Solve 10/x-0.5=12/x+2

$$\frac { 10 } { x } - 0.5 = \frac { 12 } { x + 2 }$$

Solve for x

$$x=-10$$

$$x=4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+2\right), the least common multiple of x,x+2.

$$\left(x+2\right)\times 10+x\left(x+2\right)\left(-0.5\right)=x\times 12$$

Use the distributive property to multiply x+2 by 10.

$$10x+20+x\left(x+2\right)\left(-0.5\right)=x\times 12$$

Use the distributive property to multiply x by x+2.

$$10x+20+\left(x^{2}+2x\right)\left(-0.5\right)=x\times 12$$

Use the distributive property to multiply x^{2}+2x by -0.5.

$$10x+20-0.5x^{2}-x=x\times 12$$

Combine 10x and -x to get 9x.

$$9x+20-0.5x^{2}=x\times 12$$

Subtract x\times 12 from both sides.

$$9x+20-0.5x^{2}-x\times 12=0$$

Combine 9x and -x\times 12 to get -3x.

$$-3x+20-0.5x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-0.5x^{2}-3x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -0.5 for a, -3 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-0.5\right)\times 20}}{2\left(-0.5\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-0.5\right)\times 20}}{2\left(-0.5\right)}$$

Multiply -4 times -0.5.

$$x=\frac{-\left(-3\right)±\sqrt{9+2\times 20}}{2\left(-0.5\right)}$$

Multiply 2 times 20.

$$x=\frac{-\left(-3\right)±\sqrt{9+40}}{2\left(-0.5\right)}$$

Add 9 to 40.

$$x=\frac{-\left(-3\right)±\sqrt{49}}{2\left(-0.5\right)}$$

Take the square root of 49.

$$x=\frac{-\left(-3\right)±7}{2\left(-0.5\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±7}{2\left(-0.5\right)}$$

Multiply 2 times -0.5.

$$x=\frac{3±7}{-1}$$

Now solve the equation x=\frac{3±7}{-1} when ± is plus. Add 3 to 7.

$$x=\frac{10}{-1}$$

Divide 10 by -1.

$$x=-10$$

Now solve the equation x=\frac{3±7}{-1} when ± is minus. Subtract 7 from 3.

$$x=-\frac{4}{-1}$$

Divide -4 by -1.

$$x=4$$

The equation is now solved.

$$x=-10 x=4$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+2\right), the least common multiple of x,x+2.

$$\left(x+2\right)\times 10+x\left(x+2\right)\left(-0.5\right)=x\times 12$$

Use the distributive property to multiply x+2 by 10.

$$10x+20+x\left(x+2\right)\left(-0.5\right)=x\times 12$$

Use the distributive property to multiply x by x+2.

$$10x+20+\left(x^{2}+2x\right)\left(-0.5\right)=x\times 12$$

Use the distributive property to multiply x^{2}+2x by -0.5.

$$10x+20-0.5x^{2}-x=x\times 12$$

Combine 10x and -x to get 9x.

$$9x+20-0.5x^{2}=x\times 12$$

Subtract x\times 12 from both sides.

$$9x+20-0.5x^{2}-x\times 12=0$$

Combine 9x and -x\times 12 to get -3x.

$$-3x+20-0.5x^{2}=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$-3x-0.5x^{2}=-20$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-0.5x^{2}-3x=-20$$

Multiply both sides by -2.

$$\frac{-0.5x^{2}-3x}{-0.5}=-\frac{20}{-0.5}$$

Dividing by -0.5 undoes the multiplication by -0.5.

$$x^{2}+\left(-\frac{3}{-0.5}\right)x=-\frac{20}{-0.5}$$

Divide -3 by -0.5 by multiplying -3 by the reciprocal of -0.5.

$$x^{2}+6x=-\frac{20}{-0.5}$$

Divide -20 by -0.5 by multiplying -20 by the reciprocal of -0.5.

$$x^{2}+6x=40$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=40+3^{2}$$

Square 3.

$$x^{2}+6x+9=40+9$$

Add 40 to 9.

$$x^{2}+6x+9=49$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{49}$$

Simplify.

$$x+3=7 x+3=-7$$

Subtract 3 from both sides of the equation.

$$x=4 x=-10$$