Solve x/x^2+7/8x=3/x+3

$$\frac { x } { x ^ { 2 } } + \frac { 7 } { 8 x } = \frac { 3 } { x + 3 }$$

Solve for x

$$x=5$$

Steps Using Factoring

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 8\left(x+3\right)x^{2}, the least common multiple of x^{2},8x,x+3.

$$\left(8x+24\right)x+\left(x+3\right)x\times 7=8x^{2}\times 3$$

Use the distributive property to multiply 8x+24 by x.

$$8x^{2}+24x+\left(x+3\right)x\times 7=8x^{2}\times 3$$

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

$$8x^{2}+24x+\left(x^{2}+3x\right)\times 7=8x^{2}\times 3$$

Use the distributive property to multiply x^{2}+3x by 7.

$$8x^{2}+24x+7x^{2}+21x=8x^{2}\times 3$$

Combine 8x^{2} and 7x^{2} to get 15x^{2}.

$$15x^{2}+24x+21x=8x^{2}\times 3$$

Combine 24x and 21x to get 45x.

$$15x^{2}+45x=8x^{2}\times 3$$

Multiply 8 and 3 to get 24.

$$15x^{2}+45x=24x^{2}$$

Subtract 24x^{2} from both sides.

$$15x^{2}+45x-24x^{2}=0$$

Combine 15x^{2} and -24x^{2} to get -9x^{2}.

$$-9x^{2}+45x=0$$

Factor out x.

$$x\left(-9x+45\right)=0$$

To find equation solutions, solve x=0 and -9x+45=0.

$$x=0 x=5$$

Variable x cannot be equal to 0.

$$x=5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 8\left(x+3\right)x^{2}, the least common multiple of x^{2},8x,x+3.

$$\left(8x+24\right)x+\left(x+3\right)x\times 7=8x^{2}\times 3$$

Use the distributive property to multiply 8x+24 by x.

$$8x^{2}+24x+\left(x+3\right)x\times 7=8x^{2}\times 3$$

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

$$8x^{2}+24x+\left(x^{2}+3x\right)\times 7=8x^{2}\times 3$$

Use the distributive property to multiply x^{2}+3x by 7.

$$8x^{2}+24x+7x^{2}+21x=8x^{2}\times 3$$

Combine 8x^{2} and 7x^{2} to get 15x^{2}.

$$15x^{2}+24x+21x=8x^{2}\times 3$$

Combine 24x and 21x to get 45x.

$$15x^{2}+45x=8x^{2}\times 3$$

Multiply 8 and 3 to get 24.

$$15x^{2}+45x=24x^{2}$$

Subtract 24x^{2} from both sides.

$$15x^{2}+45x-24x^{2}=0$$

Combine 15x^{2} and -24x^{2} to get -9x^{2}.

$$-9x^{2}+45x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 45 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-45±\sqrt{45^{2}}}{2\left(-9\right)}$$

Take the square root of 45^{2}.

$$x=\frac{-45±45}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-45±45}{-18}$$

Now solve the equation x=\frac{-45±45}{-18} when ± is plus. Add -45 to 45.

$$x=\frac{0}{-18}$$

Divide 0 by -18.

$$x=0$$

Now solve the equation x=\frac{-45±45}{-18} when ± is minus. Subtract 45 from -45.

$$x=-\frac{90}{-18}$$

Divide -90 by -18.

$$x=5$$

The equation is now solved.

$$x=0 x=5$$

Variable x cannot be equal to 0.

$$x=5$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 8\left(x+3\right)x^{2}, the least common multiple of x^{2},8x,x+3.

$$\left(8x+24\right)x+\left(x+3\right)x\times 7=8x^{2}\times 3$$

Use the distributive property to multiply 8x+24 by x.

$$8x^{2}+24x+\left(x+3\right)x\times 7=8x^{2}\times 3$$

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

$$8x^{2}+24x+\left(x^{2}+3x\right)\times 7=8x^{2}\times 3$$

Use the distributive property to multiply x^{2}+3x by 7.

$$8x^{2}+24x+7x^{2}+21x=8x^{2}\times 3$$

Combine 8x^{2} and 7x^{2} to get 15x^{2}.

$$15x^{2}+24x+21x=8x^{2}\times 3$$

Combine 24x and 21x to get 45x.

$$15x^{2}+45x=8x^{2}\times 3$$

Multiply 8 and 3 to get 24.

$$15x^{2}+45x=24x^{2}$$

Subtract 24x^{2} from both sides.

$$15x^{2}+45x-24x^{2}=0$$

Combine 15x^{2} and -24x^{2} to get -9x^{2}.

$$-9x^{2}+45x=0$$

Divide both sides by -9.

$$\frac{-9x^{2}+45x}{-9}=\frac{0}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{45}{-9}x=\frac{0}{-9}$$

Divide 45 by -9.

$$x^{2}-5x=\frac{0}{-9}$$

Divide 0 by -9.

$$x^{2}-5x=0$$

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

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=\frac{25}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

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

Add \frac{5}{2} to both sides of the equation.

$$x=5 x=0$$

Variable x cannot be equal to 0.

$$x=5$$