To solve the given system of linear equations:
[tex]\[
\begin{cases}
x + 2y = 1 \\
-3x + y = 10
\end{cases}
\][/tex]
we will use the method of substitution or elimination. Here is a step-by-step solution using the elimination method:
1. Label the equations for reference:
[tex]\[
\text{(1) } x + 2y = 1
\][/tex]
[tex]\[
\text{(2) } -3x + y = 10
\][/tex]
2. Eliminate one variable by manipulating the equations. We can eliminate [tex]\( y \)[/tex] by making the coefficients of [tex]\( y \)[/tex] in both equations match.
Let’s multiply the second equation by 2 to match the coefficient of [tex]\( y \)[/tex] in the first equation:
[tex]\[
\text{(2) } -3x + y = 10 \quad \Rightarrow \quad -6x + 2y = 20
\][/tex]
Now, we have:
[tex]\[
\begin{cases}
x + 2y = 1 \quad \text{(Equation 1)} \\
-6x + 2y = 20 \quad \text{(Modified Equation 2)}
\end{cases}
\][/tex]
3. Subtract Equation 1 from the Modified Equation 2 to eliminate [tex]\( y \)[/tex]:
[tex]\[
(-6x + 2y) - (x + 2y) = 20 - 1
\][/tex]
Simplifying:
[tex]\[
-6x + 2y - x - 2y = 19
\][/tex]
[tex]\[
-7x = 19
\][/tex]
4. Solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{19}{-7}
\][/tex]
[tex]\[
x = -\frac{19}{7}
\][/tex]
5. Substitute the value of [tex]\( x \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]. We will use Equation 1:
[tex]\[
x + 2y = 1
\][/tex]
[tex]\[
-\frac{19}{7} + 2y = 1
\][/tex]
6. Isolate [tex]\( y \)[/tex]:
[tex]\[
2y = 1 + \frac{19}{7}
\][/tex]
[tex]\[
2y = \frac{7}{7} + \frac{19}{7}
\][/tex]
[tex]\[
2y = \frac{26}{7}
\][/tex]
[tex]\[
y = \frac{26}{7 \cdot 2}
\][/tex]
[tex]\[
y = \frac{26}{14}
\][/tex]
[tex]\[
y = \frac{13}{7}
\][/tex]
So, the solution to the system of equations is:
[tex]\[
\boxed{\left( -\frac{19}{7}, \frac{13}{7} \right)}
\][/tex]
