Answer :

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].

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