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Standardized Test Practice
62. EXTENDED RESPONSE Lenora bought several
pounds of cashews and several pounds of
almonds for a party. The cashews cost $8 per
pound, and the almonds cost $6 per pound.
Lenora bought a total of 7 pounds and paid
a total of $48. Write and solve a system of
equations to determine the pounds of cashews
and the pounds of almonds that Lenora
purchased.

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Answer:

8x+6y=48

x+y=7

multiply bottom equation by 6:

8x+6y=48

6x+6y=42

subtract equations:

2x=6

divide 6 by 2:

x=3

therefore:

y=4

Hello!

First, set some variables. Say c = # of pounds of cashews, and a = # of pounds of almonds. With these variables, you can set up a system of equations. The two equations we will be using is how much she paid, and how many pounds she bought.

How much she paid: 8c + 6a = 48

This one can be explained by c being the number of pounds of cashews, therefore multiplying it by 8 would be the amount she paid for Cashews, and the same for the almonds.

How many pounds she bought: c + a = 7

This one is fairly simple. She bought 7 pounds total, and then c + a should therefore equal the total number of pounds she bought, so c + a = 7.

Now, set up the system.

[tex]\left \{ {{8c + 6a = 48} \atop {c + a = 7}} \right.[/tex]

I'm going to be using the method of substitution. That's where you solve for one of the variables in one of the equations, and then substitute that into the other equation.

[tex]\left \{ {{8c + 6a = 48} \atop {c + a = 7}} \right.[/tex]

[tex]\left \{ {{8c + 6a = 48} \atop {c = 7 - a}} \right.[/tex]

Now, since you know c = 7 - a, you can substitute 7 - a in the other equation in for c.

[tex]\left \{ {{8c + 6a = 48} \atop {c = 7 - a}} \right.[/tex]

8 (7 - a) + 6a = 48

And now solve.

8 (7 - a) + 6a = 48

56 - 8a + 6a = 48

-2a = -8

a = 4

To get the other variable, substitute a = 4 into one of the original equations.

c + a = 7

c + 4 = 7

c = 3

Therefore, a = 4 and c = 3. This means that Lenora purchased 4 pounds of almonds and 3 pounds of cashews.

Hope this helped!