SOLUTION: Five pencils and three erasers cost 65 cents. Three pencils and two erasers cost 40 cents. Find the cost of one pencil and the cost of one eraser. So far I've got: 5p + 3e = .65

Algebra ->  Finance -> SOLUTION: Five pencils and three erasers cost 65 cents. Three pencils and two erasers cost 40 cents. Find the cost of one pencil and the cost of one eraser. So far I've got: 5p + 3e = .65      Log On


   

Question 146978: Five pencils and three erasers cost 65 cents. Three pencils and two erasers cost 40 cents. Find the cost of one pencil and the cost of one eraser.
So far I've got: 5p + 3e = .65 and 3p + 2e = .40 but don't know where to go from here to reach the answer.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Look at it this way.
You have a system of equations.
Two equations with two unknowns, p and e.
You can solve the system to find p and e.
5p + 3e = .65
3p + 2e = .40
There are several methods you can use to solve this system of equations.
We'll use the substitution method.
First off, let's multiply both sides by 100 to get rid of the decimal.
1.500p%2B300e=65
2.300p%2B200e=40
You can use equation 1 to get 300p in terms of e and substitute into eq.2,
1.500p%2B300e=65
500p=65-300e
300p=%2865-300e%29%283%2F5%29Multiply both sides by (3/5).
300p=39-180e
Now substitute that value into equation 2 and solve for e.
2300p%2B200e=40
%2839-180e%29%2B200e=40
20e=1
e=0.05
Now you can use either equation to solve for p. I chose equation 2.
300p%2B200e=40
300p%2B200%280.05%29=40
300p%2B10=40
300p=30
p=0.10
So, the pencil costs a dime and the eraser costs a nickel.
You can check your answers by using your original equations.
5p + 3e = .65
5(0.10)+3(0.05)=.65
0.50+0.15=0.65
0.65=0.65
True
3p + 2e = .40
3(0.10)+2(0.05)=0.40
0.30+0.10=0.40
0.40=0.40
True
Your answers led to true statements.
They are good answers.