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

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

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,1 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.

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

Use the distributive property to multiply x-1 by 300.

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

To find the opposite of 300x-300, find the opposite of each term.

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

Combine x\times 400 and -300x to get 100x.

$$100x+300=40x\left(x-1\right)$$

Use the distributive property to multiply 40x by x-1.

$$100x+300=40x^{2}-40x$$

Subtract 40x^{2} from both sides.

$$100x+300-40x^{2}=-40x$$

Add 40x to both sides.

$$100x+300-40x^{2}+40x=0$$

Combine 100x and 40x to get 140x.

$$140x+300-40x^{2}=0$$

Divide both sides by 20.

$$7x+15-2x^{2}=0$$

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

$$-2x^{2}+7x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=7 ab=-2\times 15=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 7.

$$a=10 b=-3$$

Rewrite -2x^{2}+7x+15 as \left(-2x^{2}+10x\right)+\left(-3x+15\right).

$$\left(-2x^{2}+10x\right)+\left(-3x+15\right)$$

Factor out 2x in the first and 3 in the second group.

$$2x\left(-x+5\right)+3\left(-x+5\right)$$

Factor out common term -x+5 by using distributive property.

$$\left(-x+5\right)\left(2x+3\right)$$

To find equation solutions, solve -x+5=0 and 2x+3=0.

$$x=5 x=-\frac{3}{2}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,1 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.

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

Use the distributive property to multiply x-1 by 300.

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

To find the opposite of 300x-300, find the opposite of each term.

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

Combine x\times 400 and -300x to get 100x.

$$100x+300=40x\left(x-1\right)$$

Use the distributive property to multiply 40x by x-1.

$$100x+300=40x^{2}-40x$$

Subtract 40x^{2} from both sides.

$$100x+300-40x^{2}=-40x$$

Add 40x to both sides.

$$100x+300-40x^{2}+40x=0$$

Combine 100x and 40x to get 140x.

$$140x+300-40x^{2}=0$$

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.

$$-40x^{2}+140x+300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -40 for a, 140 for b, and 300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-140±\sqrt{140^{2}-4\left(-40\right)\times 300}}{2\left(-40\right)}$$

Square 140.

$$x=\frac{-140±\sqrt{19600-4\left(-40\right)\times 300}}{2\left(-40\right)}$$

Multiply -4 times -40.

$$x=\frac{-140±\sqrt{19600+160\times 300}}{2\left(-40\right)}$$

Multiply 160 times 300.

$$x=\frac{-140±\sqrt{19600+48000}}{2\left(-40\right)}$$

Add 19600 to 48000.

$$x=\frac{-140±\sqrt{67600}}{2\left(-40\right)}$$

Take the square root of 67600.

$$x=\frac{-140±260}{2\left(-40\right)}$$

Multiply 2 times -40.

$$x=\frac{-140±260}{-80}$$

Now solve the equation x=\frac{-140±260}{-80} when ± is plus. Add -140 to 260.

$$x=\frac{120}{-80}$$

Reduce the fraction \frac{120}{-80} to lowest terms by extracting and canceling out 40.

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

Now solve the equation x=\frac{-140±260}{-80} when ± is minus. Subtract 260 from -140.

$$x=-\frac{400}{-80}$$

Divide -400 by -80.

$$x=5$$

The equation is now solved.

$$x=-\frac{3}{2} x=5$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,1 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.

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

Use the distributive property to multiply x-1 by 300.

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

To find the opposite of 300x-300, find the opposite of each term.

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

Combine x\times 400 and -300x to get 100x.

$$100x+300=40x\left(x-1\right)$$

Use the distributive property to multiply 40x by x-1.

$$100x+300=40x^{2}-40x$$

Subtract 40x^{2} from both sides.

$$100x+300-40x^{2}=-40x$$

Add 40x to both sides.

$$100x+300-40x^{2}+40x=0$$

Combine 100x and 40x to get 140x.

$$140x+300-40x^{2}=0$$

Subtract 300 from both sides. Anything subtracted from zero gives its negation.

$$140x-40x^{2}=-300$$

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.

$$-40x^{2}+140x=-300$$

Divide both sides by -40.

$$\frac{-40x^{2}+140x}{-40}=-\frac{300}{-40}$$

Dividing by -40 undoes the multiplication by -40.

$$x^{2}+\frac{140}{-40}x=-\frac{300}{-40}$$

Reduce the fraction \frac{140}{-40} to lowest terms by extracting and canceling out 20.

$$x^{2}-\frac{7}{2}x=-\frac{300}{-40}$$

Reduce the fraction \frac{-300}{-40} to lowest terms by extracting and canceling out 20.

$$x^{2}-\frac{7}{2}x=\frac{15}{2}$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=\frac{15}{2}+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{15}{2}+\frac{49}{16}$$

Add \frac{15}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{169}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{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}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{13}{4} x-\frac{7}{4}=-\frac{13}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=5 x=-\frac{3}{2}$$