Sure! Let's solve the given system of linear equations step-by-step:
[tex]\[
\left\{
\begin{array}{l}
2x + y = 1 \quad \text{(Equation 1)} \\
3x + 4y = 14 \quad \text{(Equation 2)}
\end{array}
\right.
\][/tex]
First, we'll solve the first equation for [tex]\( y \)[/tex]:
[tex]\[
2x + y = 1
\][/tex]
Subtract [tex]\( 2x \)[/tex] from both sides to isolate [tex]\( y \)[/tex]:
[tex]\[
y = 1 - 2x \quad \text{(Equation 3)}
\][/tex]
Next, substitute the expression for [tex]\( y \)[/tex] from Equation 3 into Equation 2:
[tex]\[
3x + 4(1 - 2x) = 14
\][/tex]
Now, distribute the [tex]\( 4 \)[/tex] inside the parentheses:
[tex]\[
3x + 4 - 8x = 14
\][/tex]
Combine like terms:
[tex]\[
3x - 8x + 4 = 14
\][/tex]
This simplifies to:
[tex]\[
-5x + 4 = 14
\][/tex]
Subtract 4 from both sides to isolate the term with [tex]\( x \)[/tex]:
[tex]\[
-5x = 10
\][/tex]
Divide both sides by [tex]\(-5\)[/tex]:
[tex]\[
x = -2
\][/tex]
Now that we have [tex]\( x \)[/tex], we can substitute it back into Equation 3 to find [tex]\( y \)[/tex]:
[tex]\[
y = 1 - 2(-2)
\][/tex]
Simplify inside the parentheses:
[tex]\[
y = 1 + 4
\][/tex]
Therefore,
[tex]\[
y = 5
\][/tex]
So, the solution to the system of equations is:
[tex]\[
x = -2, \quad y = 5
\][/tex]
We can verify the solution by substituting [tex]\( x = -2 \)[/tex] and [tex]\( y = 5 \)[/tex] back into the original equations:
For Equation 1:
[tex]\[
2(-2) + 5 = -4 + 5 = 1 \quad \text{(True)}
\][/tex]
For Equation 2:
[tex]\[
3(-2) + 4(5) = -6 + 20 = 14 \quad \text{(True)}
\][/tex]
Both equations are satisfied, so the solution [tex]\( x = -2 \)[/tex] and [tex]\( y = 5 \)[/tex] is correct.
