Solve 4+x/3-5/x=7-x/6

$$\frac { 4 + x } { 3 } - \frac { 5 } { x } = \frac { 7 - x } { 6 }$$

Solve for x

$$x = -\frac{10}{3} = -3\frac{1}{3} \approx -3.333333333$$

$$x=3$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 6x, the least common multiple of 3,x,6.

$$2x\left(4+x\right)-6\times 5=x\left(7-x\right)$$

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

$$8x+2x^{2}-6\times 5=x\left(7-x\right)$$

Multiply -6 and 5 to get -30.

$$8x+2x^{2}-30=x\left(7-x\right)$$

Use the distributive property to multiply x by 7-x.

$$8x+2x^{2}-30=7x-x^{2}$$

Subtract 7x from both sides.

$$8x+2x^{2}-30-7x=-x^{2}$$

Combine 8x and -7x to get x.

$$x+2x^{2}-30=-x^{2}$$

Add x^{2} to both sides.

$$x+2x^{2}-30+x^{2}=0$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$x+3x^{2}-30=0$$

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

$$3x^{2}+x-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=1 ab=3\left(-30\right)=-90$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -90.

$$-1,90 -2,45 -3,30 -5,18 -6,15 -9,10$$

Calculate the sum for each pair.

$$-1+90=89 -2+45=43 -3+30=27 -5+18=13 -6+15=9 -9+10=1$$

The solution is the pair that gives sum 1.

$$a=-9 b=10$$

Rewrite 3x^{2}+x-30 as \left(3x^{2}-9x\right)+\left(10x-30\right).

$$\left(3x^{2}-9x\right)+\left(10x-30\right)$$

Factor out 3x in the first and 10 in the second group.

$$3x\left(x-3\right)+10\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(3x+10\right)$$

To find equation solutions, solve x-3=0 and 3x+10=0.

$$x=3 x=-\frac{10}{3}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 6x, the least common multiple of 3,x,6.

$$2x\left(4+x\right)-6\times 5=x\left(7-x\right)$$

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

$$8x+2x^{2}-6\times 5=x\left(7-x\right)$$

Multiply -6 and 5 to get -30.

$$8x+2x^{2}-30=x\left(7-x\right)$$

Use the distributive property to multiply x by 7-x.

$$8x+2x^{2}-30=7x-x^{2}$$

Subtract 7x from both sides.

$$8x+2x^{2}-30-7x=-x^{2}$$

Combine 8x and -7x to get x.

$$x+2x^{2}-30=-x^{2}$$

Add x^{2} to both sides.

$$x+2x^{2}-30+x^{2}=0$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$x+3x^{2}-30=0$$

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.

$$3x^{2}+x-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 1 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 3\left(-30\right)}}{2\times 3}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 3\left(-30\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-1±\sqrt{1-12\left(-30\right)}}{2\times 3}$$

Multiply -12 times -30.

$$x=\frac{-1±\sqrt{1+360}}{2\times 3}$$

Add 1 to 360.

$$x=\frac{-1±\sqrt{361}}{2\times 3}$$

Take the square root of 361.

$$x=\frac{-1±19}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-1±19}{6}$$

Now solve the equation x=\frac{-1±19}{6} when ± is plus. Add -1 to 19.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{-1±19}{6} when ± is minus. Subtract 19 from -1.

$$x=-\frac{20}{6}$$

Reduce the fraction \frac{-20}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{10}{3}$$

The equation is now solved.

$$x=3 x=-\frac{10}{3}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 6x, the least common multiple of 3,x,6.

$$2x\left(4+x\right)-6\times 5=x\left(7-x\right)$$

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

$$8x+2x^{2}-6\times 5=x\left(7-x\right)$$

Multiply -6 and 5 to get -30.

$$8x+2x^{2}-30=x\left(7-x\right)$$

Use the distributive property to multiply x by 7-x.

$$8x+2x^{2}-30=7x-x^{2}$$

Subtract 7x from both sides.

$$8x+2x^{2}-30-7x=-x^{2}$$

Combine 8x and -7x to get x.

$$x+2x^{2}-30=-x^{2}$$

Add x^{2} to both sides.

$$x+2x^{2}-30+x^{2}=0$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$x+3x^{2}-30=0$$

Add 30 to both sides. Anything plus zero gives itself.

$$x+3x^{2}=30$$

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.

$$3x^{2}+x=30$$

Divide both sides by 3.

$$\frac{3x^{2}+x}{3}=\frac{30}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{1}{3}x=\frac{30}{3}$$

Divide 30 by 3.

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

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=10+\left(\frac{1}{6}\right)^{2}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=10+\frac{1}{36}$$

Add 10 to \frac{1}{36}.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{361}{36}$$

Factor x^{2}+\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{361}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{361}{36}}$$

Simplify.

$$x+\frac{1}{6}=\frac{19}{6} x+\frac{1}{6}=-\frac{19}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$x=3 x=-\frac{10}{3}$$