Solve 27+x/12x+36=3/2x-1

$$\frac { 27 + x } { 12 x + 36 } = \frac { 3 } { 2 x - 1 }$$

Solve for x

$$x = -\frac{27}{2} = -13\frac{1}{2} = -13.5$$

$$x=5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by 12\left(2x-1\right)\left(x+3\right), the least common multiple of 12x+36,2x-1.

$$\left(2x-1\right)\left(27+x\right)=\left(12x+36\right)\times 3$$

Use the distributive property to multiply 2x-1 by 27+x and combine like terms.

$$53x+2x^{2}-27=\left(12x+36\right)\times 3$$

Use the distributive property to multiply 12x+36 by 3.

$$53x+2x^{2}-27=36x+108$$

Subtract 36x from both sides.

$$53x+2x^{2}-27-36x=108$$

Combine 53x and -36x to get 17x.

$$17x+2x^{2}-27=108$$

Subtract 108 from both sides.

$$17x+2x^{2}-27-108=0$$

Subtract 108 from -27 to get -135.

$$17x+2x^{2}-135=0$$

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.

$$2x^{2}+17x-135=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 17 for b, and -135 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-17±\sqrt{17^{2}-4\times 2\left(-135\right)}}{2\times 2}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\times 2\left(-135\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-17±\sqrt{289-8\left(-135\right)}}{2\times 2}$$

Multiply -8 times -135.

$$x=\frac{-17±\sqrt{289+1080}}{2\times 2}$$

Add 289 to 1080.

$$x=\frac{-17±\sqrt{1369}}{2\times 2}$$

Take the square root of 1369.

$$x=\frac{-17±37}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-17±37}{4}$$

Now solve the equation x=\frac{-17±37}{4} when ± is plus. Add -17 to 37.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{-17±37}{4} when ± is minus. Subtract 37 from -17.

$$x=-\frac{54}{4}$$

Reduce the fraction \frac{-54}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{27}{2}$$

The equation is now solved.

$$x=5 x=-\frac{27}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by 12\left(2x-1\right)\left(x+3\right), the least common multiple of 12x+36,2x-1.

$$\left(2x-1\right)\left(27+x\right)=\left(12x+36\right)\times 3$$

Use the distributive property to multiply 2x-1 by 27+x and combine like terms.

$$53x+2x^{2}-27=\left(12x+36\right)\times 3$$

Use the distributive property to multiply 12x+36 by 3.

$$53x+2x^{2}-27=36x+108$$

Subtract 36x from both sides.

$$53x+2x^{2}-27-36x=108$$

Combine 53x and -36x to get 17x.

$$17x+2x^{2}-27=108$$

Add 27 to both sides.

$$17x+2x^{2}=108+27$$

Add 108 and 27 to get 135.

$$17x+2x^{2}=135$$

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.

$$2x^{2}+17x=135$$

Divide both sides by 2.

$$\frac{2x^{2}+17x}{2}=\frac{135}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{17}{2}x=\frac{135}{2}$$

Divide \frac{17}{2}, the coefficient of the x term, by 2 to get \frac{17}{4}. Then add the square of \frac{17}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{17}{2}x+\left(\frac{17}{4}\right)^{2}=\frac{135}{2}+\left(\frac{17}{4}\right)^{2}$$

Square \frac{17}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{17}{2}x+\frac{289}{16}=\frac{135}{2}+\frac{289}{16}$$

Add \frac{135}{2} to \frac{289}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{17}{2}x+\frac{289}{16}=\frac{1369}{16}$$

Factor x^{2}+\frac{17}{2}x+\frac{289}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{17}{4}\right)^{2}=\frac{1369}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{17}{4}\right)^{2}}=\sqrt{\frac{1369}{16}}$$

Simplify.

$$x+\frac{17}{4}=\frac{37}{4} x+\frac{17}{4}=-\frac{37}{4}$$

Subtract \frac{17}{4} from both sides of the equation.

$$x=5 x=-\frac{27}{2}$$