Solve x+5/x+3=2x+5/x+1

$$\frac { x + 5 } { x + 3 } = \frac { 2 x + 5 } { x + 1 }$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{15}i-5}{2}\approx -2.5-1.936491673i$$

$$x=\frac{-5+\sqrt{15}i}{2}\approx -2.5+1.936491673i$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply x+1 by x+5 and combine like terms.

$$x^{2}+6x+5=\left(x+3\right)\left(2x+5\right)$$

Use the distributive property to multiply x+3 by 2x+5 and combine like terms.

$$x^{2}+6x+5=2x^{2}+11x+15$$

Subtract 2x^{2} from both sides.

$$x^{2}+6x+5-2x^{2}=11x+15$$

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

$$-x^{2}+6x+5=11x+15$$

Subtract 11x from both sides.

$$-x^{2}+6x+5-11x=15$$

Combine 6x and -11x to get -5x.

$$-x^{2}-5x+5=15$$

Subtract 15 from both sides.

$$-x^{2}-5x+5-15=0$$

Subtract 15 from 5 to get -10.

$$-x^{2}-5x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -5 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)\left(-10\right)}}{2\left(-1\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\left(-10\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-5\right)±\sqrt{25+4\left(-10\right)}}{2\left(-1\right)}$$

Multiply 4 times -10.

$$x=\frac{-\left(-5\right)±\sqrt{25-40}}{2\left(-1\right)}$$

Add 25 to -40.

$$x=\frac{-\left(-5\right)±\sqrt{-15}}{2\left(-1\right)}$$

Take the square root of -15.

$$x=\frac{-\left(-5\right)±\sqrt{15}i}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{15}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{5±\sqrt{15}i}{-2}$$

Now solve the equation x=\frac{5±\sqrt{15}i}{-2} when ± is plus. Add 5 to i\sqrt{15}.

$$x=\frac{5+\sqrt{15}i}{-2}$$

Divide 5+i\sqrt{15} by -2.

$$x=\frac{-\sqrt{15}i-5}{2}$$

Now solve the equation x=\frac{5±\sqrt{15}i}{-2} when ± is minus. Subtract i\sqrt{15} from 5.

$$x=\frac{-\sqrt{15}i+5}{-2}$$

Divide 5-i\sqrt{15} by -2.

$$x=\frac{-5+\sqrt{15}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{15}i-5}{2} x=\frac{-5+\sqrt{15}i}{2}$$

Steps for Completing the Square

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

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

Use the distributive property to multiply x+1 by x+5 and combine like terms.

$$x^{2}+6x+5=\left(x+3\right)\left(2x+5\right)$$

Use the distributive property to multiply x+3 by 2x+5 and combine like terms.

$$x^{2}+6x+5=2x^{2}+11x+15$$

Subtract 2x^{2} from both sides.

$$x^{2}+6x+5-2x^{2}=11x+15$$

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

$$-x^{2}+6x+5=11x+15$$

Subtract 11x from both sides.

$$-x^{2}+6x+5-11x=15$$

Combine 6x and -11x to get -5x.

$$-x^{2}-5x+5=15$$

Subtract 5 from both sides.

$$-x^{2}-5x=15-5$$

Subtract 5 from 15 to get 10.

$$-x^{2}-5x=10$$

Divide both sides by -1.

$$\frac{-x^{2}-5x}{-1}=\frac{10}{-1}$$

Dividing by -1 undoes the multiplication by -1.

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

Divide -5 by -1.

$$x^{2}+5x=\frac{10}{-1}$$

Divide 10 by -1.

$$x^{2}+5x=-10$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=-10+\frac{25}{4}$$

Add -10 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=-\frac{15}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=-\frac{15}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{-\frac{15}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{15}i}{2} x+\frac{5}{2}=-\frac{\sqrt{15}i}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{-5+\sqrt{15}i}{2} x=\frac{-\sqrt{15}i-5}{2}$$