Solve 60/x+20+1.5=100/x

$$\frac{ 60 }{ x+20 } +1.5= \frac{ 100 }{ x }$$

Solve for x

$$x = -\frac{100}{3} = -33\frac{1}{3} \approx -33.333333333$$

$$x=40$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -20,0 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+20,x.

$$x\times 60+x\left(x+20\right)\times 1.5=\left(x+20\right)\times 100$$

Use the distributive property to multiply x by x+20.

$$x\times 60+\left(x^{2}+20x\right)\times 1.5=\left(x+20\right)\times 100$$

Use the distributive property to multiply x^{2}+20x by 1.5.

$$x\times 60+1.5x^{2}+30x=\left(x+20\right)\times 100$$

Combine x\times 60 and 30x to get 90x.

$$90x+1.5x^{2}=\left(x+20\right)\times 100$$

Use the distributive property to multiply x+20 by 100.

$$90x+1.5x^{2}=100x+2000$$

Subtract 100x from both sides.

$$90x+1.5x^{2}-100x=2000$$

Combine 90x and -100x to get -10x.

$$-10x+1.5x^{2}=2000$$

Subtract 2000 from both sides.

$$-10x+1.5x^{2}-2000=0$$

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.

$$1.5x^{2}-10x-2000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1.5 for a, -10 for b, and -2000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 1.5\left(-2000\right)}}{2\times 1.5}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 1.5\left(-2000\right)}}{2\times 1.5}$$

Multiply -4 times 1.5.

$$x=\frac{-\left(-10\right)±\sqrt{100-6\left(-2000\right)}}{2\times 1.5}$$

Multiply -6 times -2000.

$$x=\frac{-\left(-10\right)±\sqrt{100+12000}}{2\times 1.5}$$

Add 100 to 12000.

$$x=\frac{-\left(-10\right)±\sqrt{12100}}{2\times 1.5}$$

Take the square root of 12100.

$$x=\frac{-\left(-10\right)±110}{2\times 1.5}$$

The opposite of -10 is 10.

$$x=\frac{10±110}{2\times 1.5}$$

Multiply 2 times 1.5.

$$x=\frac{10±110}{3}$$

Now solve the equation x=\frac{10±110}{3} when ± is plus. Add 10 to 110.

$$x=\frac{120}{3}$$

Divide 120 by 3.

$$x=40$$

Now solve the equation x=\frac{10±110}{3} when ± is minus. Subtract 110 from 10.

$$x=-\frac{100}{3}$$

The equation is now solved.

$$x=40 x=-\frac{100}{3}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -20,0 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+20,x.

$$x\times 60+x\left(x+20\right)\times 1.5=\left(x+20\right)\times 100$$

Use the distributive property to multiply x by x+20.

$$x\times 60+\left(x^{2}+20x\right)\times 1.5=\left(x+20\right)\times 100$$

Use the distributive property to multiply x^{2}+20x by 1.5.

$$x\times 60+1.5x^{2}+30x=\left(x+20\right)\times 100$$

Combine x\times 60 and 30x to get 90x.

$$90x+1.5x^{2}=\left(x+20\right)\times 100$$

Use the distributive property to multiply x+20 by 100.

$$90x+1.5x^{2}=100x+2000$$

Subtract 100x from both sides.

$$90x+1.5x^{2}-100x=2000$$

Combine 90x and -100x to get -10x.

$$-10x+1.5x^{2}=2000$$

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.

$$1.5x^{2}-10x=2000$$

Divide both sides of the equation by 1.5, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{1.5x^{2}-10x}{1.5}=\frac{2000}{1.5}$$

Dividing by 1.5 undoes the multiplication by 1.5.

$$x^{2}+\left(-\frac{10}{1.5}\right)x=\frac{2000}{1.5}$$

Divide -10 by 1.5 by multiplying -10 by the reciprocal of 1.5.

$$x^{2}-\frac{20}{3}x=\frac{2000}{1.5}$$

Divide 2000 by 1.5 by multiplying 2000 by the reciprocal of 1.5.

$$x^{2}-\frac{20}{3}x=\frac{4000}{3}$$

Divide -\frac{20}{3}, the coefficient of the x term, by 2 to get -\frac{10}{3}. Then add the square of -\frac{10}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{20}{3}x+\left(-\frac{10}{3}\right)^{2}=\frac{4000}{3}+\left(-\frac{10}{3}\right)^{2}$$

Square -\frac{10}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=\frac{4000}{3}+\frac{100}{9}$$

Add \frac{4000}{3} to \frac{100}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=\frac{12100}{9}$$

Factor x^{2}-\frac{20}{3}x+\frac{100}{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}.

$$\left(x-\frac{10}{3}\right)^{2}=\frac{12100}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{10}{3}\right)^{2}}=\sqrt{\frac{12100}{9}}$$

Simplify.

$$x-\frac{10}{3}=\frac{110}{3} x-\frac{10}{3}=-\frac{110}{3}$$

Add \frac{10}{3} to both sides of the equation.

$$x=40 x=-\frac{100}{3}$$