Paul mixes nuts worth $1.40 per pound with oats worth $1.65 per pound to get 25 pounds of trail...

Question:

Paul mixes nuts worth $1.40 per pound with oats worth $1.65 per pound to get 25 pounds of trail mix worth $1.50 per pound.

How many pounds of nuts and how many pounds of oats did he use?

Elimination Method:

A system of two equations can be solved in several methods. One of the methods is the "elimination method". In this method, we add or subtract the equations to get a variable canceled. If the direct addition or subtraction does not lead to cancellation, then we will multiply one or both equation(s) by certain numbers to make the cancellation possible. We will solve the resultant equation for the other variable.

Answer and Explanation:

Let us assume the number of pounds of {eq}\$ 1.40 {/eq} per pound nuts is {eq}x {/eq}.

Let us assume the number of pounds of {eq}\$ 1.65 {/eq} per pound oats is {eq}y {/eq}.

There are totally {eq}25 {/eq} pounds in the trail mix.

So we get:

$$x+y=25 \,\,\,\,\,\,\,\rightarrow (1) $$

Since the cost of the mix is {eq}\$ 1.50 {/eq} per pound, we get:

$$\begin{align} 1.40x +1.65y& = (x+y) 1.50 \\[0.4cm] 1.40x +1.65y &= (25)(1.50) & [ \text{From (1)} ] \\[0.4cm] 1.40x +1.65y &= 37.5 & \rightarrow (2) \end{align} $$

Multiply both sides of (1) by {eq}-1.40 {/eq}:

$$-1.40x-1.40y = -35 \,\,\,\,\,\,\,\rightarrow (3) $$

Adding (2) and (3):

$$0.25 y=2.5 \\[0.4cm] \text{Dividing both sides by 0.25},\\[0.4cm] y= 10 $$

Substitute this in (1):

$$x+10= 25 \\[0.4cm] \text{Subtracting 10 from both sides},\\[0.4cm] x =15 $$

Therefore, the number of pounds of nuts = {eq}\color{blue}{\boxed{\mathbf{15}}} {/eq}

and the number of pounds of oats = {eq}\color{blue}{\boxed{\mathbf{10}}} {/eq}.


Learn more about this topic:

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Elimination Method in Algebra: Definition & Examples

from High School Algebra II: Help and Review

Chapter 7 / Lesson 9
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