Solve {c}{text{Solvefor}x:}{7/x+3=4/x-3-3/x^2-9}

$$\left. \begin{array} { c } { \text { Solve for } x : } \\ { \frac { 7 } { x + 3 } = \frac { 4 } { x - 3 } - \frac { 3 } { x ^ { 2 } - 9 } } \end{array} \right.$$

Solve for x

$$x = \frac{\sqrt{37} + 1}{2} \approx 3.541381265$$

$$x=\frac{1-\sqrt{37}}{2}\approx -2.541381265$$

$$x=-4$$

Solution Steps

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+3\right), the least common multiple of x-3,x^{2}-9.

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

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$\left(x^{2}-9\right)x+\left(x-3\right)\left(x+3\right)\times 3=\left(x+3\right)\times 4-3$$

Use the distributive property to multiply x^{2}-9 by x.

$$x^{3}-9x+\left(x-3\right)\left(x+3\right)\times 3=\left(x+3\right)\times 4-3$$

Use the distributive property to multiply x-3 by x+3 and combine like terms.

$$x^{3}-9x+\left(x^{2}-9\right)\times 3=\left(x+3\right)\times 4-3$$

Use the distributive property to multiply x^{2}-9 by 3.

$$x^{3}-9x+3x^{2}-27=\left(x+3\right)\times 4-3$$

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

$$x^{3}-9x+3x^{2}-27=4x+12-3$$

Subtract 3 from 12 to get 9.

$$x^{3}-9x+3x^{2}-27=4x+9$$

Subtract 4x from both sides.

$$x^{3}-9x+3x^{2}-27-4x=9$$

Combine -9x and -4x to get -13x.

$$x^{3}-13x+3x^{2}-27=9$$

Subtract 9 from both sides.

$$x^{3}-13x+3x^{2}-27-9=0$$

Subtract 9 from -27 to get -36.

$$x^{3}-13x+3x^{2}-36=0$$

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

$$x^{3}+3x^{2}-13x-36=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+3x^{2}-13x-36 by x+4 to get x^{2}-x-9. Solve the equation where the result equals to 0.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -1 for b, and -9 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{1±\sqrt{37}}{2}$$

Solve the equation x^{2}-x-9=0 when ± is plus and when ± is minus.

$$x=\frac{1-\sqrt{37}}{2} x=\frac{\sqrt{37}+1}{2}$$

List all found solutions.

$$x=-4 x=\frac{1-\sqrt{37}}{2} x=\frac{\sqrt{37}+1}{2}$$