Solve 9x+7/7x-9=4-0x/4x-7

$$\frac{ 9x+7 }{ 7x-9 } = \frac{ 4-0x }{ 4x-7 }$$

Solve for x

$$x=\frac{\sqrt{649}}{24}+\frac{7}{8}\approx 1.936478267$$

$$x=-\frac{\sqrt{649}}{24}+\frac{7}{8}\approx -0.186478267$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values \frac{9}{7},\frac{7}{4} since division by zero is not defined. Multiply both sides of the equation by \left(4x-7\right)\left(7x-9\right), the least common multiple of 7x-9,4x-7.

$$\left(4x-7\right)\left(9x+7\right)=\left(7x-9\right)\left(4-0x\right)$$

Use the distributive property to multiply 4x-7 by 9x+7 and combine like terms.

$$36x^{2}-35x-49=\left(7x-9\right)\left(4-0x\right)$$

Anything times zero gives zero.

$$36x^{2}-35x-49=\left(7x-9\right)\left(4-0\right)$$

Subtract 0 from 4 to get 4.

$$36x^{2}-35x-49=\left(7x-9\right)\times 4$$

Use the distributive property to multiply 7x-9 by 4.

$$36x^{2}-35x-49=28x-36$$

Subtract 28x from both sides.

$$36x^{2}-35x-49-28x=-36$$

Combine -35x and -28x to get -63x.

$$36x^{2}-63x-49=-36$$

Add 36 to both sides.

$$36x^{2}-63x-49+36=0$$

Add -49 and 36 to get -13.

$$36x^{2}-63x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 36 for a, -63 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-63\right)±\sqrt{\left(-63\right)^{2}-4\times 36\left(-13\right)}}{2\times 36}$$

Square -63.

$$x=\frac{-\left(-63\right)±\sqrt{3969-4\times 36\left(-13\right)}}{2\times 36}$$

Multiply -4 times 36.

$$x=\frac{-\left(-63\right)±\sqrt{3969-144\left(-13\right)}}{2\times 36}$$

Multiply -144 times -13.

$$x=\frac{-\left(-63\right)±\sqrt{3969+1872}}{2\times 36}$$

Add 3969 to 1872.

$$x=\frac{-\left(-63\right)±\sqrt{5841}}{2\times 36}$$

Take the square root of 5841.

$$x=\frac{-\left(-63\right)±3\sqrt{649}}{2\times 36}$$

The opposite of -63 is 63.

$$x=\frac{63±3\sqrt{649}}{2\times 36}$$

Multiply 2 times 36.

$$x=\frac{63±3\sqrt{649}}{72}$$

Now solve the equation x=\frac{63±3\sqrt{649}}{72} when ± is plus. Add 63 to 3\sqrt{649}.

$$x=\frac{3\sqrt{649}+63}{72}$$

Divide 63+3\sqrt{649} by 72.

$$x=\frac{\sqrt{649}}{24}+\frac{7}{8}$$

Now solve the equation x=\frac{63±3\sqrt{649}}{72} when ± is minus. Subtract 3\sqrt{649} from 63.

$$x=\frac{63-3\sqrt{649}}{72}$$

Divide 63-3\sqrt{649} by 72.

$$x=-\frac{\sqrt{649}}{24}+\frac{7}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{649}}{24}+\frac{7}{8} x=-\frac{\sqrt{649}}{24}+\frac{7}{8}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values \frac{9}{7},\frac{7}{4} since division by zero is not defined. Multiply both sides of the equation by \left(4x-7\right)\left(7x-9\right), the least common multiple of 7x-9,4x-7.

$$\left(4x-7\right)\left(9x+7\right)=\left(7x-9\right)\left(4-0x\right)$$

Use the distributive property to multiply 4x-7 by 9x+7 and combine like terms.

$$36x^{2}-35x-49=\left(7x-9\right)\left(4-0x\right)$$

Anything times zero gives zero.

$$36x^{2}-35x-49=\left(7x-9\right)\left(4-0\right)$$

Subtract 0 from 4 to get 4.

$$36x^{2}-35x-49=\left(7x-9\right)\times 4$$

Use the distributive property to multiply 7x-9 by 4.

$$36x^{2}-35x-49=28x-36$$

Subtract 28x from both sides.

$$36x^{2}-35x-49-28x=-36$$

Combine -35x and -28x to get -63x.

$$36x^{2}-63x-49=-36$$

Add 49 to both sides.

$$36x^{2}-63x=-36+49$$

Add -36 and 49 to get 13.

$$36x^{2}-63x=13$$

Divide both sides by 36.

$$\frac{36x^{2}-63x}{36}=\frac{13}{36}$$

Dividing by 36 undoes the multiplication by 36.

$$x^{2}+\left(-\frac{63}{36}\right)x=\frac{13}{36}$$

Reduce the fraction \frac{-63}{36} to lowest terms by extracting and canceling out 9.

$$x^{2}-\frac{7}{4}x=\frac{13}{36}$$

Divide -\frac{7}{4}, the coefficient of the x term, by 2 to get -\frac{7}{8}. Then add the square of -\frac{7}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{4}x+\left(-\frac{7}{8}\right)^{2}=\frac{13}{36}+\left(-\frac{7}{8}\right)^{2}$$

Square -\frac{7}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{4}x+\frac{49}{64}=\frac{13}{36}+\frac{49}{64}$$

Add \frac{13}{36} to \frac{49}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{4}x+\frac{49}{64}=\frac{649}{576}$$

Factor x^{2}-\frac{7}{4}x+\frac{49}{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}.

$$\left(x-\frac{7}{8}\right)^{2}=\frac{649}{576}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{8}\right)^{2}}=\sqrt{\frac{649}{576}}$$

Simplify.

$$x-\frac{7}{8}=\frac{\sqrt{649}}{24} x-\frac{7}{8}=-\frac{\sqrt{649}}{24}$$

Add \frac{7}{8} to both sides of the equation.

$$x=\frac{\sqrt{649}}{24}+\frac{7}{8} x=-\frac{\sqrt{649}}{24}+\frac{7}{8}$$