Chapter 9 - 9.1 - Linear and Nonlinear Systems of Equations - 9.1 Exercises - Page 635: 19

$x=1,y=1$

Here, we have $15x+8y=23 \implies x=\dfrac{1}{3}+\dfrac{2}{3} y$ ...(1) and $3x-2y=1 \implies 3x=1+2y$ ...(2) Now, plug $x=\dfrac{1}{3}+\dfrac{2}{3} y$ into the second equation. $15(\dfrac{1}{3}+\dfrac{2}{3} y)+8y=23$ or, $15y+8y=23 \implies y=1$ Use the $y$ value in the second equation and solve for $x$. $3x=1+2(1) \implies x=1$ Hence, $(x,y)=(1,1)$