# Solve using the elimination method: 3r - 5s = 6 5r + 3s = 44

## Question:

Solve using the elimination method:

{eq}3r - 5s = 6 \\ 5r + 3s = 44 {/eq}

## Elimination method:

The method of elimination is the process of eliminating one of the variables in a system of linear equations by adding or subtracting in conjunction with multiplication or division and solving the given equations.

## Answer and Explanation:

We have,

{eq}\displaystyle \begin{align*} 3r - 5s &= 6 \ \ \rightarrow \text{Equation 1}\\ 5r + 3s &= 44 \rightarrow \text{Equation 2} \end{align*} {/eq}

Multiplying the first equation by {eq}5 {/eq}, we get,

$$15r - 25s = 30 \ \ \rightarrow \text{Equation 3}$$

Multiplying the second equation by {eq}3 {/eq}, we get,

$$15r + 9s = 132 \ \ \rightarrow \text{Equation 4}$$

Now Subtract equation 3 from equation 4, we get,

\displaystyle \begin{align*} 15r + 9s - \left ( 15r - 25s \right ) &= 132 - (30) \ \ \ \ \ \ \ \ \ \ \ \ \left ( \text{Equation 4 - Equation 3} \right ) \\ 15r + 9s - 15r + 25s &= 132 - 30 \\ 9s + 25s &= 102 \\ 34 s &= 102 \\ s &= \dfrac {102}{34} \\ s &= \dfrac {34 \cdot 3}{34} \\ s &= 3 \\ \end{align*}

Substitute {eq}s = 3 {/eq} in equation 1, we get,

\displaystyle \begin{align*} \Rightarrow 3r - 5(3) &= 6 \\ \Rightarrow 3r - 5(3) &= 3 \cdot 2 \\ \Rightarrow r - 5 &= 2 \\ \Rightarrow r &= 2 + 5 \\ \Rightarrow r &= 7 \end{align*}

Solution:

$$\boxed{r = 7 \\ s = 3}$$

#### Learn more about this topic:

Elimination Method in Algebra: Definition & Examples

from High School Algebra II: Help and Review

Chapter 7 / Lesson 9
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