Question

(1) $\begin{cases}\frac{x + 1}{3}=\frac{y + 2}{2}cdots\textcircled{1}\\2x - 5y = 0cdots\textcircled{2}end{cases}$ (2) $\begin{cases}\frac{x - 3}{5}=\frac{y - 7}{2}\\7x = 3yend{cases}$ sol $\textcircled{1}\times6$ $2(x + 1)=3(y + 2)$ $2x - 3y = 4cdots\textcircled{3}$

Explanation:

Step1: Simplify the first - equation

Multiply both sides of $\frac{x + 1}{3}=\frac{y + 2}{2}$ by 6 to get $2(x + 1)=3(y + 2)$. Expand it: $2x+2 = 3y + 6$, then $2x-3y=4$ (Equation ③).

Step2: Subtract Equation ② from Equation ③

We have Equation ②: $2x - 5y=0$ and Equation ③: $2x-3y = 4$. Subtract Equation ② from Equation ③: $(2x-3y)-(2x - 5y)=4-0$.
$2x-3y - 2x + 5y=4$, which simplifies to $2y = 4$, so $y = 2$.

Step3: Substitute $y = 2$ into Equation ②

Substitute $y = 2$ into $2x-5y = 0$. We get $2x-5\times2=0$, i.e., $2x-10 = 0$. Then $2x=10$, and $x = 5$.

Answer:

$x = 5,y = 2$