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Solve the following system of equations:

[tex]\[
\left\{
\begin{array}{l}
2x + 5y = 1 \\
2x + y = 5
\end{array}
\right.
\][/tex]


Sagot :

Certainly! To solve the system of linear equations:

[tex]\[ \left\{\begin{array}{l} 2x + 5y = 1 \\ 2x + y = 5 \end{array}\right. \][/tex]

we can follow these steps:

1. Label the equations for reference:
[tex]\[ \begin{aligned} \text{(1)} & \quad 2x + 5y = 1 \\ \text{(2)} & \quad 2x + y = 5 \end{aligned} \][/tex]

2. Eliminate one variable:
We can eliminate [tex]\(x\)[/tex] by subtracting Equation (2) from Equation (1).

First, let's subtract Equation (2) from Equation (1):
[tex]\[ (2x + 5y) - (2x + y) = 1 - 5 \][/tex]

3. Simplify the resulting equation:
[tex]\[ 2x + 5y - 2x - y = -4 \][/tex]
[tex]\[ 4y = -4 \][/tex]

4. Solve for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{-4}{4} = -1 \][/tex]

5. Substitute [tex]\(y = -1\)[/tex] back into one of the original equations to find [tex]\(x\)[/tex]:
Let's use Equation (2):
[tex]\[ 2x + y = 5 \][/tex]
Substitute [tex]\(y = -1\)[/tex]:
[tex]\[ 2x - 1 = 5 \][/tex]

6. Solve for [tex]\(x\)[/tex]:
[tex]\[ 2x = 6 \][/tex]
[tex]\[ x = \frac{6}{2} = 3 \][/tex]

So, the solution to the system of equations is:
[tex]\[ x = 3, \quad y = -1 \][/tex]

Therefore, the solution to the given system of equations is [tex]\((x, y) = (3, -1)\)[/tex].
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