To solve the system of linear equations:
[tex]\[
\left\{
\begin{array}{l}
-x + y = 1 \\
2x + 5y = 1
\end{array}
\right.
\][/tex]
we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations simultaneously. Here's a step-by-step solution:
1. Rewrite the first equation for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[
y = x + 1
\][/tex]
2. Substitute this expression for [tex]\( y \)[/tex] into the second equation:
[tex]\[
2x + 5(x + 1) = 1
\][/tex]
This simplifies to:
[tex]\[
2x + 5x + 5 = 1
\][/tex]
Combining like terms, we get:
[tex]\[
7x + 5 = 1
\][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[
7x + 5 = 1
\][/tex]
Subtract 5 from both sides:
[tex]\[
7x = -4
\][/tex]
Divide by 7:
[tex]\[
x = -\frac{4}{7}
\][/tex]
4. Substitute the value of [tex]\( x \)[/tex] back into the expression for [tex]\( y \)[/tex] from step 1:
[tex]\[
y = x + 1 = -\frac{4}{7} + 1
\][/tex]
Convert 1 into a fraction with a denominator of 7:
[tex]\[
y = -\frac{4}{7} + \frac{7}{7}
\][/tex]
Combine the fractions:
[tex]\[
y = \frac{3}{7}
\][/tex]
Therefore, the solution to the system of equations is:
[tex]\[
x = -\frac{4}{7}, \quad y = \frac{3}{7}
\][/tex]
So, the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations are [tex]\(\left( -\frac{4}{7}, \frac{3}{7} \right)\)[/tex].
