Solve 400/x+400div5*2/x+400div5*3/x-20=11

$$\frac{ 400 }{ x } + \frac{ 400 \div 5 \times 2 }{ x } + \frac{ 400 \div 5 \times 3 }{ x-20 } =11$$

Solve for x

$$x=80$$

$$x = \frac{140}{11} = 12\frac{8}{11} \approx 12.727272727$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,20 since division by zero is not defined. Multiply both sides of the equation by x\left(x-20\right), the least common multiple of x,x-20.

$$\left(x-20\right)\times 400+\left(x-20\right)\times \frac{400}{5}\times 2+x\times \frac{400}{5}\times 3=11x\left(x-20\right)$$

Use the distributive property to multiply x-20 by 400.

$$400x-8000+\left(x-20\right)\times \frac{400}{5}\times 2+x\times \frac{400}{5}\times 3=11x\left(x-20\right)$$

Divide 400 by 5 to get 80.

$$400x-8000+\left(x-20\right)\times 80\times 2+x\times \frac{400}{5}\times 3=11x\left(x-20\right)$$

Multiply 80 and 2 to get 160.

$$400x-8000+\left(x-20\right)\times 160+x\times \frac{400}{5}\times 3=11x\left(x-20\right)$$

Use the distributive property to multiply x-20 by 160.

$$400x-8000+160x-3200+x\times \frac{400}{5}\times 3=11x\left(x-20\right)$$

Combine 400x and 160x to get 560x.

$$560x-8000-3200+x\times \frac{400}{5}\times 3=11x\left(x-20\right)$$

Subtract 3200 from -8000 to get -11200.

$$560x-11200+x\times \frac{400}{5}\times 3=11x\left(x-20\right)$$

Divide 400 by 5 to get 80.

$$560x-11200+x\times 80\times 3=11x\left(x-20\right)$$

Multiply 80 and 3 to get 240.

$$560x-11200+x\times 240=11x\left(x-20\right)$$

Combine 560x and x\times 240 to get 800x.

$$800x-11200=11x\left(x-20\right)$$

Use the distributive property to multiply 11x by x-20.

$$800x-11200=11x^{2}-220x$$

Subtract 11x^{2} from both sides.

$$800x-11200-11x^{2}=-220x$$

Add 220x to both sides.

$$800x-11200-11x^{2}+220x=0$$

Combine 800x and 220x to get 1020x.

$$1020x-11200-11x^{2}=0$$

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.

$$-11x^{2}+1020x-11200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -11 for a, 1020 for b, and -11200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1020±\sqrt{1020^{2}-4\left(-11\right)\left(-11200\right)}}{2\left(-11\right)}$$

Square 1020.

$$x=\frac{-1020±\sqrt{1040400-4\left(-11\right)\left(-11200\right)}}{2\left(-11\right)}$$

Multiply -4 times -11.

$$x=\frac{-1020±\sqrt{1040400+44\left(-11200\right)}}{2\left(-11\right)}$$

Multiply 44 times -11200.

$$x=\frac{-1020±\sqrt{1040400-492800}}{2\left(-11\right)}$$

Add 1040400 to -492800.

$$x=\frac{-1020±\sqrt{547600}}{2\left(-11\right)}$$

Take the square root of 547600.

$$x=\frac{-1020±740}{2\left(-11\right)}$$

Multiply 2 times -11.

$$x=\frac{-1020±740}{-22}$$

Now solve the equation x=\frac{-1020±740}{-22} when ± is plus. Add -1020 to 740.

$$x=-\frac{280}{-22}$$

Reduce the fraction \frac{-280}{-22} to lowest terms by extracting and canceling out 2.

$$x=\frac{140}{11}$$

Now solve the equation x=\frac{-1020±740}{-22} when ± is minus. Subtract 740 from -1020.

$$x=-\frac{1760}{-22}$$

Divide -1760 by -22.

$$x=80$$

The equation is now solved.

$$x=\frac{140}{11} x=80$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,20 since division by zero is not defined. Multiply both sides of the equation by x\left(x-20\right), the least common multiple of x,x-20.

$$\left(x-20\right)\times 400+\left(x-20\right)\times \frac{400}{5}\times 2+x\times \frac{400}{5}\times 3=11x\left(x-20\right)$$

Use the distributive property to multiply x-20 by 400.

$$400x-8000+\left(x-20\right)\times \frac{400}{5}\times 2+x\times \frac{400}{5}\times 3=11x\left(x-20\right)$$

Divide 400 by 5 to get 80.

$$400x-8000+\left(x-20\right)\times 80\times 2+x\times \frac{400}{5}\times 3=11x\left(x-20\right)$$

Multiply 80 and 2 to get 160.

$$400x-8000+\left(x-20\right)\times 160+x\times \frac{400}{5}\times 3=11x\left(x-20\right)$$

Use the distributive property to multiply x-20 by 160.

$$400x-8000+160x-3200+x\times \frac{400}{5}\times 3=11x\left(x-20\right)$$

Combine 400x and 160x to get 560x.

$$560x-8000-3200+x\times \frac{400}{5}\times 3=11x\left(x-20\right)$$

Subtract 3200 from -8000 to get -11200.

$$560x-11200+x\times \frac{400}{5}\times 3=11x\left(x-20\right)$$

Divide 400 by 5 to get 80.

$$560x-11200+x\times 80\times 3=11x\left(x-20\right)$$

Multiply 80 and 3 to get 240.

$$560x-11200+x\times 240=11x\left(x-20\right)$$

Combine 560x and x\times 240 to get 800x.

$$800x-11200=11x\left(x-20\right)$$

Use the distributive property to multiply 11x by x-20.

$$800x-11200=11x^{2}-220x$$

Subtract 11x^{2} from both sides.

$$800x-11200-11x^{2}=-220x$$

Add 220x to both sides.

$$800x-11200-11x^{2}+220x=0$$

Combine 800x and 220x to get 1020x.

$$1020x-11200-11x^{2}=0$$

Add 11200 to both sides. Anything plus zero gives itself.

$$1020x-11x^{2}=11200$$

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.

$$-11x^{2}+1020x=11200$$

Divide both sides by -11.

$$\frac{-11x^{2}+1020x}{-11}=\frac{11200}{-11}$$

Dividing by -11 undoes the multiplication by -11.

$$x^{2}+\frac{1020}{-11}x=\frac{11200}{-11}$$

Divide 1020 by -11.

$$x^{2}-\frac{1020}{11}x=\frac{11200}{-11}$$

Divide 11200 by -11.

$$x^{2}-\frac{1020}{11}x=-\frac{11200}{11}$$

Divide -\frac{1020}{11}, the coefficient of the x term, by 2 to get -\frac{510}{11}. Then add the square of -\frac{510}{11} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1020}{11}x+\left(-\frac{510}{11}\right)^{2}=-\frac{11200}{11}+\left(-\frac{510}{11}\right)^{2}$$

Square -\frac{510}{11} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1020}{11}x+\frac{260100}{121}=-\frac{11200}{11}+\frac{260100}{121}$$

Add -\frac{11200}{11} to \frac{260100}{121} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1020}{11}x+\frac{260100}{121}=\frac{136900}{121}$$

Factor x^{2}-\frac{1020}{11}x+\frac{260100}{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}.

$$\left(x-\frac{510}{11}\right)^{2}=\frac{136900}{121}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{510}{11}\right)^{2}}=\sqrt{\frac{136900}{121}}$$

Simplify.

$$x-\frac{510}{11}=\frac{370}{11} x-\frac{510}{11}=-\frac{370}{11}$$

Add \frac{510}{11} to both sides of the equation.

$$x=80 x=\frac{140}{11}$$