Solve 4/x-3-2/x+1=2(x+5)/(x-3)(x+1)

$$\frac { 4 } { x - 3 } - \frac { 2 } { x + 1 } = \frac { 2 ( x + 5 ) } { ( x - 3 ) ( x + 1 ) }$$

Solve for x (complex solution)

$$x\in \mathrm{C}\setminus 3,-1$$

Solution Steps

Variable x cannot be equal to any of the values -1,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+1\right), the least common multiple of x-3,x+1,\left(x-3\right)\left(x+1\right).

$$\left(x+1\right)\times 4-\left(x-3\right)\times 2=2\left(x+5\right)$$

Use the distributive property to multiply x+1 by 4.

$$4x+4-\left(x-3\right)\times 2=2\left(x+5\right)$$

Use the distributive property to multiply x-3 by 2.

$$4x+4-\left(2x-6\right)=2\left(x+5\right)$$

To find the opposite of 2x-6, find the opposite of each term.

$$4x+4-2x+6=2\left(x+5\right)$$

Combine 4x and -2x to get 2x.

$$2x+4+6=2\left(x+5\right)$$

Add 4 and 6 to get 10.

$$2x+10=2\left(x+5\right)$$

Use the distributive property to multiply 2 by x+5.

$$2x+10=2x+10$$

Subtract 2x from both sides.

$$2x+10-2x=10$$

Combine 2x and -2x to get 0.

$$10=10$$

Compare 10 and 10.

$$\text{true}$$

This is true for any x.

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

Variable x cannot be equal to any of the values -1,3.

$$x\in \mathrm{C}\setminus -1,3$$

Solve for x

$$x\in \mathrm{R}\setminus 3,-1$$

Solution Steps

Variable x cannot be equal to any of the values -1,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+1\right), the least common multiple of x-3,x+1,\left(x-3\right)\left(x+1\right).

$$\left(x+1\right)\times 4-\left(x-3\right)\times 2=2\left(x+5\right)$$

Use the distributive property to multiply x+1 by 4.

$$4x+4-\left(x-3\right)\times 2=2\left(x+5\right)$$

Use the distributive property to multiply x-3 by 2.

$$4x+4-\left(2x-6\right)=2\left(x+5\right)$$

To find the opposite of 2x-6, find the opposite of each term.

$$4x+4-2x+6=2\left(x+5\right)$$

Combine 4x and -2x to get 2x.

$$2x+4+6=2\left(x+5\right)$$

Add 4 and 6 to get 10.

$$2x+10=2\left(x+5\right)$$

Use the distributive property to multiply 2 by x+5.

$$2x+10=2x+10$$

Subtract 2x from both sides.

$$2x+10-2x=10$$

Combine 2x and -2x to get 0.

$$10=10$$

Compare 10 and 10.

$$\text{true}$$

This is true for any x.

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

Variable x cannot be equal to any of the values -1,3.

$$x\in \mathrm{R}\setminus -1,3$$