Solve 300/x+1+40=400/x

$$\frac { 300 } { x + 1 } + 40 = \frac { 400 } { x }$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x=4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+1\right), the least common multiple of x+1,x.

$$x\times 300+x\left(x+1\right)\times 40=\left(x+1\right)\times 400$$

Use the distributive property to multiply x by x+1.

$$x\times 300+\left(x^{2}+x\right)\times 40=\left(x+1\right)\times 400$$

Use the distributive property to multiply x^{2}+x by 40.

$$x\times 300+40x^{2}+40x=\left(x+1\right)\times 400$$

Combine x\times 300 and 40x to get 340x.

$$340x+40x^{2}=\left(x+1\right)\times 400$$

Use the distributive property to multiply x+1 by 400.

$$340x+40x^{2}=400x+400$$

Subtract 400x from both sides.

$$340x+40x^{2}-400x=400$$

Combine 340x and -400x to get -60x.

$$-60x+40x^{2}=400$$

Subtract 400 from both sides.

$$-60x+40x^{2}-400=0$$

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.

$$40x^{2}-60x-400=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 40 for a, -60 for b, and -400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\times 40\left(-400\right)}}{2\times 40}$$

Square -60.

$$x=\frac{-\left(-60\right)±\sqrt{3600-4\times 40\left(-400\right)}}{2\times 40}$$

Multiply -4 times 40.

$$x=\frac{-\left(-60\right)±\sqrt{3600-160\left(-400\right)}}{2\times 40}$$

Multiply -160 times -400.

$$x=\frac{-\left(-60\right)±\sqrt{3600+64000}}{2\times 40}$$

Add 3600 to 64000.

$$x=\frac{-\left(-60\right)±\sqrt{67600}}{2\times 40}$$

Take the square root of 67600.

$$x=\frac{-\left(-60\right)±260}{2\times 40}$$

The opposite of -60 is 60.

$$x=\frac{60±260}{2\times 40}$$

Multiply 2 times 40.

$$x=\frac{60±260}{80}$$

Now solve the equation x=\frac{60±260}{80} when ± is plus. Add 60 to 260.

$$x=\frac{320}{80}$$

Divide 320 by 80.

$$x=4$$

Now solve the equation x=\frac{60±260}{80} when ± is minus. Subtract 260 from 60.

$$x=-\frac{200}{80}$$

Reduce the fraction \frac{-200}{80} to lowest terms by extracting and canceling out 40.

$$x=-\frac{5}{2}$$

The equation is now solved.

$$x=4 x=-\frac{5}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+1\right), the least common multiple of x+1,x.

$$x\times 300+x\left(x+1\right)\times 40=\left(x+1\right)\times 400$$

Use the distributive property to multiply x by x+1.

$$x\times 300+\left(x^{2}+x\right)\times 40=\left(x+1\right)\times 400$$

Use the distributive property to multiply x^{2}+x by 40.

$$x\times 300+40x^{2}+40x=\left(x+1\right)\times 400$$

Combine x\times 300 and 40x to get 340x.

$$340x+40x^{2}=\left(x+1\right)\times 400$$

Use the distributive property to multiply x+1 by 400.

$$340x+40x^{2}=400x+400$$

Subtract 400x from both sides.

$$340x+40x^{2}-400x=400$$

Combine 340x and -400x to get -60x.

$$-60x+40x^{2}=400$$

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.

$$40x^{2}-60x=400$$

Divide both sides by 40.

$$\frac{40x^{2}-60x}{40}=\frac{400}{40}$$

Dividing by 40 undoes the multiplication by 40.

$$x^{2}+\left(-\frac{60}{40}\right)x=\frac{400}{40}$$

Reduce the fraction \frac{-60}{40} to lowest terms by extracting and canceling out 20.

$$x^{2}-\frac{3}{2}x=\frac{400}{40}$$

Divide 400 by 40.

$$x^{2}-\frac{3}{2}x=10$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=10+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=10+\frac{9}{16}$$

Add 10 to \frac{9}{16}.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{169}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{13}{4} x-\frac{3}{4}=-\frac{13}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=4 x=-\frac{5}{2}$$