Solve {left(x-2/5x+2right)}^-1={left(x/x+40right)}^-1

$${ \left( \frac{ x-2 }{ 5x+2 } \right) }^{ -1 } = { \left( \frac{ x }{ x+40 } \right) }^{ -1 }$$

Solve for x

$$x=4$$

$$x=5$$

Steps Using Factoring By Grouping

To raise \frac{x-2}{5x+2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(x-2\right)^{-1}}{\left(5x+2\right)^{-1}}=\left(\frac{x}{x+40}\right)^{-1}$$

To raise \frac{x}{x+40} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(x-2\right)^{-1}}{\left(5x+2\right)^{-1}}=\frac{x^{-1}}{\left(x+40\right)^{-1}}$$

Subtract \frac{x^{-1}}{\left(x+40\right)^{-1}} from both sides.

$$\frac{\left(x-2\right)^{-1}}{\left(5x+2\right)^{-1}}-\frac{x^{-1}}{\left(x+40\right)^{-1}}=0$$

Reorder the terms.

$$\frac{1}{\frac{1}{5x+2}\left(x-2\right)}-\frac{1}{\frac{1}{x+40}x}=0$$

Express \frac{1}{5x+2}\left(x-2\right) as a single fraction.

$$\frac{1}{\frac{x-2}{5x+2}}-\frac{1}{\frac{1}{x+40}x}=0$$

Variable x cannot be equal to -\frac{2}{5} since division by zero is not defined. Divide 1 by \frac{x-2}{5x+2} by multiplying 1 by the reciprocal of \frac{x-2}{5x+2}.

$$\frac{5x+2}{x-2}-\frac{1}{\frac{1}{x+40}x}=0$$

Express \frac{1}{x+40}x as a single fraction.

$$\frac{5x+2}{x-2}-\frac{1}{\frac{x}{x+40}}=0$$

Variable x cannot be equal to -40 since division by zero is not defined. Divide 1 by \frac{x}{x+40} by multiplying 1 by the reciprocal of \frac{x}{x+40}.

$$\frac{5x+2}{x-2}-\frac{x+40}{x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-2 and x is x\left(x-2\right). Multiply \frac{5x+2}{x-2} times \frac{x}{x}. Multiply \frac{x+40}{x} times \frac{x-2}{x-2}.

$$\frac{\left(5x+2\right)x}{x\left(x-2\right)}-\frac{\left(x+40\right)\left(x-2\right)}{x\left(x-2\right)}=0$$

Since \frac{\left(5x+2\right)x}{x\left(x-2\right)} and \frac{\left(x+40\right)\left(x-2\right)}{x\left(x-2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(5x+2\right)x-\left(x+40\right)\left(x-2\right)}{x\left(x-2\right)}=0$$

Do the multiplications in \left(5x+2\right)x-\left(x+40\right)\left(x-2\right).

$$\frac{5x^{2}+2x-x^{2}+2x-40x+80}{x\left(x-2\right)}=0$$

Combine like terms in 5x^{2}+2x-x^{2}+2x-40x+80.

$$\frac{4x^{2}-36x+80}{x\left(x-2\right)}=0$$

Use the distributive property to multiply x by x-2.

$$\frac{4x^{2}-36x+80}{x^{2}-2x}=0$$

Variable x cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by x\left(x-2\right).

$$4x^{2}-36x+80=0$$

Divide both sides by 4.

$$x^{2}-9x+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 20=20$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -9.

$$a=-5 b=-4$$

Rewrite x^{2}-9x+20 as \left(x^{2}-5x\right)+\left(-4x+20\right).

$$\left(x^{2}-5x\right)+\left(-4x+20\right)$$

Factor out x in the first and -4 in the second group.

$$x\left(x-5\right)-4\left(x-5\right)$$

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

$$\left(x-5\right)\left(x-4\right)$$

To find equation solutions, solve x-5=0 and x-4=0.

$$x=5 x=4$$

Steps Using the Quadratic Formula

To raise \frac{x-2}{5x+2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(x-2\right)^{-1}}{\left(5x+2\right)^{-1}}=\left(\frac{x}{x+40}\right)^{-1}$$

To raise \frac{x}{x+40} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(x-2\right)^{-1}}{\left(5x+2\right)^{-1}}=\frac{x^{-1}}{\left(x+40\right)^{-1}}$$

Subtract \frac{x^{-1}}{\left(x+40\right)^{-1}} from both sides.

$$\frac{\left(x-2\right)^{-1}}{\left(5x+2\right)^{-1}}-\frac{x^{-1}}{\left(x+40\right)^{-1}}=0$$

Reorder the terms.

$$\frac{1}{\frac{1}{5x+2}\left(x-2\right)}-\frac{1}{\frac{1}{x+40}x}=0$$

Express \frac{1}{5x+2}\left(x-2\right) as a single fraction.

$$\frac{1}{\frac{x-2}{5x+2}}-\frac{1}{\frac{1}{x+40}x}=0$$

Variable x cannot be equal to -\frac{2}{5} since division by zero is not defined. Divide 1 by \frac{x-2}{5x+2} by multiplying 1 by the reciprocal of \frac{x-2}{5x+2}.

$$\frac{5x+2}{x-2}-\frac{1}{\frac{1}{x+40}x}=0$$

Express \frac{1}{x+40}x as a single fraction.

$$\frac{5x+2}{x-2}-\frac{1}{\frac{x}{x+40}}=0$$

Variable x cannot be equal to -40 since division by zero is not defined. Divide 1 by \frac{x}{x+40} by multiplying 1 by the reciprocal of \frac{x}{x+40}.

$$\frac{5x+2}{x-2}-\frac{x+40}{x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-2 and x is x\left(x-2\right). Multiply \frac{5x+2}{x-2} times \frac{x}{x}. Multiply \frac{x+40}{x} times \frac{x-2}{x-2}.

$$\frac{\left(5x+2\right)x}{x\left(x-2\right)}-\frac{\left(x+40\right)\left(x-2\right)}{x\left(x-2\right)}=0$$

Since \frac{\left(5x+2\right)x}{x\left(x-2\right)} and \frac{\left(x+40\right)\left(x-2\right)}{x\left(x-2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(5x+2\right)x-\left(x+40\right)\left(x-2\right)}{x\left(x-2\right)}=0$$

Do the multiplications in \left(5x+2\right)x-\left(x+40\right)\left(x-2\right).

$$\frac{5x^{2}+2x-x^{2}+2x-40x+80}{x\left(x-2\right)}=0$$

Combine like terms in 5x^{2}+2x-x^{2}+2x-40x+80.

$$\frac{4x^{2}-36x+80}{x\left(x-2\right)}=0$$

Use the distributive property to multiply x by x-2.

$$\frac{4x^{2}-36x+80}{x^{2}-2x}=0$$

Variable x cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by x\left(x-2\right).

$$4x^{2}-36x+80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -36 for b, and 80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 4\times 80}}{2\times 4}$$

Square -36.

$$x=\frac{-\left(-36\right)±\sqrt{1296-4\times 4\times 80}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-36\right)±\sqrt{1296-16\times 80}}{2\times 4}$$

Multiply -16 times 80.

$$x=\frac{-\left(-36\right)±\sqrt{1296-1280}}{2\times 4}$$

Add 1296 to -1280.

$$x=\frac{-\left(-36\right)±\sqrt{16}}{2\times 4}$$

Take the square root of 16.

$$x=\frac{-\left(-36\right)±4}{2\times 4}$$

The opposite of -36 is 36.

$$x=\frac{36±4}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{36±4}{8}$$

Now solve the equation x=\frac{36±4}{8} when ± is plus. Add 36 to 4.

$$x=\frac{40}{8}$$

Divide 40 by 8.

$$x=5$$

Now solve the equation x=\frac{36±4}{8} when ± is minus. Subtract 4 from 36.

$$x=\frac{32}{8}$$

Divide 32 by 8.

$$x=4$$

The equation is now solved.

$$x=5 x=4$$

Steps for Completing the Square

To raise \frac{x-2}{5x+2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(x-2\right)^{-1}}{\left(5x+2\right)^{-1}}=\left(\frac{x}{x+40}\right)^{-1}$$

To raise \frac{x}{x+40} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(x-2\right)^{-1}}{\left(5x+2\right)^{-1}}=\frac{x^{-1}}{\left(x+40\right)^{-1}}$$

Subtract \frac{x^{-1}}{\left(x+40\right)^{-1}} from both sides.

$$\frac{\left(x-2\right)^{-1}}{\left(5x+2\right)^{-1}}-\frac{x^{-1}}{\left(x+40\right)^{-1}}=0$$

Reorder the terms.

$$\frac{1}{\frac{1}{5x+2}\left(x-2\right)}-\frac{1}{\frac{1}{x+40}x}=0$$

Express \frac{1}{5x+2}\left(x-2\right) as a single fraction.

$$\frac{1}{\frac{x-2}{5x+2}}-\frac{1}{\frac{1}{x+40}x}=0$$

Variable x cannot be equal to -\frac{2}{5} since division by zero is not defined. Divide 1 by \frac{x-2}{5x+2} by multiplying 1 by the reciprocal of \frac{x-2}{5x+2}.

$$\frac{5x+2}{x-2}-\frac{1}{\frac{1}{x+40}x}=0$$

Express \frac{1}{x+40}x as a single fraction.

$$\frac{5x+2}{x-2}-\frac{1}{\frac{x}{x+40}}=0$$

Variable x cannot be equal to -40 since division by zero is not defined. Divide 1 by \frac{x}{x+40} by multiplying 1 by the reciprocal of \frac{x}{x+40}.

$$\frac{5x+2}{x-2}-\frac{x+40}{x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-2 and x is x\left(x-2\right). Multiply \frac{5x+2}{x-2} times \frac{x}{x}. Multiply \frac{x+40}{x} times \frac{x-2}{x-2}.

$$\frac{\left(5x+2\right)x}{x\left(x-2\right)}-\frac{\left(x+40\right)\left(x-2\right)}{x\left(x-2\right)}=0$$

Since \frac{\left(5x+2\right)x}{x\left(x-2\right)} and \frac{\left(x+40\right)\left(x-2\right)}{x\left(x-2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(5x+2\right)x-\left(x+40\right)\left(x-2\right)}{x\left(x-2\right)}=0$$

Do the multiplications in \left(5x+2\right)x-\left(x+40\right)\left(x-2\right).

$$\frac{5x^{2}+2x-x^{2}+2x-40x+80}{x\left(x-2\right)}=0$$

Combine like terms in 5x^{2}+2x-x^{2}+2x-40x+80.

$$\frac{4x^{2}-36x+80}{x\left(x-2\right)}=0$$

Use the distributive property to multiply x by x-2.

$$\frac{4x^{2}-36x+80}{x^{2}-2x}=0$$

Variable x cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by x\left(x-2\right).

$$4x^{2}-36x+80=0$$

Subtract 80 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}-36x=-80$$

Divide both sides by 4.

$$\frac{4x^{2}-36x}{4}=-\frac{80}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{36}{4}\right)x=-\frac{80}{4}$$

Divide -36 by 4.

$$x^{2}-9x=-\frac{80}{4}$$

Divide -80 by 4.

$$x^{2}-9x=-20$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-20+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-20+\frac{81}{4}$$

Add -20 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{1}{4}$$

Factor x^{2}-9x+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{1}{2} x-\frac{9}{2}=-\frac{1}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=5 x=4$$