Solve (7x+1/x^2-9+x/x-3-3/x+3):3x+6/x^2-9=frac{x+5}{3}

$$( \frac { 7 x + 1 } { x ^ { 2 } - 9 } + \frac { x } { x - 3 } - \frac { 3 } { x + 3 } ) : \frac { 3 x + 6 } { x ^ { 2 } - 9 } = \frac { x + 5 } { 3 }$$

Solve for x (complex solution)

$$x\in \mathrm{C}\setminus -3,3,-2$$

Short Solution Steps

Multiply both sides of the equation by 3.

$$\left(2+x\right)^{-1}\left(x^{2}-9\right)\left(\frac{7x+1}{x^{2}-9}+\frac{x}{x-3}-\frac{3}{x+3}\right)=x+5$$

Use the distributive property to multiply \left(2+x\right)^{-1} by x^{2}-9.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(\frac{7x+1}{x^{2}-9}+\frac{x}{x-3}-\frac{3}{x+3}\right)=x+5$$

Factor x^{2}-9.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(\frac{7x+1}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x-3}-\frac{3}{x+3}\right)=x+5$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-3\right)\left(x+3\right) and x-3 is \left(x-3\right)\left(x+3\right). Multiply \frac{x}{x-3} times \frac{x+3}{x+3}.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(\frac{7x+1}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x+3}\right)=x+5$$

Since \frac{7x+1}{\left(x-3\right)\left(x+3\right)} and \frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(\frac{7x+1+x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x+3}\right)=x+5$$

Do the multiplications in 7x+1+x\left(x+3\right).

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(\frac{7x+1+x^{2}+3x}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x+3}\right)=x+5$$

Combine like terms in 7x+1+x^{2}+3x.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(\frac{10x+1+x^{2}}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x+3}\right)=x+5$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-3\right)\left(x+3\right) and x+3 is \left(x-3\right)\left(x+3\right). Multiply \frac{3}{x+3} times \frac{x-3}{x-3}.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(\frac{10x+1+x^{2}}{\left(x-3\right)\left(x+3\right)}-\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right)=x+5$$

Since \frac{10x+1+x^{2}}{\left(x-3\right)\left(x+3\right)} and \frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\times \frac{10x+1+x^{2}-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=x+5$$

Do the multiplications in 10x+1+x^{2}-3\left(x-3\right).

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\times \frac{10x+1+x^{2}-3x+9}{\left(x-3\right)\left(x+3\right)}=x+5$$

Combine like terms in 10x+1+x^{2}-3x+9.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\times \frac{7x+10+x^{2}}{\left(x-3\right)\left(x+3\right)}=x+5$$

Express \left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\times \frac{7x+10+x^{2}}{\left(x-3\right)\left(x+3\right)} as a single fraction.

$$\frac{\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(7x+10+x^{2}\right)}{\left(x-3\right)\left(x+3\right)}=x+5$$

Use the distributive property to multiply \left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1} by 7x+10+x^{2} and combine like terms.

$$\frac{7\left(2+x\right)^{-1}x^{3}+\left(2+x\right)^{-1}x^{2}+\left(2+x\right)^{-1}x^{4}-63\left(2+x\right)^{-1}x-90\left(2+x\right)^{-1}}{\left(x-3\right)\left(x+3\right)}=x+5$$

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$\frac{7\left(2+x\right)^{-1}x^{3}+\left(2+x\right)^{-1}x^{2}+\left(2+x\right)^{-1}x^{4}-63\left(2+x\right)^{-1}x-90\left(2+x\right)^{-1}}{x^{2}-9}=x+5$$

Subtract x from both sides.

$$\frac{7\left(2+x\right)^{-1}x^{3}+\left(2+x\right)^{-1}x^{2}+\left(2+x\right)^{-1}x^{4}-63\left(2+x\right)^{-1}x-90\left(2+x\right)^{-1}}{x^{2}-9}-x=5$$

Subtract 5 from both sides.

$$\frac{7\left(2+x\right)^{-1}x^{3}+\left(2+x\right)^{-1}x^{2}+\left(2+x\right)^{-1}x^{4}-63\left(2+x\right)^{-1}x-90\left(2+x\right)^{-1}}{x^{2}-9}-x-5=0$$

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+3\right).

$$7\left(2+x\right)^{-1}x^{3}+\left(2+x\right)^{-1}x^{2}+\left(2+x\right)^{-1}x^{4}-63\left(2+x\right)^{-1}x-90\left(2+x\right)^{-1}-x\left(x-3\right)\left(x+3\right)+\left(x-3\right)\left(x+3\right)\left(-5\right)=0$$

Reorder the terms.

$$-x\left(x-3\right)\left(x+3\right)+\frac{1}{x+2}x^{4}+7\times \frac{1}{x+2}x^{3}-5\left(x-3\right)\left(x+3\right)+\frac{1}{x+2}x^{2}-63\times \frac{1}{x+2}x-90\times \frac{1}{x+2}=0$$

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by x+2.

$$-x\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{4}+7\times 1x^{3}-5\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Do the multiplications.

$$-x\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{4}+7x^{3}-5\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Use the distributive property to multiply -x by x-3.

$$\left(-x^{2}+3x\right)\left(x+3\right)\left(x+2\right)+1x^{4}+7x^{3}-5\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Use the distributive property to multiply -x^{2}+3x by x+3 and combine like terms.

$$\left(-x^{3}+9x\right)\left(x+2\right)+1x^{4}+7x^{3}-5\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Use the distributive property to multiply -x^{3}+9x by x+2.

$$-x^{4}-2x^{3}+9x^{2}+18x+1x^{4}+7x^{3}-5\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Combine -x^{4} and 1x^{4} to get 0.

$$-2x^{3}+9x^{2}+18x+7x^{3}-5\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Combine -2x^{3} and 7x^{3} to get 5x^{3}.

$$5x^{3}+9x^{2}+18x-5\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Use the distributive property to multiply -5 by x-3.

$$5x^{3}+9x^{2}+18x+\left(-5x+15\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Use the distributive property to multiply -5x+15 by x+3 and combine like terms.

$$5x^{3}+9x^{2}+18x+\left(-5x^{2}+45\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Use the distributive property to multiply -5x^{2}+45 by x+2.

$$5x^{3}+9x^{2}+18x-5x^{3}-10x^{2}+45x+90+1x^{2}-63x-90=0$$

Combine 5x^{3} and -5x^{3} to get 0.

$$9x^{2}+18x-10x^{2}+45x+90+1x^{2}-63x-90=0$$

Combine 9x^{2} and -10x^{2} to get -x^{2}.

$$-x^{2}+18x+45x+90+1x^{2}-63x-90=0$$

Combine 18x and 45x to get 63x.

$$-x^{2}+63x+90+1x^{2}-63x-90=0$$

Combine -x^{2} and 1x^{2} to get 0.

$$63x+90-63x-90=0$$

Combine 63x and -63x to get 0.

$$90-90=0$$

Subtract 90 from 90 to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Variable x cannot be equal to any of the values -2,-3,3.

$$x\in \mathrm{C}\setminus -3,-2,3$$

Solve for x

$$x\in \mathrm{R}\setminus 3,-3,-2$$

Short Solution Steps

Multiply both sides of the equation by 3.

$$\left(2+x\right)^{-1}\left(x^{2}-9\right)\left(\frac{7x+1}{x^{2}-9}+\frac{x}{x-3}-\frac{3}{x+3}\right)=x+5$$

Use the distributive property to multiply \left(2+x\right)^{-1} by x^{2}-9.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(\frac{7x+1}{x^{2}-9}+\frac{x}{x-3}-\frac{3}{x+3}\right)=x+5$$

Factor x^{2}-9.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(\frac{7x+1}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x-3}-\frac{3}{x+3}\right)=x+5$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-3\right)\left(x+3\right) and x-3 is \left(x-3\right)\left(x+3\right). Multiply \frac{x}{x-3} times \frac{x+3}{x+3}.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(\frac{7x+1}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x+3}\right)=x+5$$

Since \frac{7x+1}{\left(x-3\right)\left(x+3\right)} and \frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(\frac{7x+1+x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x+3}\right)=x+5$$

Do the multiplications in 7x+1+x\left(x+3\right).

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(\frac{7x+1+x^{2}+3x}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x+3}\right)=x+5$$

Combine like terms in 7x+1+x^{2}+3x.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(\frac{10x+1+x^{2}}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x+3}\right)=x+5$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-3\right)\left(x+3\right) and x+3 is \left(x-3\right)\left(x+3\right). Multiply \frac{3}{x+3} times \frac{x-3}{x-3}.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(\frac{10x+1+x^{2}}{\left(x-3\right)\left(x+3\right)}-\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right)=x+5$$

Since \frac{10x+1+x^{2}}{\left(x-3\right)\left(x+3\right)} and \frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\times \frac{10x+1+x^{2}-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=x+5$$

Do the multiplications in 10x+1+x^{2}-3\left(x-3\right).

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\times \frac{10x+1+x^{2}-3x+9}{\left(x-3\right)\left(x+3\right)}=x+5$$

Combine like terms in 10x+1+x^{2}-3x+9.

$$\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\times \frac{7x+10+x^{2}}{\left(x-3\right)\left(x+3\right)}=x+5$$

Express \left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\times \frac{7x+10+x^{2}}{\left(x-3\right)\left(x+3\right)} as a single fraction.

$$\frac{\left(\left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1}\right)\left(7x+10+x^{2}\right)}{\left(x-3\right)\left(x+3\right)}=x+5$$

Use the distributive property to multiply \left(2+x\right)^{-1}x^{2}-9\left(2+x\right)^{-1} by 7x+10+x^{2} and combine like terms.

$$\frac{7\left(2+x\right)^{-1}x^{3}+\left(2+x\right)^{-1}x^{2}+\left(2+x\right)^{-1}x^{4}-63\left(2+x\right)^{-1}x-90\left(2+x\right)^{-1}}{\left(x-3\right)\left(x+3\right)}=x+5$$

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$\frac{7\left(2+x\right)^{-1}x^{3}+\left(2+x\right)^{-1}x^{2}+\left(2+x\right)^{-1}x^{4}-63\left(2+x\right)^{-1}x-90\left(2+x\right)^{-1}}{x^{2}-9}=x+5$$

Subtract x from both sides.

$$\frac{7\left(2+x\right)^{-1}x^{3}+\left(2+x\right)^{-1}x^{2}+\left(2+x\right)^{-1}x^{4}-63\left(2+x\right)^{-1}x-90\left(2+x\right)^{-1}}{x^{2}-9}-x=5$$

Subtract 5 from both sides.

$$\frac{7\left(2+x\right)^{-1}x^{3}+\left(2+x\right)^{-1}x^{2}+\left(2+x\right)^{-1}x^{4}-63\left(2+x\right)^{-1}x-90\left(2+x\right)^{-1}}{x^{2}-9}-x-5=0$$

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+3\right).

$$7\left(2+x\right)^{-1}x^{3}+\left(2+x\right)^{-1}x^{2}+\left(2+x\right)^{-1}x^{4}-63\left(2+x\right)^{-1}x-90\left(2+x\right)^{-1}-x\left(x-3\right)\left(x+3\right)+\left(x-3\right)\left(x+3\right)\left(-5\right)=0$$

Reorder the terms.

$$-x\left(x-3\right)\left(x+3\right)+\frac{1}{x+2}x^{4}+7\times \frac{1}{x+2}x^{3}-5\left(x-3\right)\left(x+3\right)+\frac{1}{x+2}x^{2}-63\times \frac{1}{x+2}x-90\times \frac{1}{x+2}=0$$

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by x+2.

$$-x\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{4}+7\times 1x^{3}-5\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Do the multiplications.

$$-x\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{4}+7x^{3}-5\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Use the distributive property to multiply -x by x-3.

$$\left(-x^{2}+3x\right)\left(x+3\right)\left(x+2\right)+1x^{4}+7x^{3}-5\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Use the distributive property to multiply -x^{2}+3x by x+3 and combine like terms.

$$\left(-x^{3}+9x\right)\left(x+2\right)+1x^{4}+7x^{3}-5\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Use the distributive property to multiply -x^{3}+9x by x+2.

$$-x^{4}-2x^{3}+9x^{2}+18x+1x^{4}+7x^{3}-5\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Combine -x^{4} and 1x^{4} to get 0.

$$-2x^{3}+9x^{2}+18x+7x^{3}-5\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Combine -2x^{3} and 7x^{3} to get 5x^{3}.

$$5x^{3}+9x^{2}+18x-5\left(x-3\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Use the distributive property to multiply -5 by x-3.

$$5x^{3}+9x^{2}+18x+\left(-5x+15\right)\left(x+3\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Use the distributive property to multiply -5x+15 by x+3 and combine like terms.

$$5x^{3}+9x^{2}+18x+\left(-5x^{2}+45\right)\left(x+2\right)+1x^{2}-63x-90=0$$

Use the distributive property to multiply -5x^{2}+45 by x+2.

$$5x^{3}+9x^{2}+18x-5x^{3}-10x^{2}+45x+90+1x^{2}-63x-90=0$$

Combine 5x^{3} and -5x^{3} to get 0.

$$9x^{2}+18x-10x^{2}+45x+90+1x^{2}-63x-90=0$$

Combine 9x^{2} and -10x^{2} to get -x^{2}.

$$-x^{2}+18x+45x+90+1x^{2}-63x-90=0$$

Combine 18x and 45x to get 63x.

$$-x^{2}+63x+90+1x^{2}-63x-90=0$$

Combine -x^{2} and 1x^{2} to get 0.

$$63x+90-63x-90=0$$

Combine 63x and -63x to get 0.

$$90-90=0$$

Subtract 90 from 90 to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$

Variable x cannot be equal to any of the values -2,-3,3.

$$x\in \mathrm{R}\setminus -3,-2,3$$