How much was she charged for each shirt, each pair of slacks, and each sports coat?

Question

Rachel took clothes to the cleaners three times last month. First, she brought 4 shirts and 1 pair of slacks and paid $15.45. Then she brought 7 shirts, 4 pairs of slacks, and 2 sports coats and paid $50.87. Finally, she bought 3 shirts and 1 sports coat and paid $16. 96. How much was she charged for each shirt, each pair of slacks, and each sports coat?

Answer 1

The first thing I do when I see these word problems is ask myself, "What am I going to do with this information? Is it all part of one equation, or am I dealing with a set of equations here?" In this case, there are 3 different instances, or days, where Rachel is charged, so we can deduce that there are a set of equations to be made.

Sets of equations are useful in creating a relationship between multiple sets of data. To find undisclosed information, such as how much Rachel is charged for a particular article of clothing, you need multiple, un-identical sets of data. If there are 3 variables, then you need 3 equations to be able to solve it.

From there, I want to pair up the relevant information. In this case, it's fairly clear which information should be paired up, as they list it off (e.g. "First," "Then," "Finally").

Sets of equations:

let t = shirts (t-shirts), s = slacks, and c = sports coats

{ 4t + s = $15.45

{ 7t + 4s + 2c = $50.87

{ 3t + c = $16.96

Immediately, I notice that there is a shirt, t, in every equation. So, for convenience, I'll substitute my other variables for t.

I start off by isolating the variables that are not t in the two two-variable equations:

4t + s = 15.45

s = 15.45 - 4t

3t + c = 16.96

c = 16.96 - 3t

I then proceed to substitute those variables for their corresponding "relationship" to t into the three-variable equation. I quote "relationship" because that's an important concept to understand about this kind of math. When solving for multiple variables, the only way to do so is by finding relationships between the variables, such that one can create an equation that consists of only one variable. However complex it may be, it will likely still be solvable.

(s) (c)

7t + 4(15.45 - 4t) + 2(16.96 - 3t) = 50.87

I proceed to distribute my coefficients, 4 and 2.

4(15.45 - 4t) 2(16.96 - 3t)

7t + 61.8 - 16t + 33.92 - 6t = 50.87

Combine like terms.

7t + 61.8 - 16t + 33.92 - 6t = 50.87

-15t + 95.78 = 50.87

Isolate t.

t = 44.91 = $2.99

15

Now that we have t, we can plug it into the two equations with only two variables and solve for the corresponding variable.

s = 15.45 - 4(2.99)

s = 15.45 - 11.96

s = $3.49

c = 16.96 - 3(2.99)

c = 16.96 - 8.97

c = $7.99

Mathematically, the solution to this set of equations is (t, s, c):

(2.99, 3.49, 7.99)

However, this is a word problem and many teachers like their answers to match the reality of the word problem.

An acceptable answer will likely be:

Rachel was charged $2.99/ea. for her shirts, $3.49/ea. for her slacks, and $7.99/ea.for her sports coats.

Don't forget to say "each" (or /ea.)! It's an important detail in math. It's like saying "y = 3" is the same thing as "y = x". One's constant and one's variable!

Hope that helps!