Solve f^-1(x)(5x-42/10-x)-3/(frac{5x-42{10-x})+5}+9

$$\frac { f ^ { - 1 } ( x ) ( \frac { 5 x - 42 } { 10 - x } ) - 3 } { ( \frac { 5 x - 42 } { 10 - x } ) + 5 } + 9$$

Evaluate

$$\frac{5x^{2}}{8f}+\frac{3x}{8}-\frac{21x}{4f}+\frac{21}{4}$$

Short Solution Steps

Express f^{-1}\times \frac{5x-42}{10-x} as a single fraction.

$$\frac{\frac{f^{-1}\left(5x-42\right)}{10-x}x-3}{\frac{5x-42}{10-x}+5}+9$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 5 times \frac{10-x}{10-x}.

$$\frac{\frac{f^{-1}\left(5x-42\right)}{10-x}x-3}{\frac{5x-42}{10-x}+\frac{5\left(10-x\right)}{10-x}}+9$$

Since \frac{5x-42}{10-x} and \frac{5\left(10-x\right)}{10-x} have the same denominator, add them by adding their numerators.

$$\frac{\frac{f^{-1}\left(5x-42\right)}{10-x}x-3}{\frac{5x-42+5\left(10-x\right)}{10-x}}+9$$

Do the multiplications in 5x-42+5\left(10-x\right).

$$\frac{\frac{f^{-1}\left(5x-42\right)}{10-x}x-3}{\frac{5x-42+50-5x}{10-x}}+9$$

Combine like terms in 5x-42+50-5x.

$$\frac{\frac{f^{-1}\left(5x-42\right)}{10-x}x-3}{\frac{8}{10-x}}+9$$

Express \frac{f^{-1}\left(5x-42\right)}{10-x}x as a single fraction.

$$\frac{\frac{f^{-1}\left(5x-42\right)x}{10-x}-3}{\frac{8}{10-x}}+9$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{10-x}{10-x}.

$$\frac{\frac{f^{-1}\left(5x-42\right)x}{10-x}-\frac{3\left(10-x\right)}{10-x}}{\frac{8}{10-x}}+9$$

Since \frac{f^{-1}\left(5x-42\right)x}{10-x} and \frac{3\left(10-x\right)}{10-x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{f^{-1}\left(5x-42\right)x-3\left(10-x\right)}{10-x}}{\frac{8}{10-x}}+9$$

Do the multiplications in f^{-1}\left(5x-42\right)x-3\left(10-x\right).

$$\frac{\frac{5\times \frac{1}{f}x^{2}-42\times \frac{1}{f}x-30+3x}{10-x}}{\frac{8}{10-x}}+9$$

Combine like terms in 5\times \frac{1}{f}x^{2}-42\times \frac{1}{f}x-30+3x.

$$\frac{\frac{-30+3x+\left(-42x+5x^{2}\right)\times \frac{1}{f}}{10-x}}{\frac{8}{10-x}}+9$$

Divide \frac{-30+3x+\left(-42x+5x^{2}\right)\times \frac{1}{f}}{10-x} by \frac{8}{10-x} by multiplying \frac{-30+3x+\left(-42x+5x^{2}\right)\times \frac{1}{f}}{10-x} by the reciprocal of \frac{8}{10-x}.

$$\frac{\left(-30+3x+\left(-42x+5x^{2}\right)\times \frac{1}{f}\right)\left(10-x\right)}{\left(10-x\right)\times 8}+9$$

Express \left(-42x+5x^{2}\right)\times \frac{1}{f} as a single fraction.

$$\frac{\left(-30+3x+\frac{-42x+5x^{2}}{f}\right)\left(10-x\right)}{\left(10-x\right)\times 8}+9$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -30+3x times \frac{f}{f}.

$$\frac{\left(\frac{\left(-30+3x\right)f}{f}+\frac{-42x+5x^{2}}{f}\right)\left(10-x\right)}{\left(10-x\right)\times 8}+9$$

Since \frac{\left(-30+3x\right)f}{f} and \frac{-42x+5x^{2}}{f} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\left(-30+3x\right)f-42x+5x^{2}}{f}\left(10-x\right)}{\left(10-x\right)\times 8}+9$$

Do the multiplications in \left(-30+3x\right)f-42x+5x^{2}.

$$\frac{\frac{-30f+3xf-42x+5x^{2}}{f}\left(10-x\right)}{\left(10-x\right)\times 8}+9$$

Express \frac{-30f+3xf-42x+5x^{2}}{f}\left(10-x\right) as a single fraction.

$$\frac{\frac{\left(-30f+3xf-42x+5x^{2}\right)\left(10-x\right)}{f}}{\left(10-x\right)\times 8}+9$$

Express \frac{\frac{\left(-30f+3xf-42x+5x^{2}\right)\left(10-x\right)}{f}}{\left(10-x\right)\times 8} as a single fraction.

$$\frac{\left(-30f+3xf-42x+5x^{2}\right)\left(10-x\right)}{f\left(10-x\right)\times 8}+9$$

Cancel out -x+10 in both numerator and denominator.

$$\frac{5x^{2}+3fx-42x-30f}{8f}+9$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9 times \frac{8f}{8f}.

$$\frac{5x^{2}+3fx-42x-30f}{8f}+\frac{9\times 8f}{8f}$$

Since \frac{5x^{2}+3fx-42x-30f}{8f} and \frac{9\times 8f}{8f} have the same denominator, add them by adding their numerators.

$$\frac{5x^{2}+3fx-42x-30f+9\times 8f}{8f}$$

Do the multiplications in 5x^{2}+3fx-42x-30f+9\times 8f.

$$\frac{5x^{2}+3fx-42x-30f+72f}{8f}$$

Combine like terms in 5x^{2}+3fx-42x-30f+72f.

$$\frac{3fx+5x^{2}-42x+42f}{8f}$$

Expand

$$\frac{5x^{2}}{8f}+\frac{3x}{8}-\frac{21x}{4f}+\frac{21}{4}$$

Short Solution Steps

Express f^{-1}\times \frac{5x-42}{10-x} as a single fraction.

$$\frac{\frac{f^{-1}\left(5x-42\right)}{10-x}x-3}{\frac{5x-42}{10-x}+5}+9$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 5 times \frac{10-x}{10-x}.

$$\frac{\frac{f^{-1}\left(5x-42\right)}{10-x}x-3}{\frac{5x-42}{10-x}+\frac{5\left(10-x\right)}{10-x}}+9$$

Since \frac{5x-42}{10-x} and \frac{5\left(10-x\right)}{10-x} have the same denominator, add them by adding their numerators.

$$\frac{\frac{f^{-1}\left(5x-42\right)}{10-x}x-3}{\frac{5x-42+5\left(10-x\right)}{10-x}}+9$$

Do the multiplications in 5x-42+5\left(10-x\right).

$$\frac{\frac{f^{-1}\left(5x-42\right)}{10-x}x-3}{\frac{5x-42+50-5x}{10-x}}+9$$

Combine like terms in 5x-42+50-5x.

$$\frac{\frac{f^{-1}\left(5x-42\right)}{10-x}x-3}{\frac{8}{10-x}}+9$$

Express \frac{f^{-1}\left(5x-42\right)}{10-x}x as a single fraction.

$$\frac{\frac{f^{-1}\left(5x-42\right)x}{10-x}-3}{\frac{8}{10-x}}+9$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{10-x}{10-x}.

$$\frac{\frac{f^{-1}\left(5x-42\right)x}{10-x}-\frac{3\left(10-x\right)}{10-x}}{\frac{8}{10-x}}+9$$

Since \frac{f^{-1}\left(5x-42\right)x}{10-x} and \frac{3\left(10-x\right)}{10-x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{f^{-1}\left(5x-42\right)x-3\left(10-x\right)}{10-x}}{\frac{8}{10-x}}+9$$

Do the multiplications in f^{-1}\left(5x-42\right)x-3\left(10-x\right).

$$\frac{\frac{5\times \frac{1}{f}x^{2}-42\times \frac{1}{f}x-30+3x}{10-x}}{\frac{8}{10-x}}+9$$

Combine like terms in 5\times \frac{1}{f}x^{2}-42\times \frac{1}{f}x-30+3x.

$$\frac{\frac{-30+3x+\left(-42x+5x^{2}\right)\times \frac{1}{f}}{10-x}}{\frac{8}{10-x}}+9$$

Divide \frac{-30+3x+\left(-42x+5x^{2}\right)\times \frac{1}{f}}{10-x} by \frac{8}{10-x} by multiplying \frac{-30+3x+\left(-42x+5x^{2}\right)\times \frac{1}{f}}{10-x} by the reciprocal of \frac{8}{10-x}.

$$\frac{\left(-30+3x+\left(-42x+5x^{2}\right)\times \frac{1}{f}\right)\left(10-x\right)}{\left(10-x\right)\times 8}+9$$

Express \left(-42x+5x^{2}\right)\times \frac{1}{f} as a single fraction.

$$\frac{\left(-30+3x+\frac{-42x+5x^{2}}{f}\right)\left(10-x\right)}{\left(10-x\right)\times 8}+9$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -30+3x times \frac{f}{f}.

$$\frac{\left(\frac{\left(-30+3x\right)f}{f}+\frac{-42x+5x^{2}}{f}\right)\left(10-x\right)}{\left(10-x\right)\times 8}+9$$

Since \frac{\left(-30+3x\right)f}{f} and \frac{-42x+5x^{2}}{f} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\left(-30+3x\right)f-42x+5x^{2}}{f}\left(10-x\right)}{\left(10-x\right)\times 8}+9$$

Do the multiplications in \left(-30+3x\right)f-42x+5x^{2}.

$$\frac{\frac{-30f+3xf-42x+5x^{2}}{f}\left(10-x\right)}{\left(10-x\right)\times 8}+9$$

Express \frac{-30f+3xf-42x+5x^{2}}{f}\left(10-x\right) as a single fraction.

$$\frac{\frac{\left(-30f+3xf-42x+5x^{2}\right)\left(10-x\right)}{f}}{\left(10-x\right)\times 8}+9$$

Express \frac{\frac{\left(-30f+3xf-42x+5x^{2}\right)\left(10-x\right)}{f}}{\left(10-x\right)\times 8} as a single fraction.

$$\frac{\left(-30f+3xf-42x+5x^{2}\right)\left(10-x\right)}{f\left(10-x\right)\times 8}+9$$

Cancel out -x+10 in both numerator and denominator.

$$\frac{5x^{2}+3fx-42x-30f}{8f}+9$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9 times \frac{8f}{8f}.

$$\frac{5x^{2}+3fx-42x-30f}{8f}+\frac{9\times 8f}{8f}$$

Since \frac{5x^{2}+3fx-42x-30f}{8f} and \frac{9\times 8f}{8f} have the same denominator, add them by adding their numerators.

$$\frac{5x^{2}+3fx-42x-30f+9\times 8f}{8f}$$

Do the multiplications in 5x^{2}+3fx-42x-30f+9\times 8f.

$$\frac{5x^{2}+3fx-42x-30f+72f}{8f}$$

Combine like terms in 5x^{2}+3fx-42x-30f+72f.

$$\frac{3fx+5x^{2}-42x+42f}{8f}$$