Certainly! Let's solve the given system of linear equations step-by-step:
[tex]\[
\left\{\begin{array}{r}
x + 3y = 20 \\
x + y = 12
\end{array}\right.
\][/tex]
### Step 1: Express one variable in terms of the other
First, we can solve the second equation for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[
x + y = 12 \implies x = 12 - y
\][/tex]
### Step 2: Substitute the expression into the first equation
Next, we'll substitute [tex]\( x = 12 - y \)[/tex] into the first equation:
[tex]\[
x + 3y = 20
\][/tex]
[tex]\[
(12 - y) + 3y = 20
\][/tex]
### Step 3: Simplify the resulting equation
Combine like terms:
[tex]\[
12 - y + 3y = 20
\][/tex]
[tex]\[
12 + 2y = 20
\][/tex]
### Step 4: Solve for [tex]\( y \)[/tex]
Isolate [tex]\( y \)[/tex] by subtracting 12 from both sides:
[tex]\[
2y = 20 - 12
\][/tex]
[tex]\[
2y = 8
\][/tex]
[tex]\[
y = 4
\][/tex]
### Step 5: Substitute [tex]\( y = 4 \)[/tex] back into the expression for [tex]\( x \)[/tex]
Now substitute [tex]\( y = 4 \)[/tex] back into the equation [tex]\( x = 12 - y \)[/tex]:
[tex]\[
x = 12 - 4
\][/tex]
[tex]\[
x = 8
\][/tex]
### Conclusion
The solution to the system of equations is:
[tex]\[
x = 8, \quad y = 4
\][/tex]
Therefore, the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations are [tex]\( x = 8 \)[/tex] and [tex]\( y = 4 \)[/tex].
