This Lesson (Word problem to solve combined system of linear equations and a "price equation") was created by by ikleyn(52906): View Source, Show About ikleyn:
Word problem to solve combined system of linear equations and a "price equation"
Problem 1
Andrew bought some apples and pears. The ratio of the number of apples bought to the number of pears bought was 7:4.
He spent $61.20. He paid $22.80 more for the apples than the pears. Each apple was $0.30 more than each pears.
(a) How much did he spend on the pears?
(b) How many pears did he buy?
Solution
Let A be the amount he spent for apples.
Let P be the amount he spent for pears.
From the problem, we have these two equations
A + P = 61.20, (1)
A - P = 22.80. (2)
To find P, subtract equation (2) from equation (1). You will get
2P = 61.20 - 22.80
2P = 38.4, P = 38.4/2 = 19.2.
Thus, Andrew spent $19.20 for pears. It is the answer for question (a).
From this, we conclude that Andrew spent $61.20 - $19.20 = $42.00 for apples.
Now we start solving (b).
Andrew has 7n apples and 4n pears. The number 'n' is unknown, and we want to find it.
The price for one apple is = .
The price for one pear is = .
The difference equation for price is
- = 0.3 dollars.
Simplify and find 'n'
= 0.3,
n = = 4.
So, Andrew bought 7n = 7*4 = 28 apples and 4n = 4*4 = 16 pears.
It is the answer to question (b).
