Question

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

$(2x - 3y)-(2x - 5y)=4 - 0$. Expand the left - hand side: $2x-3y-2x + 5y=4$, which simplifies to $2y=4$.

Step3: Solve for y

Divide both sides of $2y = 4$ by 2: $y=\frac{4}{2}=2$.

Step4: Solve for x

Substitute $y = 2$ into Equation 2: $2x-5\times2=0$, so $2x=10$, and $x = 5$.

Answer:

$x = 5,y = 2$