Question

solve the system of equations \\(\

$$\begin{cases} 3x + 2y = 10 \\\\ 3x - y = 22 \\end{cases}$$

\\)

Explanation:

Step1: Subtract the two equations to eliminate \(x\)

We have the system of equations:
\[

$$\begin{cases} 3x + 2y = 10 & (1)\\ 3x - y = 22 & (2) \end{cases}$$

\]
Subtract equation \((2)\) from equation \((1)\):
\[
(3x + 2y)-(3x - y)=10 - 22
\]
Simplify the left - hand side: \(3x+2y - 3x + y=-12\), which gives \(3y=-12\).

Step2: Solve for \(y\)

From \(3y=-12\), divide both sides by 3: \(y=\frac{-12}{3}=-4\).

Step3: Substitute \(y = - 4\) into one of the original equations to solve for \(x\)

Substitute \(y=-4\) into equation \((2)\): \(3x-(-4)=22\), that is \(3x + 4 = 22\).
Subtract 4 from both sides: \(3x=22 - 4=18\).
Divide both sides by 3: \(x=\frac{18}{3}=6\).

Answer:

The solution of the system of equations is \(x = 6,y=-4\)