Solve 4/x-3=5/2x+3

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

Solve for x

$$x=-2$$

$$x=1$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -\frac{3}{2},0 since division by zero is not defined. Multiply both sides of the equation by x\left(2x+3\right), the least common multiple of x,2x+3.

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

Use the distributive property to multiply 2x+3 by 4.

$$8x+12+x\left(2x+3\right)\left(-3\right)=x\times 5$$

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

$$8x+12+\left(2x^{2}+3x\right)\left(-3\right)=x\times 5$$

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

$$8x+12-6x^{2}-9x=x\times 5$$

Combine 8x and -9x to get -x.

$$-x+12-6x^{2}=x\times 5$$

Subtract x\times 5 from both sides.

$$-x+12-6x^{2}-x\times 5=0$$

Combine -x and -x\times 5 to get -6x.

$$-6x+12-6x^{2}=0$$

Divide both sides by 6.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-2=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-2$$

Rewrite -x^{2}-x+2 as \left(-x^{2}+x\right)+\left(-2x+2\right).

$$\left(-x^{2}+x\right)+\left(-2x+2\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(-x+1\right)+2\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(x+2\right)$$

To find equation solutions, solve -x+1=0 and x+2=0.

$$x=1 x=-2$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{3}{2},0 since division by zero is not defined. Multiply both sides of the equation by x\left(2x+3\right), the least common multiple of x,2x+3.

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

Use the distributive property to multiply 2x+3 by 4.

$$8x+12+x\left(2x+3\right)\left(-3\right)=x\times 5$$

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

$$8x+12+\left(2x^{2}+3x\right)\left(-3\right)=x\times 5$$

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

$$8x+12-6x^{2}-9x=x\times 5$$

Combine 8x and -9x to get -x.

$$-x+12-6x^{2}=x\times 5$$

Subtract x\times 5 from both sides.

$$-x+12-6x^{2}-x\times 5=0$$

Combine -x and -x\times 5 to get -6x.

$$-6x+12-6x^{2}=0$$

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.

$$-6x^{2}-6x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, -6 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-6\right)\times 12}}{2\left(-6\right)}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-6\right)\times 12}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-6\right)±\sqrt{36+24\times 12}}{2\left(-6\right)}$$

Multiply 24 times 12.

$$x=\frac{-\left(-6\right)±\sqrt{36+288}}{2\left(-6\right)}$$

Add 36 to 288.

$$x=\frac{-\left(-6\right)±\sqrt{324}}{2\left(-6\right)}$$

Take the square root of 324.

$$x=\frac{-\left(-6\right)±18}{2\left(-6\right)}$$

The opposite of -6 is 6.

$$x=\frac{6±18}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{6±18}{-12}$$

Now solve the equation x=\frac{6±18}{-12} when ± is plus. Add 6 to 18.

$$x=\frac{24}{-12}$$

Divide 24 by -12.

$$x=-2$$

Now solve the equation x=\frac{6±18}{-12} when ± is minus. Subtract 18 from 6.

$$x=-\frac{12}{-12}$$

Divide -12 by -12.

$$x=1$$

The equation is now solved.

$$x=-2 x=1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{3}{2},0 since division by zero is not defined. Multiply both sides of the equation by x\left(2x+3\right), the least common multiple of x,2x+3.

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

Use the distributive property to multiply 2x+3 by 4.

$$8x+12+x\left(2x+3\right)\left(-3\right)=x\times 5$$

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

$$8x+12+\left(2x^{2}+3x\right)\left(-3\right)=x\times 5$$

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

$$8x+12-6x^{2}-9x=x\times 5$$

Combine 8x and -9x to get -x.

$$-x+12-6x^{2}=x\times 5$$

Subtract x\times 5 from both sides.

$$-x+12-6x^{2}-x\times 5=0$$

Combine -x and -x\times 5 to get -6x.

$$-6x+12-6x^{2}=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$-6x-6x^{2}=-12$$

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.

$$-6x^{2}-6x=-12$$

Divide both sides by -6.

$$\frac{-6x^{2}-6x}{-6}=-\frac{12}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\left(-\frac{6}{-6}\right)x=-\frac{12}{-6}$$

Divide -6 by -6.

$$x^{2}+x=-\frac{12}{-6}$$

Divide -12 by -6.

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

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=2+\frac{1}{4}$$

Add 2 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{9}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=1 x=-2$$