Statement Solve by the substitution method: {2x−3y=164x+5y=10\begin{cases} 2x - 3y = 16\\ 4x + 5y = 10 \end{cases}{2x−3y=164x+5y=10 Solution From the first equation isolate xxx: 2x=16+3y ⟹ x=16+3y2.2x = 16 + 3y \implies x = \frac{16+3y}{2}.2x=16+3y⟹x=216+3y. Substitute into the second: 4⋅16+3y2+5y=10 ⟹ 2(16+3y)+5y=10 ⟹ 32+6y+5y=10.4\cdot\frac{16+3y}{2} + 5y = 10 \implies 2(16+3y) + 5y = 10 \implies 32 + 6y + 5y = 10.4⋅216+3y+5y=10⟹2(16+3y)+5y=10⟹32+6y+5y=10. Hence 11y=−2211y = -2211y=−22, so y=−2y = -2y=−2. Then x=16+3(−2)2=102=5.x = \frac{16 + 3(-2)}{2} = \frac{10}{2} = 5.x=216+3(−2)=210=5. Check: 2⋅5−3⋅(−2)=10+6=162\cdot 5 - 3\cdot(-2) = 10+6 = 162⋅5−3⋅(−2)=10+6=16 and 4⋅5+5⋅(−2)=20−10=104\cdot 5 + 5\cdot(-2) = 20-10 = 104⋅5+5⋅(−2)=20−10=10. Correct. x=5,y=−2\boxed{x = 5, \quad y = -2}x=5,y=−2