.

$$\left(x-5\right)\left(5x-5\right)=\left(2x+5\right)\left(2x-11\right)$$

Use the distributive property to multiply x-5 by 5x-5 and combine like terms.

$$5x^{2}-30x+25=\left(2x+5\right)\left(2x-11\right)$$

Use the distributive property to multiply 2x+5 by 2x-11 and combine like terms.

$$5x^{2}-30x+25=4x^{2}-12x-55$$

Subtract 4x^{2} from both sides.

$$5x^{2}-30x+25-4x^{2}=-12x-55$$

Combine 5x^{2} and -4x^{2} to get x^{2}.

$$x^{2}-30x+25=-12x-55$$

Add 12x to both sides.

$$x^{2}-30x+25+12x=-55$$

Combine -30x and 12x to get -18x.

$$x^{2}-18x+25=-55$$

Subtract 25 from both sides.

$$x^{2}-18x=-55-25$$

Subtract 25 from -55 to get -80.

$$x^{2}-18x=-80$$

Divide -18, the coefficient of the x term, by 2 to get -9. Then add the square of -9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-18x+\left(-9\right)^{2}=-80+\left(-9\right)^{2}$$

Square -9.

$$x^{2}-18x+81=-80+81$$

Add -80 to 81.

$$x^{2}-18x+81=1$$

Factor x^{2}-18x+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-9\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-9\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-9=1 x-9=-1$$

Add 9 to both sides of the equation.

$$x=10 x=8$$