Solve 220/x-10-160/x=5

$$\frac { 220 } { x - 10 } - \frac { 160 } { x } = 5$$

Solve for x

$$x=-10$$

$$x=32$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,10 since division by zero is not defined. Multiply both sides of the equation by x\left(x-10\right), the least common multiple of x-10,x.

$$x\times 220-\left(x-10\right)\times 160=5x\left(x-10\right)$$

Use the distributive property to multiply x-10 by 160.

$$x\times 220-\left(160x-1600\right)=5x\left(x-10\right)$$

To find the opposite of 160x-1600, find the opposite of each term.

$$x\times 220-160x+1600=5x\left(x-10\right)$$

Combine x\times 220 and -160x to get 60x.

$$60x+1600=5x\left(x-10\right)$$

Use the distributive property to multiply 5x by x-10.

$$60x+1600=5x^{2}-50x$$

Subtract 5x^{2} from both sides.

$$60x+1600-5x^{2}=-50x$$

Add 50x to both sides.

$$60x+1600-5x^{2}+50x=0$$

Combine 60x and 50x to get 110x.

$$110x+1600-5x^{2}=0$$

Divide both sides by 5.

$$22x+320-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+22x+320=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+320. To find a and b, set up a system to be solved.

$$a+b=22 ab=-320=-320$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -320.

$$-1,320 -2,160 -4,80 -5,64 -8,40 -10,32 -16,20$$

Calculate the sum for each pair.

$$-1+320=319 -2+160=158 -4+80=76 -5+64=59 -8+40=32 -10+32=22 -16+20=4$$

The solution is the pair that gives sum 22.

$$a=32 b=-10$$

Rewrite -x^{2}+22x+320 as \left(-x^{2}+32x\right)+\left(-10x+320\right).

$$\left(-x^{2}+32x\right)+\left(-10x+320\right)$$

Factor out -x in the first and -10 in the second group.

$$-x\left(x-32\right)-10\left(x-32\right)$$

Factor out common term x-32 by using distributive property.

$$\left(x-32\right)\left(-x-10\right)$$

To find equation solutions, solve x-32=0 and -x-10=0.

$$x=32 x=-10$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,10 since division by zero is not defined. Multiply both sides of the equation by x\left(x-10\right), the least common multiple of x-10,x.

$$x\times 220-\left(x-10\right)\times 160=5x\left(x-10\right)$$

Use the distributive property to multiply x-10 by 160.

$$x\times 220-\left(160x-1600\right)=5x\left(x-10\right)$$

To find the opposite of 160x-1600, find the opposite of each term.

$$x\times 220-160x+1600=5x\left(x-10\right)$$

Combine x\times 220 and -160x to get 60x.

$$60x+1600=5x\left(x-10\right)$$

Use the distributive property to multiply 5x by x-10.

$$60x+1600=5x^{2}-50x$$

Subtract 5x^{2} from both sides.

$$60x+1600-5x^{2}=-50x$$

Add 50x to both sides.

$$60x+1600-5x^{2}+50x=0$$

Combine 60x and 50x to get 110x.

$$110x+1600-5x^{2}=0$$

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.

$$-5x^{2}+110x+1600=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 110 for b, and 1600 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-110±\sqrt{110^{2}-4\left(-5\right)\times 1600}}{2\left(-5\right)}$$

Square 110.

$$x=\frac{-110±\sqrt{12100-4\left(-5\right)\times 1600}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-110±\sqrt{12100+20\times 1600}}{2\left(-5\right)}$$

Multiply 20 times 1600.

$$x=\frac{-110±\sqrt{12100+32000}}{2\left(-5\right)}$$

Add 12100 to 32000.

$$x=\frac{-110±\sqrt{44100}}{2\left(-5\right)}$$

Take the square root of 44100.

$$x=\frac{-110±210}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-110±210}{-10}$$

Now solve the equation x=\frac{-110±210}{-10} when ± is plus. Add -110 to 210.

$$x=\frac{100}{-10}$$

Divide 100 by -10.

$$x=-10$$

Now solve the equation x=\frac{-110±210}{-10} when ± is minus. Subtract 210 from -110.

$$x=-\frac{320}{-10}$$

Divide -320 by -10.

$$x=32$$

The equation is now solved.

$$x=-10 x=32$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,10 since division by zero is not defined. Multiply both sides of the equation by x\left(x-10\right), the least common multiple of x-10,x.

$$x\times 220-\left(x-10\right)\times 160=5x\left(x-10\right)$$

Use the distributive property to multiply x-10 by 160.

$$x\times 220-\left(160x-1600\right)=5x\left(x-10\right)$$

To find the opposite of 160x-1600, find the opposite of each term.

$$x\times 220-160x+1600=5x\left(x-10\right)$$

Combine x\times 220 and -160x to get 60x.

$$60x+1600=5x\left(x-10\right)$$

Use the distributive property to multiply 5x by x-10.

$$60x+1600=5x^{2}-50x$$

Subtract 5x^{2} from both sides.

$$60x+1600-5x^{2}=-50x$$

Add 50x to both sides.

$$60x+1600-5x^{2}+50x=0$$

Combine 60x and 50x to get 110x.

$$110x+1600-5x^{2}=0$$

Subtract 1600 from both sides. Anything subtracted from zero gives its negation.

$$110x-5x^{2}=-1600$$

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.

$$-5x^{2}+110x=-1600$$

Divide both sides by -5.

$$\frac{-5x^{2}+110x}{-5}=-\frac{1600}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{110}{-5}x=-\frac{1600}{-5}$$

Divide 110 by -5.

$$x^{2}-22x=-\frac{1600}{-5}$$

Divide -1600 by -5.

$$x^{2}-22x=320$$

Divide -22, the coefficient of the x term, by 2 to get -11. Then add the square of -11 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-22x+\left(-11\right)^{2}=320+\left(-11\right)^{2}$$

Square -11.

$$x^{2}-22x+121=320+121$$

Add 320 to 121.

$$x^{2}-22x+121=441$$

Factor x^{2}-22x+121. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-11\right)^{2}=441$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-11\right)^{2}}=\sqrt{441}$$

Simplify.

$$x-11=21 x-11=-21$$

Add 11 to both sides of the equation.

$$x=32 x=-10$$