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

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

Solve for x

$$x=3$$

$$x=0$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,\frac{3}{5} since division by zero is not defined. Multiply both sides of the equation by \left(5x-3\right)\left(x+3\right), the least common multiple of 5x-3,x+3.

$$\left(x+3\right)\left(7x+3\right)=\left(5x-3\right)\left(5x-3\right)$$

Multiply 5x-3 and 5x-3 to get \left(5x-3\right)^{2}.

$$\left(x+3\right)\left(7x+3\right)=\left(5x-3\right)^{2}$$

Use the distributive property to multiply x+3 by 7x+3 and combine like terms.

$$7x^{2}+24x+9=\left(5x-3\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-3\right)^{2}.

$$7x^{2}+24x+9=25x^{2}-30x+9$$

Subtract 25x^{2} from both sides.

$$7x^{2}+24x+9-25x^{2}=-30x+9$$

Combine 7x^{2} and -25x^{2} to get -18x^{2}.

$$-18x^{2}+24x+9=-30x+9$$

Add 30x to both sides.

$$-18x^{2}+24x+9+30x=9$$

Combine 24x and 30x to get 54x.

$$-18x^{2}+54x+9=9$$

Subtract 9 from both sides.

$$-18x^{2}+54x+9-9=0$$

Subtract 9 from 9 to get 0.

$$-18x^{2}+54x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -18 for a, 54 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-54±\sqrt{54^{2}}}{2\left(-18\right)}$$

Take the square root of 54^{2}.

$$x=\frac{-54±54}{2\left(-18\right)}$$

Multiply 2 times -18.

$$x=\frac{-54±54}{-36}$$

Now solve the equation x=\frac{-54±54}{-36} when ± is plus. Add -54 to 54.

$$x=\frac{0}{-36}$$

Divide 0 by -36.

$$x=0$$

Now solve the equation x=\frac{-54±54}{-36} when ± is minus. Subtract 54 from -54.

$$x=-\frac{108}{-36}$$

Divide -108 by -36.

$$x=3$$

The equation is now solved.

$$x=0 x=3$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,\frac{3}{5} since division by zero is not defined. Multiply both sides of the equation by \left(5x-3\right)\left(x+3\right), the least common multiple of 5x-3,x+3.

$$\left(x+3\right)\left(7x+3\right)=\left(5x-3\right)\left(5x-3\right)$$

Multiply 5x-3 and 5x-3 to get \left(5x-3\right)^{2}.

$$\left(x+3\right)\left(7x+3\right)=\left(5x-3\right)^{2}$$

Use the distributive property to multiply x+3 by 7x+3 and combine like terms.

$$7x^{2}+24x+9=\left(5x-3\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-3\right)^{2}.

$$7x^{2}+24x+9=25x^{2}-30x+9$$

Subtract 25x^{2} from both sides.

$$7x^{2}+24x+9-25x^{2}=-30x+9$$

Combine 7x^{2} and -25x^{2} to get -18x^{2}.

$$-18x^{2}+24x+9=-30x+9$$

Add 30x to both sides.

$$-18x^{2}+24x+9+30x=9$$

Combine 24x and 30x to get 54x.

$$-18x^{2}+54x+9=9$$

Subtract 9 from both sides.

$$-18x^{2}+54x=9-9$$

Subtract 9 from 9 to get 0.

$$-18x^{2}+54x=0$$

Divide both sides by -18.

$$\frac{-18x^{2}+54x}{-18}=\frac{0}{-18}$$

Dividing by -18 undoes the multiplication by -18.

$$x^{2}+\frac{54}{-18}x=\frac{0}{-18}$$

Divide 54 by -18.

$$x^{2}-3x=\frac{0}{-18}$$

Divide 0 by -18.

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

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Factor x^{2}-3x+\frac{9}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=3 x=0$$