Solve 4/x+1/10=10/90-x

$$\frac{ 4 }{ x } + \frac{ 1 }{ 10 } = \frac{ 10 }{ 90-x }$$

Solve for x

$$x=-90$$

$$x=40$$

Steps Using Factoring

Variable x cannot be equal to any of the values 0,90 since division by zero is not defined. Multiply both sides of the equation by 10x\left(x-90\right), the least common multiple of x,10,90-x.

$$\left(10x-900\right)\times 4+10x\left(x-90\right)\times \frac{1}{10}=-10x\times 10$$

Use the distributive property to multiply 10x-900 by 4.

$$40x-3600+10x\left(x-90\right)\times \frac{1}{10}=-10x\times 10$$

Multiply 10 and \frac{1}{10} to get 1.

$$40x-3600+x\left(x-90\right)=-10x\times 10$$

Use the distributive property to multiply x by x-90.

$$40x-3600+x^{2}-90x=-10x\times 10$$

Combine 40x and -90x to get -50x.

$$-50x-3600+x^{2}=-10x\times 10$$

Multiply -10 and 10 to get -100.

$$-50x-3600+x^{2}=-100x$$

Add 100x to both sides.

$$-50x-3600+x^{2}+100x=0$$

Combine -50x and 100x to get 50x.

$$50x-3600+x^{2}=0$$

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

$$x^{2}+50x-3600=0$$

To solve the equation, factor x^{2}+50x-3600 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=50 ab=-3600$$

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 -3600.

$$-1,3600 -2,1800 -3,1200 -4,900 -5,720 -6,600 -8,450 -9,400 -10,360 -12,300 -15,240 -16,225 -18,200 -20,180 -24,150 -25,144 -30,120 -36,100 -40,90 -45,80 -48,75 -50,72 -60,60$$

Calculate the sum for each pair.

$$-1+3600=3599 -2+1800=1798 -3+1200=1197 -4+900=896 -5+720=715 -6+600=594 -8+450=442 -9+400=391 -10+360=350 -12+300=288 -15+240=225 -16+225=209 -18+200=182 -20+180=160 -24+150=126 -25+144=119 -30+120=90 -36+100=64 -40+90=50 -45+80=35 -48+75=27 -50+72=22 -60+60=0$$

The solution is the pair that gives sum 50.

$$a=-40 b=90$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-40\right)\left(x+90\right)$$

To find equation solutions, solve x-40=0 and x+90=0.

$$x=40 x=-90$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,90 since division by zero is not defined. Multiply both sides of the equation by 10x\left(x-90\right), the least common multiple of x,10,90-x.

$$\left(10x-900\right)\times 4+10x\left(x-90\right)\times \frac{1}{10}=-10x\times 10$$

Use the distributive property to multiply 10x-900 by 4.

$$40x-3600+10x\left(x-90\right)\times \frac{1}{10}=-10x\times 10$$

Multiply 10 and \frac{1}{10} to get 1.

$$40x-3600+x\left(x-90\right)=-10x\times 10$$

Use the distributive property to multiply x by x-90.

$$40x-3600+x^{2}-90x=-10x\times 10$$

Combine 40x and -90x to get -50x.

$$-50x-3600+x^{2}=-10x\times 10$$

Multiply -10 and 10 to get -100.

$$-50x-3600+x^{2}=-100x$$

Add 100x to both sides.

$$-50x-3600+x^{2}+100x=0$$

Combine -50x and 100x to get 50x.

$$50x-3600+x^{2}=0$$

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

$$x^{2}+50x-3600=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-3600. To find a and b, set up a system to be solved.

$$a+b=50 ab=1\left(-3600\right)=-3600$$

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 -3600.

$$-1,3600 -2,1800 -3,1200 -4,900 -5,720 -6,600 -8,450 -9,400 -10,360 -12,300 -15,240 -16,225 -18,200 -20,180 -24,150 -25,144 -30,120 -36,100 -40,90 -45,80 -48,75 -50,72 -60,60$$

Calculate the sum for each pair.

$$-1+3600=3599 -2+1800=1798 -3+1200=1197 -4+900=896 -5+720=715 -6+600=594 -8+450=442 -9+400=391 -10+360=350 -12+300=288 -15+240=225 -16+225=209 -18+200=182 -20+180=160 -24+150=126 -25+144=119 -30+120=90 -36+100=64 -40+90=50 -45+80=35 -48+75=27 -50+72=22 -60+60=0$$

The solution is the pair that gives sum 50.

$$a=-40 b=90$$

Rewrite x^{2}+50x-3600 as \left(x^{2}-40x\right)+\left(90x-3600\right).

$$\left(x^{2}-40x\right)+\left(90x-3600\right)$$

Factor out x in the first and 90 in the second group.

$$x\left(x-40\right)+90\left(x-40\right)$$

Factor out common term x-40 by using distributive property.

$$\left(x-40\right)\left(x+90\right)$$

To find equation solutions, solve x-40=0 and x+90=0.

$$x=40 x=-90$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,90 since division by zero is not defined. Multiply both sides of the equation by 10x\left(x-90\right), the least common multiple of x,10,90-x.

$$\left(10x-900\right)\times 4+10x\left(x-90\right)\times \frac{1}{10}=-10x\times 10$$

Use the distributive property to multiply 10x-900 by 4.

$$40x-3600+10x\left(x-90\right)\times \frac{1}{10}=-10x\times 10$$

Multiply 10 and \frac{1}{10} to get 1.

$$40x-3600+x\left(x-90\right)=-10x\times 10$$

Use the distributive property to multiply x by x-90.

$$40x-3600+x^{2}-90x=-10x\times 10$$

Combine 40x and -90x to get -50x.

$$-50x-3600+x^{2}=-10x\times 10$$

Multiply -10 and 10 to get -100.

$$-50x-3600+x^{2}=-100x$$

Add 100x to both sides.

$$-50x-3600+x^{2}+100x=0$$

Combine -50x and 100x to get 50x.

$$50x-3600+x^{2}=0$$

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.

$$x^{2}+50x-3600=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 50 for b, and -3600 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-50±\sqrt{50^{2}-4\left(-3600\right)}}{2}$$

Square 50.

$$x=\frac{-50±\sqrt{2500-4\left(-3600\right)}}{2}$$

Multiply -4 times -3600.

$$x=\frac{-50±\sqrt{2500+14400}}{2}$$

Add 2500 to 14400.

$$x=\frac{-50±\sqrt{16900}}{2}$$

Take the square root of 16900.

$$x=\frac{-50±130}{2}$$

Now solve the equation x=\frac{-50±130}{2} when ± is plus. Add -50 to 130.

$$x=\frac{80}{2}$$

Divide 80 by 2.

$$x=40$$

Now solve the equation x=\frac{-50±130}{2} when ± is minus. Subtract 130 from -50.

$$x=-\frac{180}{2}$$

Divide -180 by 2.

$$x=-90$$

The equation is now solved.

$$x=40 x=-90$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,90 since division by zero is not defined. Multiply both sides of the equation by 10x\left(x-90\right), the least common multiple of x,10,90-x.

$$\left(10x-900\right)\times 4+10x\left(x-90\right)\times \frac{1}{10}=-10x\times 10$$

Use the distributive property to multiply 10x-900 by 4.

$$40x-3600+10x\left(x-90\right)\times \frac{1}{10}=-10x\times 10$$

Multiply 10 and \frac{1}{10} to get 1.

$$40x-3600+x\left(x-90\right)=-10x\times 10$$

Use the distributive property to multiply x by x-90.

$$40x-3600+x^{2}-90x=-10x\times 10$$

Combine 40x and -90x to get -50x.

$$-50x-3600+x^{2}=-10x\times 10$$

Multiply -10 and 10 to get -100.

$$-50x-3600+x^{2}=-100x$$

Add 100x to both sides.

$$-50x-3600+x^{2}+100x=0$$

Combine -50x and 100x to get 50x.

$$50x-3600+x^{2}=0$$

Add 3600 to both sides. Anything plus zero gives itself.

$$50x+x^{2}=3600$$

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.

$$x^{2}+50x=3600$$

Divide 50, the coefficient of the x term, by 2 to get 25. Then add the square of 25 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+50x+25^{2}=3600+25^{2}$$

Square 25.

$$x^{2}+50x+625=3600+625$$

Add 3600 to 625.

$$x^{2}+50x+625=4225$$

Factor x^{2}+50x+625. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+25\right)^{2}=4225$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+25\right)^{2}}=\sqrt{4225}$$

Simplify.

$$x+25=65 x+25=-65$$

Subtract 25 from both sides of the equation.

$$x=40 x=-90$$