Chapter 3 - Section 3.2 - Problem Solving and Business Applications Using Systems of Equations - Exercise Set - Page 205: 15

$x=8000,y=6000$

Given: $0.09x+0.03y=900$ and $0.10x+0.01y=860$ from the first equation, we have $x=10000-\dfrac{y}{3}$ After plugging the equation of in the variable of $x$ in the second equation, we get $0.1(10000-\dfrac{y}{3})+0.01y=860 \implies \dfrac{-7}{300}y=-140$ or, $y=6000$ Now, $0.09x+0.03(6000)=900$ or, $x=8000$ Thus, $x=8000,y=6000$