Solve x+10/8x+9=x+10/5x-9

$$\frac{ x+10 }{ 8x+9 } = \frac{ x+10 }{ 5x-9 }$$

Solve for x

$$x=-6$$

$$x=-10$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{9}{8},\frac{9}{5} since division by zero is not defined. Multiply both sides of the equation by \left(5x-9\right)\left(8x+9\right), the least common multiple of 8x+9,5x-9.

$$\left(5x-9\right)\left(x+10\right)=\left(8x+9\right)\left(x+10\right)$$

Use the distributive property to multiply 5x-9 by x+10 and combine like terms.

$$5x^{2}+41x-90=\left(8x+9\right)\left(x+10\right)$$

Use the distributive property to multiply 8x+9 by x+10 and combine like terms.

$$5x^{2}+41x-90=8x^{2}+89x+90$$

Subtract 8x^{2} from both sides.

$$5x^{2}+41x-90-8x^{2}=89x+90$$

Combine 5x^{2} and -8x^{2} to get -3x^{2}.

$$-3x^{2}+41x-90=89x+90$$

Subtract 89x from both sides.

$$-3x^{2}+41x-90-89x=90$$

Combine 41x and -89x to get -48x.

$$-3x^{2}-48x-90=90$$

Subtract 90 from both sides.

$$-3x^{2}-48x-90-90=0$$

Subtract 90 from -90 to get -180.

$$-3x^{2}-48x-180=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -48 for b, and -180 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-48\right)±\sqrt{\left(-48\right)^{2}-4\left(-3\right)\left(-180\right)}}{2\left(-3\right)}$$

Square -48.

$$x=\frac{-\left(-48\right)±\sqrt{2304-4\left(-3\right)\left(-180\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-48\right)±\sqrt{2304+12\left(-180\right)}}{2\left(-3\right)}$$

Multiply 12 times -180.

$$x=\frac{-\left(-48\right)±\sqrt{2304-2160}}{2\left(-3\right)}$$

Add 2304 to -2160.

$$x=\frac{-\left(-48\right)±\sqrt{144}}{2\left(-3\right)}$$

Take the square root of 144.

$$x=\frac{-\left(-48\right)±12}{2\left(-3\right)}$$

The opposite of -48 is 48.

$$x=\frac{48±12}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{48±12}{-6}$$

Now solve the equation x=\frac{48±12}{-6} when ± is plus. Add 48 to 12.

$$x=\frac{60}{-6}$$

Divide 60 by -6.

$$x=-10$$

Now solve the equation x=\frac{48±12}{-6} when ± is minus. Subtract 12 from 48.

$$x=\frac{36}{-6}$$

Divide 36 by -6.

$$x=-6$$

The equation is now solved.

$$x=-10 x=-6$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{9}{8},\frac{9}{5} since division by zero is not defined. Multiply both sides of the equation by \left(5x-9\right)\left(8x+9\right), the least common multiple of 8x+9,5x-9.

$$\left(5x-9\right)\left(x+10\right)=\left(8x+9\right)\left(x+10\right)$$

Use the distributive property to multiply 5x-9 by x+10 and combine like terms.

$$5x^{2}+41x-90=\left(8x+9\right)\left(x+10\right)$$

Use the distributive property to multiply 8x+9 by x+10 and combine like terms.

$$5x^{2}+41x-90=8x^{2}+89x+90$$

Subtract 8x^{2} from both sides.

$$5x^{2}+41x-90-8x^{2}=89x+90$$

Combine 5x^{2} and -8x^{2} to get -3x^{2}.

$$-3x^{2}+41x-90=89x+90$$

Subtract 89x from both sides.

$$-3x^{2}+41x-90-89x=90$$

Combine 41x and -89x to get -48x.

$$-3x^{2}-48x-90=90$$

Add 90 to both sides.

$$-3x^{2}-48x=90+90$$

Add 90 and 90 to get 180.

$$-3x^{2}-48x=180$$

Divide both sides by -3.

$$\frac{-3x^{2}-48x}{-3}=\frac{180}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{48}{-3}\right)x=\frac{180}{-3}$$

Divide -48 by -3.

$$x^{2}+16x=\frac{180}{-3}$$

Divide 180 by -3.

$$x^{2}+16x=-60$$

Divide 16, the coefficient of the x term, by 2 to get 8. Then add the square of 8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+16x+8^{2}=-60+8^{2}$$

Square 8.

$$x^{2}+16x+64=-60+64$$

Add -60 to 64.

$$x^{2}+16x+64=4$$

Factor x^{2}+16x+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+8\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+8\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+8=2 x+8=-2$$

Subtract 8 from both sides of the equation.

$$x=-6 x=-10$$