9514 1404 393
Answer:
- 6 pounds almonds
- 3 pounds walnuts
- 1 pound cashews
Step-by-step explanation:
We can let a, w, c represent pounds of almonds, walnuts, and cashews, respectively. Then the given relations are ...
a +w +c = 10 . . . . . total pounds of nut mix
a = 2w . . . . . . . . . twice as many almonds as walnuts
1.25a +2.75w +3.15c = 18.90 . . . . total cost of the mix
__
Solving the system of equations can proceed by any of the methods you have learned for systems of two equations in two unknowns. Substitution, elimination, and matrix methods all work. Graphing methods are also available.
We can easily reduce this system to two equations in two unknowns by using the second equation to substitute for 'a'.
2w +w +c = 10 ⇒ 3w +c = 10
1.25(2w) +2.75w +3.15c = 18.90 ⇒ 5.25w +3.15c = 18.90
Solving the first of these equations for c, then substituting into the second of these equations gives ...
c = 10 -3w
5.25w +3.15(10 -3w) = 18.90
-4.20w +31.50 = 18.90 . . . . simplify
12.60 = 4.20w . . . . . . . . . . . add 4.20w -18.90
3 = w . . . . . . . . . . . . . . . . . . . divide by 4.20
c = 10 -3(3) = 1 . . . . . . . . . . . use the expression for c
a = 2(3) = 6 . . . . . . . . . . . . . use the expression for a
The nut mix used 6 pounds of almonds, 3 pounds of walnuts, and 1 pound of cashews.
_____
Additional comment
The attached graph shows one way to obtain a graphical solution. The first and third equations were solved for c, and those expressions were set equal to obtain the graph for the red line. The blue line is the second equation. Here, we have used (x, y) for (a, w). The solution for these variables is (a, w) = (6, 3). Using the first expression for c gives c = 10 -(6+3) = 1. This is the same solution as above.
