Answer:
p=0.07 and e=0.03
Step-by-step explanation:
This is a typical problem of 2-equation system with 2 unknown variables. Lets first think about it.
Julio bought 4 pencils at a price of p each and 3 erasers at a price of e each at a cost of 0.37 (from here I will omit the $ symbol for simplicity). So we could write an equation of the form:
4p + 3e = 0.37
David paid 0.33 for 3 pencils and 4 erasers, paying the same prices as Julio:
3p + 4e = 0.33
So, we have a 2-equation system with two unknowns p and e:
4p + 3e = 0.37 [eq 1]
3p + 4e = 0.33 [eq 2]
Take equation 1 and multiply it by 3 and take equation 2 and multiply it by -4. As me multiply every term of each equation the equation maintains:
4p + 3e = 0.37 (x3) 12p + 9e = 1.11
3p + 4e = 0.33 (x -4) -12p - 16e = -1.32
Now, sum both equations, summing term by term, i.e., the p's together and the e's together:
(12p - 12p) + (9e - 16e) = 11.1 -1.32
-7e = -0.21
Divide both sides by (-7):
-7e/(-7) = -0.21/(-7)
e = 0.03
So, the cost of 1 eraser is $0.03. Now, replace it value in eq 1 or eq2 to find p:
4p + 3e = 0.37
4p + 3(0.03) = 0.37
4p + 0.09 = 0.37
Subtract 0.09 in both sides:
4p = 0.37 - 0.09
4p = 0.28
Divide both sides by 4:
4p/4 = 0.28/4
p = 0.07
Lets verify our result, replacing p and e found in equation 2:
3(0.07) + 4(0.03) = 0.21 + 0.12 = 0.33
Verified!
