Solve 4x^2-9/9x^2-6x+1:2x-3/3x-1+4-x/1-3x

$$\frac { 4 x ^ { 2 } - 9 } { 9 x ^ { 2 } - 6 x + 1 } : \frac { 2 x - 3 } { 3 x - 1 } + \frac { 4 - x } { 1 - 3 x }$$

Evaluate

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Short Solution Steps

Divide \frac{4x^{2}-9}{9x^{2}-6x+1} by \frac{2x-3}{3x-1} by multiplying \frac{4x^{2}-9}{9x^{2}-6x+1} by the reciprocal of \frac{2x-3}{3x-1}.

$$\frac{\left(4x^{2}-9\right)\left(3x-1\right)}{\left(9x^{2}-6x+1\right)\left(2x-3\right)}+\frac{4-x}{1-3x}$$

Factor the expressions that are not already factored in \frac{\left(4x^{2}-9\right)\left(3x-1\right)}{\left(9x^{2}-6x+1\right)\left(2x-3\right)}.

$$\frac{\left(2x-3\right)\left(3x-1\right)\left(2x+3\right)}{\left(2x-3\right)\left(3x-1\right)^{2}}+\frac{4-x}{1-3x}$$

Cancel out \left(2x-3\right)\left(3x-1\right) in both numerator and denominator.

$$\frac{2x+3}{3x-1}+\frac{4-x}{1-3x}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3x-1 and 1-3x is 3x-1. Multiply \frac{4-x}{1-3x} times \frac{-1}{-1}.

$$\frac{2x+3}{3x-1}+\frac{-\left(4-x\right)}{3x-1}$$

Since \frac{2x+3}{3x-1} and \frac{-\left(4-x\right)}{3x-1} have the same denominator, add them by adding their numerators.

$$\frac{2x+3-\left(4-x\right)}{3x-1}$$

Do the multiplications in 2x+3-\left(4-x\right).

$$\frac{2x+3-4+x}{3x-1}$$

Combine like terms in 2x+3-4+x.

$$\frac{3x-1}{3x-1}$$

Cancel out 3x-1 in both numerator and denominator.

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Factor

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