Solve (2y-5)/(y+5)=(3y+21)/(2y-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

$$\frac{ 2y-5 }{ y+5 } = \frac{ 3y+21 }{ 2y-1 }$$

Solve for y

$y=-2$
$y=50$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $y$ cannot be equal to any of the values $-5,\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $\left(2y-1\right)\left(y+5\right)$, the least common multiple of $y+5,2y-1$.

$$\left(2y-1\right)\left(2y-5\right)=\left(y+5\right)\left(3y+21\right)$$

Use the distributive property to multiply $2y-1$ by $2y-5$ and combine like terms.

$$4y^{2}-12y+5=\left(y+5\right)\left(3y+21\right)$$

Use the distributive property to multiply $y+5$ by $3y+21$ and combine like terms.

$$4y^{2}-12y+5=3y^{2}+36y+105$$

Subtract $3y^{2}$ from both sides.

$$4y^{2}-12y+5-3y^{2}=36y+105$$

Combine $4y^{2}$ and $-3y^{2}$ to get $y^{2}$.

$$y^{2}-12y+5=36y+105$$

Subtract $36y$ from both sides.

$$y^{2}-12y+5-36y=105$$

Combine $-12y$ and $-36y$ to get $-48y$.

$$y^{2}-48y+5=105$$

Subtract $105$ from both sides.

$$y^{2}-48y+5-105=0$$

Subtract $105$ from $5$ to get $-100$.

$$y^{2}-48y-100=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-48$ for $b$, and $-100$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$y=\frac{-\left(-48\right)±\sqrt{\left(-48\right)^{2}-4\left(-100\right)}}{2}$$

Square $-48$.

$$y=\frac{-\left(-48\right)±\sqrt{2304-4\left(-100\right)}}{2}$$

Multiply $-4$ times $-100$.

$$y=\frac{-\left(-48\right)±\sqrt{2304+400}}{2}$$

Add $2304$ to $400$.

$$y=\frac{-\left(-48\right)±\sqrt{2704}}{2}$$

Take the square root of $2704$.

$$y=\frac{-\left(-48\right)±52}{2}$$

The opposite of $-48$ is $48$.

$$y=\frac{48±52}{2}$$

Now solve the equation $y=\frac{48±52}{2}$ when $±$ is plus. Add $48$ to $52$.

$$y=\frac{100}{2}$$

Divide $100$ by $2$.

$$y=50$$

Now solve the equation $y=\frac{48±52}{2}$ when $±$ is minus. Subtract $52$ from $48$.

$$y=-\frac{4}{2}$$

Divide $-4$ by $2$.

$$y=-2$$

The equation is now solved.

$$y=50$$ $$y=-2$$

Steps for Completing the Square

Variable $y$ cannot be equal to any of the values $-5,\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $\left(2y-1\right)\left(y+5\right)$, the least common multiple of $y+5,2y-1$.

$$\left(2y-1\right)\left(2y-5\right)=\left(y+5\right)\left(3y+21\right)$$

Use the distributive property to multiply $2y-1$ by $2y-5$ and combine like terms.

$$4y^{2}-12y+5=\left(y+5\right)\left(3y+21\right)$$

Use the distributive property to multiply $y+5$ by $3y+21$ and combine like terms.

$$4y^{2}-12y+5=3y^{2}+36y+105$$

Subtract $3y^{2}$ from both sides.

$$4y^{2}-12y+5-3y^{2}=36y+105$$

Combine $4y^{2}$ and $-3y^{2}$ to get $y^{2}$.

$$y^{2}-12y+5=36y+105$$

Subtract $36y$ from both sides.

$$y^{2}-12y+5-36y=105$$

Combine $-12y$ and $-36y$ to get $-48y$.

$$y^{2}-48y+5=105$$

Subtract $5$ from both sides.

$$y^{2}-48y=105-5$$

Subtract $5$ from $105$ to get $100$.

$$y^{2}-48y=100$$

Divide $-48$, the coefficient of the $x$ term, by $2$ to get $-24$. Then add the square of $-24$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-48y+\left(-24\right)^{2}=100+\left(-24\right)^{2}$$

Square $-24$.

$$y^{2}-48y+576=100+576$$

Add $100$ to $576$.

$$y^{2}-48y+576=676$$

Factor $y^{2}-48y+576$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(y-24\right)^{2}=676$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-24\right)^{2}}=\sqrt{676}$$

Simplify.

$$y-24=26$$ $$y-24=-26$$

Add $24$ to both sides of the equation.

$$y=50$$ $$y=-2$$