# Solve using the elimination method: -4x + 9y = 9 x - 3y = -6

## Question:

Solve using the elimination method:

{eq}-4x + 9y = 9 \\ \ \ x - 3y = -6 {/eq}

## Solution of the Linear Equations:

{eq}\\ {/eq}

The solution or the set of values of {eq}x {/eq} and {eq}y {/eq} for the given set of linear equations will be determined using the method of elimination. In this method, we eliminate one of the variables out of {eq}x {/eq} and {eq}y {/eq} using some scalar multiplication and basic arithmetic operations. Once we have the value of one variable, we can evaluate the value of other variables using any of the given equations.

## Answer and Explanation:

{eq}\\ {/eq}

{eq}-4x + 9y = 9 \; \; \cdots \cdots \; \; (1) \\ x - 3y = - 6 \; \; \cdots \cdots \; \; (2) {/eq}

Now multiply equation (2) with a scalar of {eq}4 {/eq} in order to make the coefficients of {eq}x {/eq} equal but of opposite sign in both the equations:

{eq}4 (x - 3y) 4 \times (-6) \\ 4x - 12y = -24 \; \; \cdots \cdots \; \; (3) {/eq}

Now add equation (1) and equation (3) in order to eliminate the variable {eq}x {/eq}:

{eq}9y - 12y = 9 - 24 \\ -3y = -15 \\ \Longrightarrow y = \dfrac {15}{3} = 5 {/eq}

Now put the value of {eq}y = 5 {/eq} in equation (1) in order to get the value of {eq}x {/eq}:

{eq}-4x + 9 \times 5 = 9 \\ -4x = 9 - 45 \\ -4x = - 36 \\ \Longrightarrow x = \dfrac {36}{4} = 9 {/eq}

Finally, we have the values of the variables {eq}x {/eq} and {eq}y {/eq} given below:

{eq}\Longrightarrow \boxed {(x, \; y) = (9, \; 5)} {/eq}

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from High School Algebra II: Help and Review

Chapter 7 / Lesson 9