# Jessica has two bags of nuts one with 70% peanuts, and the other with 21% peanuts. To make a 11...

## Question:

Jessica has two bags of nuts: one with 70% peanuts, and the other with 21% peanuts. To make a 11 lb bag of 36% peanuts, how much of each should Jessica use?

## Elimination Method:

(i) The elimination method is used to solve a system of two equations in two unknowns (variables).

(ii) In this method, we add or subtract the given equations and eliminate one variable.

(iii) We solve the resultant equation in one variable using the algebraic operations.

(iv) We can then substitute the value of this known variable in one of the given two equations and solve for the other variable.

The number of lbs of {eq}70\% {/eq} and {eq}21\% {/eq} peanuts is {eq}x {/eq} and {eq}y {/eq} respectively.

Since the total amount is {eq}11 {/eq} lbs, we get the equation:

$$x+y= 11\,\,\,\,\,\,\,\rightarrow (1)$$

Using the given information:

$$70\% \text{ of }x+ 21\% \text{ of }y = 36\% \text{ of }11\\[0.4cm] 0.70x+ 0.21y= 0.36(11)\\[0.4cm] 0.70x+ 0.21y= 3.96\,\,\,\,\,\,\,\rightarrow (2)$$

Multiply both sides of (1) by {eq}-0.70 {/eq}, then we get:

$$-0.70x-0.70y= -7.7\,\,\,\,\,\,\,\rightarrow (3)$$

$$-0.49y= -3.74\\[0.4cm] \text{Dividing both sides by 0.20}, \\[0.4cm] y= \dfrac{-3.74}{-0.49} =\frac{374}{49}$$

Substitute this in (1):

$$x+ \frac{374}{49}=11\\[0.4cm] \text{Subtracting } \frac{374}{49} \text{ from both sides}, \\[0.4cm] x= \frac{165}{49}$$

Therefore, the number of lbs of {eq}70\% {/eq} peanuts = {eq}\frac{165}{49} \approx \color{blue}{\boxed{\mathbf{3.37 }}} {/eq}

and the number of lbs of {eq}21\% {/eq} peanuts = {eq}\frac{374}{49} \approx \color{blue}{\boxed{\mathbf{7.63}}} {/eq}.

The final answers are rounded to two decimals.

Elimination Method in Algebra: Definition & Examples

from High School Algebra II: Help and Review

Chapter 7 / Lesson 9
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