Solve y(1/(x + 1) + 1/(x - 1)) and (2/(2x + 1) + 2/(2x - 1)) | Equatix - Math Solver

Fractions & Decimals

Ratios & Proportions

Basic Number Theory

Integers & Absolute Values

Factors & Multiples

Prime Factorization

Basic Equations & Inequalities

Coordinate Plane & Graphing

Linear Equations & Inequalities

Quadratic Equations

Polynomials & Factoring

Functions & Graphs

Systems of Equations

Trigonometric Functions

Trigonometric Identities & Equations

Right Triangle Trigonometry

Graphing Trigonometric Functions

Limits & Continuity

Derivatives & Applications

Integrals & Applications

Series & Sequences

Multivariable Calculus

Functions & Graphs

Polynomial & Rational Functions

Exponential & Logarithmic Functions

Sequences & Series

Descriptive Statistics

Normal Distribution

Hypothesis Testing

Regression Analysis

Linear Programming

Sets & Counting

Eigenvalues & Eigenvectors

Linear Transformations

Atomic Structure

Thermochemistry

Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$y ( \frac { 1 } { x + 1 } + \frac { 1 } { x - 1 } ) ( \frac { 2 } { 2 x + 1 } + \frac { 2 } { 2 x - 1 } )$$

Evaluate

$\frac{16yx^{2}}{\left(4x^{2}-1\right)\left(x^{2}-1\right)}$
$|x|\neq 1\text{ and }|x|\neq \frac{1}{2}$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $x+1$ and $x-1$ is $\left(x-1\right)\left(x+1\right)$. Multiply $\frac{1}{x+1}$ times $\frac{x-1}{x-1}$. Multiply $\frac{1}{x-1}$ times $\frac{x+1}{x+1}$.

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

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

$$y\times \frac{x-1+x+1}{\left(x-1\right)\left(x+1\right)}\left(\frac{2}{2x+1}+\frac{2}{2x-1}\right)$$

Combine like terms in $x-1+x+1$.

$$y\times \frac{2x}{\left(x-1\right)\left(x+1\right)}\left(\frac{2}{2x+1}+\frac{2}{2x-1}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $2x+1$ and $2x-1$ is $\left(2x-1\right)\left(2x+1\right)$. Multiply $\frac{2}{2x+1}$ times $\frac{2x-1}{2x-1}$. Multiply $\frac{2}{2x-1}$ times $\frac{2x+1}{2x+1}$.

$$y\times \frac{2x}{\left(x-1\right)\left(x+1\right)}\left(\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\right)$$

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

$$y\times \frac{2x}{\left(x-1\right)\left(x+1\right)}\times \frac{2\left(2x-1\right)+2\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}$$

Do the multiplications in $2\left(2x-1\right)+2\left(2x+1\right)$.

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

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

$$y\times \frac{2x}{\left(x-1\right)\left(x+1\right)}\times \frac{8x}{\left(2x-1\right)\left(2x+1\right)}$$

Express $y\times \frac{2x}{\left(x-1\right)\left(x+1\right)}$ as a single fraction.

$$\frac{y\times 2x}{\left(x-1\right)\left(x+1\right)}\times \frac{8x}{\left(2x-1\right)\left(2x+1\right)}$$

Multiply $\frac{y\times 2x}{\left(x-1\right)\left(x+1\right)}$ times $\frac{8x}{\left(2x-1\right)\left(2x+1\right)}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{y\times 2x\times 8x}{\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(2x+1\right)}$$

Multiply $x$ and $x$ to get $x^{2}$.

$$\frac{y\times 2x^{2}\times 8}{\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(2x+1\right)}$$

Multiply $2$ and $8$ to get $16$.

$$\frac{y\times 16x^{2}}{\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(2x+1\right)}$$

Apply the distributive property by multiplying each term of $x-1$ by each term of $x+1$.

$$\frac{y\times 16x^{2}}{\left(x^{2}+x-x-1\right)\left(2x-1\right)\left(2x+1\right)}$$

Combine $x$ and $-x$ to get $0$.

$$\frac{y\times 16x^{2}}{\left(x^{2}-1\right)\left(2x-1\right)\left(2x+1\right)}$$

Apply the distributive property by multiplying each term of $x^{2}-1$ by each term of $2x-1$.

$$\frac{y\times 16x^{2}}{\left(2x^{3}-x^{2}-2x+1\right)\left(2x+1\right)}$$

Apply the distributive property by multiplying each term of $2x^{3}-x^{2}-2x+1$ by each term of $2x+1$.

$$\frac{y\times 16x^{2}}{4x^{4}+2x^{3}-2x^{3}-x^{2}-4x^{2}-2x+2x+1}$$

Combine $2x^{3}$ and $-2x^{3}$ to get $0$.

$$\frac{y\times 16x^{2}}{4x^{4}-x^{2}-4x^{2}-2x+2x+1}$$

Combine $-x^{2}$ and $-4x^{2}$ to get $-5x^{2}$.

$$\frac{y\times 16x^{2}}{4x^{4}-5x^{2}-2x+2x+1}$$

Combine $-2x$ and $2x$ to get $0$.

$$\frac{y\times 16x^{2}}{4x^{4}-5x^{2}+1}$$

Expand

$\frac{16yx^{2}}{\left(4x^{2}-1\right)\left(x^{2}-1\right)}$
$|x|\neq 1\text{ and }|x|\neq \frac{1}{2}$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $x+1$ and $x-1$ is $\left(x-1\right)\left(x+1\right)$. Multiply $\frac{1}{x+1}$ times $\frac{x-1}{x-1}$. Multiply $\frac{1}{x-1}$ times $\frac{x+1}{x+1}$.

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

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

$$y\times \frac{x-1+x+1}{\left(x-1\right)\left(x+1\right)}\left(\frac{2}{2x+1}+\frac{2}{2x-1}\right)$$

Combine like terms in $x-1+x+1$.

$$y\times \frac{2x}{\left(x-1\right)\left(x+1\right)}\left(\frac{2}{2x+1}+\frac{2}{2x-1}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $2x+1$ and $2x-1$ is $\left(2x-1\right)\left(2x+1\right)$. Multiply $\frac{2}{2x+1}$ times $\frac{2x-1}{2x-1}$. Multiply $\frac{2}{2x-1}$ times $\frac{2x+1}{2x+1}$.

$$y\times \frac{2x}{\left(x-1\right)\left(x+1\right)}\left(\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\right)$$

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

$$y\times \frac{2x}{\left(x-1\right)\left(x+1\right)}\times \frac{2\left(2x-1\right)+2\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}$$

Do the multiplications in $2\left(2x-1\right)+2\left(2x+1\right)$.

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

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

$$y\times \frac{2x}{\left(x-1\right)\left(x+1\right)}\times \frac{8x}{\left(2x-1\right)\left(2x+1\right)}$$

Express $y\times \frac{2x}{\left(x-1\right)\left(x+1\right)}$ as a single fraction.

$$\frac{y\times 2x}{\left(x-1\right)\left(x+1\right)}\times \frac{8x}{\left(2x-1\right)\left(2x+1\right)}$$

Multiply $\frac{y\times 2x}{\left(x-1\right)\left(x+1\right)}$ times $\frac{8x}{\left(2x-1\right)\left(2x+1\right)}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{y\times 2x\times 8x}{\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(2x+1\right)}$$

Multiply $x$ and $x$ to get $x^{2}$.

$$\frac{y\times 2x^{2}\times 8}{\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(2x+1\right)}$$

Multiply $2$ and $8$ to get $16$.

$$\frac{y\times 16x^{2}}{\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(2x+1\right)}$$

Apply the distributive property by multiplying each term of $x-1$ by each term of $x+1$.

$$\frac{y\times 16x^{2}}{\left(x^{2}+x-x-1\right)\left(2x-1\right)\left(2x+1\right)}$$

Combine $x$ and $-x$ to get $0$.

$$\frac{y\times 16x^{2}}{\left(x^{2}-1\right)\left(2x-1\right)\left(2x+1\right)}$$

Apply the distributive property by multiplying each term of $x^{2}-1$ by each term of $2x-1$.

$$\frac{y\times 16x^{2}}{\left(2x^{3}-x^{2}-2x+1\right)\left(2x+1\right)}$$

Apply the distributive property by multiplying each term of $2x^{3}-x^{2}-2x+1$ by each term of $2x+1$.

$$\frac{y\times 16x^{2}}{4x^{4}+2x^{3}-2x^{3}-x^{2}-4x^{2}-2x+2x+1}$$

Combine $2x^{3}$ and $-2x^{3}$ to get $0$.

$$\frac{y\times 16x^{2}}{4x^{4}-x^{2}-4x^{2}-2x+2x+1}$$

Combine $-x^{2}$ and $-4x^{2}$ to get $-5x^{2}$.

$$\frac{y\times 16x^{2}}{4x^{4}-5x^{2}-2x+2x+1}$$

Combine $-2x$ and $2x$ to get $0$.

$$\frac{y\times 16x^{2}}{4x^{4}-5x^{2}+1}$$