Asked by Hailey Mccarthy on May 25, 2024
Verified
Simplify the rational expression. 30x2+11x−7100x2−49\frac { 30 x ^ { 2 } + 11 x - 7 } { 100 x ^ { 2 } - 49 }100x2−4930x2+11x−7
A) 3x−110x+7,x≠107\frac { 3 x - 1 } { 10 x + 7 } , x \neq \frac { 10 } { 7 }10x+73x−1,x=710
B) 10x−73,x≠0\frac { 10 x - 7 } { 3 } , x \neq 0310x−7,x=0
C) 7−10x3,x≠0\frac { 7 - 10 x } { 3 } , x \neq 037−10x,x=0
D) 10x+710x−7,x≠710\frac { 10 x + 7 } { 10 x - 7 } , x \neq \frac { 7 } { 10 }10x−710x+7,x=107
E) 3x−110x−7,x≠710\frac { 3 x - 1 } { 10 x - 7 } , x \neq \frac { 7 } { 10 }10x−73x−1,x=107
Rational Expression
A ratio of two polynomials.
Simplify
The process of reducing an expression or equation to its simplest form, often by combining like terms or applying mathematical operations.
- Facilitate the simplification of polynomials in rational expressions.
Verified Answer
SM
Sabrina MoralesMay 28, 2024
Final Answer :
E
Explanation :
We can factor both the numerator and denominator to simplify the expression.
30x2+11x−7100x2−49=(6x−1)(5x+7)(10x+7)(10x−7)=6x−110x−7⋅5x+710x+7=3x−15x+7⋅5x+710x+7=3x−110x−7, for x≠710\begin{align*}\frac{30x^2+11x-7}{100x^2-49} &= \frac{(6x-1)(5x+7)}{(10x+7)(10x-7)} \\&= \frac{6x-1}{10x-7} \cdot \frac{5x+7}{10x+7} \\&= \frac{3x-1}{5x+7} \cdot \frac{5x+7}{10x+7} \\&= \frac{3x-1}{10x-7}, \text{ for } x \neq \frac{7}{10}\end{align*}100x2−4930x2+11x−7=(10x+7)(10x−7)(6x−1)(5x+7)=10x−76x−1⋅10x+75x+7=5x+73x−1⋅10x+75x+7=10x−73x−1, for x=107
Therefore, the best choice is E.
30x2+11x−7100x2−49=(6x−1)(5x+7)(10x+7)(10x−7)=6x−110x−7⋅5x+710x+7=3x−15x+7⋅5x+710x+7=3x−110x−7, for x≠710\begin{align*}\frac{30x^2+11x-7}{100x^2-49} &= \frac{(6x-1)(5x+7)}{(10x+7)(10x-7)} \\&= \frac{6x-1}{10x-7} \cdot \frac{5x+7}{10x+7} \\&= \frac{3x-1}{5x+7} \cdot \frac{5x+7}{10x+7} \\&= \frac{3x-1}{10x-7}, \text{ for } x \neq \frac{7}{10}\end{align*}100x2−4930x2+11x−7=(10x+7)(10x−7)(6x−1)(5x+7)=10x−76x−1⋅10x+75x+7=5x+73x−1⋅10x+75x+7=10x−73x−1, for x=107
Therefore, the best choice is E.
Learning Objectives
- Facilitate the simplification of polynomials in rational expressions.