Chapter 10 - Systems of Equations and Inequalities - Section 10.1 Systems of Linear Equations: Substitution and Elimination - 10.1 Assess Your Understanding - Page 736: 74

$r=5 \% $ and $Y=1200$

We need to solve the given system of equations as follows: $$0.05Y−1000r=10~~~~(1) \\ 0.05Y+800r=100~~~~(2)$$ We will multiply equation $(1)$ by $-1$. $$-0.05Y+1000r=-10~~~~(3)$$ Add equations $(2)$ and $(3)$ to eliminate Y. $$0.05Y+800r-0.05Y+1000r =100-10 \\ 1800 r =90 \\ r=0.05= 5 \%$$ Now, back substitute the value of $r$ into Equation $(1)$ to solve for $Y$: $$0.05Y−1000(0.05)=10 \\ 0.05Y -50=10 \\ Y=1200$$ Therefore, our results are: $r=5 \% $ and $Y=1200$