Solve 6/x-7/2x+3=2

$$\frac { 6 } { x } - \frac { 7 } { 2 x + 3 } = 2$$

Solve for x

$$x = -\frac{9}{4} = -2\frac{1}{4} = -2.25$$

$$x=2$$

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.

$$\left(2x+3\right)\times 6-x\times 7=2x\left(2x+3\right)$$

Use the distributive property to multiply 2x+3 by 6.

$$12x+18-x\times 7=2x\left(2x+3\right)$$

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

$$12x+18-x\times 7=4x^{2}+6x$$

Subtract 4x^{2} from both sides.

$$12x+18-x\times 7-4x^{2}=6x$$

Subtract 6x from both sides.

$$12x+18-x\times 7-4x^{2}-6x=0$$

Combine 12x and -6x to get 6x.

$$6x+18-x\times 7-4x^{2}=0$$

Multiply -1 and 7 to get -7.

$$6x+18-7x-4x^{2}=0$$

Combine 6x and -7x to get -x.

$$-x+18-4x^{2}=0$$

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

$$-4x^{2}-x+18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx+18. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-4\times 18=-72$$

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. List all such integer pairs that give product -72.

$$1,-72 2,-36 3,-24 4,-18 6,-12 8,-9$$

Calculate the sum for each pair.

$$1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1$$

The solution is the pair that gives sum -1.

$$a=8 b=-9$$

Rewrite -4x^{2}-x+18 as \left(-4x^{2}+8x\right)+\left(-9x+18\right).

$$\left(-4x^{2}+8x\right)+\left(-9x+18\right)$$

Factor out 4x in the first and 9 in the second group.

$$4x\left(-x+2\right)+9\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(4x+9\right)$$

To find equation solutions, solve -x+2=0 and 4x+9=0.

$$x=2 x=-\frac{9}{4}$$

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.

$$\left(2x+3\right)\times 6-x\times 7=2x\left(2x+3\right)$$

Use the distributive property to multiply 2x+3 by 6.

$$12x+18-x\times 7=2x\left(2x+3\right)$$

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

$$12x+18-x\times 7=4x^{2}+6x$$

Subtract 4x^{2} from both sides.

$$12x+18-x\times 7-4x^{2}=6x$$

Subtract 6x from both sides.

$$12x+18-x\times 7-4x^{2}-6x=0$$

Combine 12x and -6x to get 6x.

$$6x+18-x\times 7-4x^{2}=0$$

Multiply -1 and 7 to get -7.

$$6x+18-7x-4x^{2}=0$$

Combine 6x and -7x to get -x.

$$-x+18-4x^{2}=0$$

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.

$$-4x^{2}-x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -1 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-4\right)\times 18}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-1\right)±\sqrt{1+16\times 18}}{2\left(-4\right)}$$

Multiply 16 times 18.

$$x=\frac{-\left(-1\right)±\sqrt{1+288}}{2\left(-4\right)}$$

Add 1 to 288.

$$x=\frac{-\left(-1\right)±\sqrt{289}}{2\left(-4\right)}$$

Take the square root of 289.

$$x=\frac{-\left(-1\right)±17}{2\left(-4\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±17}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{1±17}{-8}$$

Now solve the equation x=\frac{1±17}{-8} when ± is plus. Add 1 to 17.

$$x=\frac{18}{-8}$$

Reduce the fraction \frac{18}{-8} to lowest terms by extracting and canceling out 2.

$$x=-\frac{9}{4}$$

Now solve the equation x=\frac{1±17}{-8} when ± is minus. Subtract 17 from 1.

$$x=-\frac{16}{-8}$$

Divide -16 by -8.

$$x=2$$

The equation is now solved.

$$x=-\frac{9}{4} x=2$$

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.

$$\left(2x+3\right)\times 6-x\times 7=2x\left(2x+3\right)$$

Use the distributive property to multiply 2x+3 by 6.

$$12x+18-x\times 7=2x\left(2x+3\right)$$

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

$$12x+18-x\times 7=4x^{2}+6x$$

Subtract 4x^{2} from both sides.

$$12x+18-x\times 7-4x^{2}=6x$$

Subtract 6x from both sides.

$$12x+18-x\times 7-4x^{2}-6x=0$$

Combine 12x and -6x to get 6x.

$$6x+18-x\times 7-4x^{2}=0$$

Subtract 18 from both sides. Anything subtracted from zero gives its negation.

$$6x-x\times 7-4x^{2}=-18$$

Multiply -1 and 7 to get -7.

$$6x-7x-4x^{2}=-18$$

Combine 6x and -7x to get -x.

$$-x-4x^{2}=-18$$

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.

$$-4x^{2}-x=-18$$

Divide both sides by -4.

$$\frac{-4x^{2}-x}{-4}=-\frac{18}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{1}{-4}\right)x=-\frac{18}{-4}$$

Divide -1 by -4.

$$x^{2}+\frac{1}{4}x=-\frac{18}{-4}$$

Reduce the fraction \frac{-18}{-4} to lowest terms by extracting and canceling out 2.

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

Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=\frac{9}{2}+\left(\frac{1}{8}\right)^{2}$$

Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{9}{2}+\frac{1}{64}$$

Add \frac{9}{2} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{289}{64}$$

Factor x^{2}+\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{289}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{289}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{17}{8} x+\frac{1}{8}=-\frac{17}{8}$$

Subtract \frac{1}{8} from both sides of the equation.

$$x=2 x=-\frac{9}{4}$$