system of equation word problems

Question

Molly's Custom Kitchen Supplies sells handmade forks and spoons. It costs the store $1.70 to buy the supplies to make a fork and $1.30 to buy the supplies to make a spoon. The store sells forks for $5.60 and spoons for $5.40. Last April Molly's Custom Kitchen Supplies spent $37.90 on materials for forks and spoons. They sold the finished products for a total of $147.20. How

many forks and how many spoons did they make last April?

Answer 1

1.7F + 1.3S = 37.9

5.6F + 5.4S = 147.2

9.52F + 7.28S = 212.24

9.52F + 9.18S = 250.24

1.90S = 38

S=20 spoons

1.7F = 37.9-1.2(20) = 13.9

5.6F =147.2-5.4(20) = 39.2

F = 39.2/5.6 = 7 forks

Answer 2

Hi, Silvio!

We can solve this problem by making two equations, one to show the expenses for Molly's business, the other to show the profits. First, let's look at the expenses.

I don't know how many forks (f) and spoons (s) were made, but I know the cost for each one. The cost to make f number of forks will be f * the cost of a single fork, or 1.7f. The cost to make s number of spoons will be s * the cost of a single spoon, or 1.3s. Finally, we are told that, in April, Molly spent $37.90 on supplies for both forks and spoons. Taking all of this together, we get the equation:

1.7f + 1.3s = 37.9

Similarly, I don't know how many forks and knives were sold, but I know the profit from each. Earnings from the sale of f forks will be f * the earnings from a single fork, or 5.6f. Earnings from the sale of s spoons will be s * the earnings from a single spoon, or 5.4s. Finally, we are told that, in April, Molly sold $147.20 from both forks and spoons. Taking all of this together, we get the equation:

5.6f + 5.4s = 147.2

Now that we have these two equations, we can use substitution or elimination to solve for one of the variables. I prefer substitution, so I'll model that here, but you can also use elimination!

It doesn't matter which equation we use or which variable we solve for. I'll use the first equation and solve for f.

1.7f + 1.3s = 37.9

To get the f term alone on the left side of the equation, I'll subtract 1.3s from both sides.

1.7f = 37.9 - 1.3s

Next, to get f alone, I'll divide both sides of the equation by 1.7

f = (37.9 - 1.3s) / 1.7

Now that I have a solution for f, I can plug the right side of that solution into any place where I have an f in the second of my original equations.

5.6f + 5.4s = 147.2

5.6((37.9 - 1.3s) / 1.7) + 5.4s = 147.2

First, I'll distribute out the 5.6 to the numerator of the fraction by which it is multiplied.

(212.24 - 7.28s) / 1.7 + 5.4s = 147.2

Next, I'll divide both parts of the numerator by 1.7. Since I know the question uses money, I'll round to the second decimal place.

124.85 - 4.28s + 5.4s = 147.2

I'll combine like terms.

124.85 + 1.12s = 147.2

To get the s term alone on the left side of the equation, I'll subtract 124.85 from both sides.

1.12s = 22.35

Finally, to get the s term alone, I'll divide both sides of the equation by 1.12. Since this number shows how many spoons were sold, I'll round to the nearest whole number (you can't sell half a spoon!).

s = 19.96, rounded to 20

Now that we know the value for s, we can plug it into either equation to find f. I'll choose to use the first.

1.7f + 1.3s = 37.9

1.7f + 1.3(20) = 37.9

1.7f + 26 = 37.9

I'll subtract 26 from both sides to get the f term alone.

1.7f = 11.9

I'll divide both sides by 1.7 to get f alone.

f = 7

Therefore, Molly made and sold 20 spoons and 7 forks in the month of April.

I hope this helps!

Answer 3

Hi Silvio D

Since you have a cost and total for Supplies and you have sell prices and total for Finished Products. You can set up the equations accordingly, use elimination to find the information you need.

Let f = forks

Let s = spoons

Equation 1 for Supplies

1.70f + 1.30s = 37.9

Equation 2 for Finished Goods

5.60f + 5.40s = 147.2

Please give the elimination a try.

I got 7 forks and 20 spoons.