Skip to main content
Evaluate
Tick mark Image
Expand
Tick mark Image
Graph

Similar Problems from Web Search

Share

\frac{\frac{\left(x-1\right)\left(a+1\right)}{a+1}-\frac{4a-1}{a+1}}{\frac{a^{2}-8a+b}{a+1}}
To add or subtract expressions, expand them to make their denominators the same. Multiply x-1 times \frac{a+1}{a+1}.
\frac{\frac{\left(x-1\right)\left(a+1\right)-\left(4a-1\right)}{a+1}}{\frac{a^{2}-8a+b}{a+1}}
Since \frac{\left(x-1\right)\left(a+1\right)}{a+1} and \frac{4a-1}{a+1} have the same denominator, subtract them by subtracting their numerators.
\frac{\frac{xa+x-a-1-4a+1}{a+1}}{\frac{a^{2}-8a+b}{a+1}}
Do the multiplications in \left(x-1\right)\left(a+1\right)-\left(4a-1\right).
\frac{\frac{xa+x-5a}{a+1}}{\frac{a^{2}-8a+b}{a+1}}
Combine like terms in xa+x-a-1-4a+1.
\frac{\left(xa+x-5a\right)\left(a+1\right)}{\left(a+1\right)\left(a^{2}-8a+b\right)}
Divide \frac{xa+x-5a}{a+1} by \frac{a^{2}-8a+b}{a+1} by multiplying \frac{xa+x-5a}{a+1} by the reciprocal of \frac{a^{2}-8a+b}{a+1}.
\frac{ax+x-5a}{a^{2}-8a+b}
Cancel out a+1 in both numerator and denominator.
\frac{\frac{\left(x-1\right)\left(a+1\right)}{a+1}-\frac{4a-1}{a+1}}{\frac{a^{2}-8a+b}{a+1}}
To add or subtract expressions, expand them to make their denominators the same. Multiply x-1 times \frac{a+1}{a+1}.
\frac{\frac{\left(x-1\right)\left(a+1\right)-\left(4a-1\right)}{a+1}}{\frac{a^{2}-8a+b}{a+1}}
Since \frac{\left(x-1\right)\left(a+1\right)}{a+1} and \frac{4a-1}{a+1} have the same denominator, subtract them by subtracting their numerators.
\frac{\frac{xa+x-a-1-4a+1}{a+1}}{\frac{a^{2}-8a+b}{a+1}}
Do the multiplications in \left(x-1\right)\left(a+1\right)-\left(4a-1\right).
\frac{\frac{xa+x-5a}{a+1}}{\frac{a^{2}-8a+b}{a+1}}
Combine like terms in xa+x-a-1-4a+1.
\frac{\left(xa+x-5a\right)\left(a+1\right)}{\left(a+1\right)\left(a^{2}-8a+b\right)}
Divide \frac{xa+x-5a}{a+1} by \frac{a^{2}-8a+b}{a+1} by multiplying \frac{xa+x-5a}{a+1} by the reciprocal of \frac{a^{2}-8a+b}{a+1}.
\frac{ax+x-5a}{a^{2}-8a+b}
Cancel out a+1 in both numerator and denominator.