SOLUTION: I have been trying to figure this out for two days. I really need help!!(This is Applications of Linear Equations) An art dealer sold two artworks for $1520 thereby making a profi

Question 80797This question is from textbook Elementary & Intermediate Algebra: Concepts & Applications
: I have been trying to figure this out for two days. I really need help!!(This is Applications of Linear Equations)
An art dealer sold two artworks for $1520 thereby making a profit of 25% on the first and 10% profit on the other, whereas if he had approached any exhibition he would have sold them together for $1535 with a profit of 10% on the first and 25% on the other artwork. Find the actual cost of each artwork. This question is from textbook Elementary & Intermediate Algebra: Concepts & Applications

Answer by dolly(163) (Show Source): You can put this solution on YOUR website!
Let the cost of the first art work = $x
Profit = 25% = 0.25
So sale price = x + 0.25x
= 1.25x
Let the cost of the second art work = $y
Profit = 10% = 0.10
So sale price = y + 0.10y
= 1.10y
Total sale price given = $1520
==> 1.25x + 1.10y = 1520 ------------(1)
Case (ii)
Profit on first work = 0.10
so sale price = 1.10x
Profit on second work = 0.25
so sale price = 1.25y
Total sale price given = $1535
==> 1.10x + 1.25y = 1535 ------------(2)
Adding 1 and 2 we get, 2.35x + 2.35y = 3050
==> x + y = 1300 [dividing by 2.35 throughout]------(3)
Subtracting 1 and 2 we get, 0.15x - 0.15y = - 15
==> x - y = -100 [dividing by 0.15 throughout]------(4)
Adding 3 and 4, 2x = 1200
==> x = 600
plugging in (3), we get y = 700
Thus the art works cost $600 and $ 700 respectively.

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