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First, solve the system for [tex]\( x \)[/tex] and [tex]\( y \)[/tex].

[tex]\[
\begin{array}{l}
2x - 5y = -26 \\
y = 1 - x
\end{array}
\][/tex]

The value of [tex]\( y \)[/tex] that is part of the solution is [tex]\( y = \)[/tex] [tex]$\qquad$[/tex].

The solution is [tex]$\square$[/tex].

Question ID: 71172

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Sagot :

To solve the system of equations:

[tex]\[ \begin{cases} 2x - 5y = -26 \\ y = 1 - x \end{cases} \][/tex]

we can use the substitution method. Here are the detailed steps:

1. Express one variable in terms of the other: The second equation is already solved for [tex]\( y \)[/tex]:

[tex]\[ y = 1 - x \][/tex]

2. Substitute this expression into the first equation: Substitute [tex]\( y = 1 - x \)[/tex] into [tex]\( 2x - 5y = -26 \)[/tex]:

[tex]\[ 2x - 5(1 - x) = -26 \][/tex]

3. Simplify and solve for [tex]\( x \)[/tex]:
[tex]\[ 2x - 5 + 5x = -26 \][/tex]
Combining like terms:
[tex]\[ 7x - 5 = -26 \][/tex]
Add 5 to both sides:
[tex]\[ 7x = -21 \][/tex]
Divide by 7:
[tex]\[ x = -3 \][/tex]

4. Find [tex]\( y \)[/tex] using the value of [tex]\( x \)[/tex]: Substitute [tex]\( x = -3 \)[/tex] back into [tex]\( y = 1 - x \)[/tex]:

[tex]\[ y = 1 - (-3) = 1 + 3 = 4 \][/tex]

So, the value of [tex]\( y \)[/tex] that is part of the solution is [tex]\( y = 4 \)[/tex].

The solution to the system of equations is:

[tex]\[ (-3, 4) \][/tex]

Hence, the value of [tex]\( y \)[/tex] that is part of the solution is [tex]\( y = 4 \)[/tex].

The solution is [tex]\( (-3, 4) \)[/tex].