Answer :
Let's solve the given system of linear equations step by step:
[tex]\[ \begin{cases} x + y = 2 \quad \text{(Equation 1)} \\ 2x - y = 1 \quad \text{(Equation 2)} \end{cases} \][/tex]
### Step 1: Add the equations
First, we add Equation 1 and Equation 2 together to eliminate [tex]\( y \)[/tex]:
[tex]\[ (x + y) + (2x - y) = 2 + 1 \][/tex]
This simplifies to:
[tex]\[ x + y + 2x - y = 3 \][/tex]
[tex]\[ 3x = 3 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{3}{3} \][/tex]
[tex]\[ x = 1 \][/tex]
### Step 2: Substitute [tex]\( x = 1 \)[/tex] into one of the original equations
Now, we substitute [tex]\( x = 1 \)[/tex] into Equation 1 to solve for [tex]\( y \)[/tex]:
[tex]\[ x + y = 2 \][/tex]
[tex]\[ 1 + y = 2 \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = 2 - 1 \][/tex]
[tex]\[ y = 1 \][/tex]
### Solution
Thus, the solution to the system of equations is:
[tex]\[ x = 1, \quad y = 1 \][/tex]
So the final answer is:
[tex]\[ (x, y) = (1, 1) \][/tex]
[tex]\[ \begin{cases} x + y = 2 \quad \text{(Equation 1)} \\ 2x - y = 1 \quad \text{(Equation 2)} \end{cases} \][/tex]
### Step 1: Add the equations
First, we add Equation 1 and Equation 2 together to eliminate [tex]\( y \)[/tex]:
[tex]\[ (x + y) + (2x - y) = 2 + 1 \][/tex]
This simplifies to:
[tex]\[ x + y + 2x - y = 3 \][/tex]
[tex]\[ 3x = 3 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{3}{3} \][/tex]
[tex]\[ x = 1 \][/tex]
### Step 2: Substitute [tex]\( x = 1 \)[/tex] into one of the original equations
Now, we substitute [tex]\( x = 1 \)[/tex] into Equation 1 to solve for [tex]\( y \)[/tex]:
[tex]\[ x + y = 2 \][/tex]
[tex]\[ 1 + y = 2 \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = 2 - 1 \][/tex]
[tex]\[ y = 1 \][/tex]
### Solution
Thus, the solution to the system of equations is:
[tex]\[ x = 1, \quad y = 1 \][/tex]
So the final answer is:
[tex]\[ (x, y) = (1, 1) \][/tex]
