Solve 3-x/(x+1)(x+2)=15/5

$$\frac { 3 - x } { ( x + 1 ) ( x + 2 ) } = \frac { 15 } { 5 }$$

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=-3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,-1 since division by zero is not defined. Multiply both sides of the equation by 5\left(x+1\right)\left(x+2\right), the least common multiple of \left(x+1\right)\left(x+2\right),5.

$$5\left(3-x\right)=\left(x+1\right)\left(x+2\right)\times 15$$

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

$$15-5x=\left(x+1\right)\left(x+2\right)\times 15$$

Use the distributive property to multiply x+1 by x+2 and combine like terms.

$$15-5x=\left(x^{2}+3x+2\right)\times 15$$

Use the distributive property to multiply x^{2}+3x+2 by 15.

$$15-5x=15x^{2}+45x+30$$

Subtract 15x^{2} from both sides.

$$15-5x-15x^{2}=45x+30$$

Subtract 45x from both sides.

$$15-5x-15x^{2}-45x=30$$

Combine -5x and -45x to get -50x.

$$15-50x-15x^{2}=30$$

Subtract 30 from both sides.

$$15-50x-15x^{2}-30=0$$

Subtract 30 from 15 to get -15.

$$-15-50x-15x^{2}=0$$

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.

$$-15x^{2}-50x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -15 for a, -50 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\left(-15\right)\left(-15\right)}}{2\left(-15\right)}$$

Square -50.

$$x=\frac{-\left(-50\right)±\sqrt{2500-4\left(-15\right)\left(-15\right)}}{2\left(-15\right)}$$

Multiply -4 times -15.

$$x=\frac{-\left(-50\right)±\sqrt{2500+60\left(-15\right)}}{2\left(-15\right)}$$

Multiply 60 times -15.

$$x=\frac{-\left(-50\right)±\sqrt{2500-900}}{2\left(-15\right)}$$

Add 2500 to -900.

$$x=\frac{-\left(-50\right)±\sqrt{1600}}{2\left(-15\right)}$$

Take the square root of 1600.

$$x=\frac{-\left(-50\right)±40}{2\left(-15\right)}$$

The opposite of -50 is 50.

$$x=\frac{50±40}{2\left(-15\right)}$$

Multiply 2 times -15.

$$x=\frac{50±40}{-30}$$

Now solve the equation x=\frac{50±40}{-30} when ± is plus. Add 50 to 40.

$$x=\frac{90}{-30}$$

Divide 90 by -30.

$$x=-3$$

Now solve the equation x=\frac{50±40}{-30} when ± is minus. Subtract 40 from 50.

$$x=\frac{10}{-30}$$

Reduce the fraction \frac{10}{-30} to lowest terms by extracting and canceling out 10.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,-1 since division by zero is not defined. Multiply both sides of the equation by 5\left(x+1\right)\left(x+2\right), the least common multiple of \left(x+1\right)\left(x+2\right),5.

$$5\left(3-x\right)=\left(x+1\right)\left(x+2\right)\times 15$$

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

$$15-5x=\left(x+1\right)\left(x+2\right)\times 15$$

Use the distributive property to multiply x+1 by x+2 and combine like terms.

$$15-5x=\left(x^{2}+3x+2\right)\times 15$$

Use the distributive property to multiply x^{2}+3x+2 by 15.

$$15-5x=15x^{2}+45x+30$$

Subtract 15x^{2} from both sides.

$$15-5x-15x^{2}=45x+30$$

Subtract 45x from both sides.

$$15-5x-15x^{2}-45x=30$$

Combine -5x and -45x to get -50x.

$$15-50x-15x^{2}=30$$

Subtract 15 from both sides.

$$-50x-15x^{2}=30-15$$

Subtract 15 from 30 to get 15.

$$-50x-15x^{2}=15$$

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.

$$-15x^{2}-50x=15$$

Divide both sides by -15.

$$\frac{-15x^{2}-50x}{-15}=\frac{15}{-15}$$

Dividing by -15 undoes the multiplication by -15.

$$x^{2}+\left(-\frac{50}{-15}\right)x=\frac{15}{-15}$$

Reduce the fraction \frac{-50}{-15} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{10}{3}x=\frac{15}{-15}$$

Divide 15 by -15.

$$x^{2}+\frac{10}{3}x=-1$$

Divide \frac{10}{3}, the coefficient of the x term, by 2 to get \frac{5}{3}. Then add the square of \frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=-1+\frac{25}{9}$$

Add -1 to \frac{25}{9}.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{16}{9}$$

Factor x^{2}+\frac{10}{3}x+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$x+\frac{5}{3}=\frac{4}{3} x+\frac{5}{3}=-\frac{4}{3}$$

Subtract \frac{5}{3} from both sides of the equation.

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