Certainly! Let's solve the given system of equations step-by-step:
The given system of equations is:
[tex]\[ \left\{ \begin{array}{l}
2x + 2y = 6 \\
3x + y = 1
\end{array} \right. \][/tex]
### Step 1: Simplify the First Equation
First, we can simplify the first equation by dividing every term by 2:
[tex]\[ 2x + 2y = 6 \implies x + y = 3 \][/tex]
So now we have:
[tex]\[ \left\{ \begin{array}{l}
x + y = 3 \\
3x + y = 1
\end{array} \right. \][/tex]
### Step 2: Substitute One Equation into the Other
We can solve for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] using the simpler equation [tex]\( x + y = 3 \)[/tex]:
[tex]\[ y = 3 - x \][/tex]
### Step 3: Substitute into the Second Equation
Next, we substitute [tex]\( y = 3 - x \)[/tex] into the second equation [tex]\( 3x + y = 1 \)[/tex]:
[tex]\[ 3x + (3 - x) = 1 \][/tex]
### Step 4: Solve for [tex]\( x \)[/tex]
Combine like terms:
[tex]\[ 3x + 3 - x = 1 \][/tex]
[tex]\[ 2x + 3 = 1 \][/tex]
Isolate [tex]\( x \)[/tex] by subtracting 3 from both sides:
[tex]\[ 2x = 1 - 3 \][/tex]
[tex]\[ 2x = -2 \][/tex]
Now, divide both sides by 2:
[tex]\[ x = -1 \][/tex]
### Step 5: Solve for [tex]\( y \)[/tex]
Finally, substitute [tex]\( x = -1 \)[/tex] back into the equation [tex]\( y = 3 - x \)[/tex]:
[tex]\[ y = 3 - (-1) \][/tex]
[tex]\[ y = 3 + 1 \][/tex]
[tex]\[ y = 4 \][/tex]
### Step 6: Solution
The solution to the system of equations is:
[tex]\[ x = -1 \][/tex]
[tex]\[ y = 4 \][/tex]
Therefore, the solution to the system [tex]\(\left\{ \begin{array}{l}
2x + 2y = 6 \\
3x + y = 1
\end{array} \right.\)[/tex] is:
[tex]\[ (x, y) = (-1, 4) \][/tex]
