Let's solve the system of linear equations:
[tex]\[ \left\{\begin{array}{l}
x + 2y = 4 \\
2x - y = \frac{1}{2}
\end{array}\right. \][/tex]
### Step-by-Step Solution:
#### Step 1: Solve the first equation for [tex]\( x \)[/tex]:
[tex]\[ x + 2y = 4 \][/tex]
[tex]\[ x = 4 - 2y \][/tex]
#### Step 2: Substitute the expression for [tex]\( x \)[/tex] into the second equation:
[tex]\[ 2(4 - 2y) - y = \frac{1}{2} \][/tex]
#### Step 3: Expand and simplify the equation:
[tex]\[ 8 - 4y - y = \frac{1}{2} \][/tex]
[tex]\[ 8 - 5y = \frac{1}{2} \][/tex]
#### Step 4: Isolate [tex]\( y \)[/tex] by moving the constant term to the other side:
[tex]\[ 8 - \frac{1}{2} = 5y \][/tex]
[tex]\[ \frac{16}{2} - \frac{1}{2} = 5y \][/tex]
[tex]\[ \frac{15}{2} = 5y \][/tex]
#### Step 5: Solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{15}{2} \times \frac{1}{5} \][/tex]
[tex]\[ y = \frac{3}{2} \][/tex]
[tex]\[ y = 1.5 \][/tex]
#### Step 6: Substitute [tex]\( y = 1.5 \)[/tex] back into the expression for [tex]\( x \)[/tex]:
[tex]\[ x = 4 - 2(1.5) \][/tex]
[tex]\[ x = 4 - 3 \][/tex]
[tex]\[ x = 1 \][/tex]
Thus, the solution to the system of equations is:
[tex]\[ x = 1 \][/tex]
[tex]\[ y = 1.5 \][/tex]
### Step 7: Verify the solution by substituting [tex]\( x \)[/tex] and [tex]\( y \)[/tex] into the original equations:
For the first equation [tex]\( x + 2y = 4 \)[/tex]:
[tex]\[ 1 + 2(1.5) = 4 \][/tex]
[tex]\[ 1 + 3 = 4 \][/tex]
[tex]\[ 4 = 4 \][/tex] (Correct!)
For the second equation [tex]\( 2x - y = \frac{1}{2} \)[/tex]:
[tex]\[ 2(1) - 1.5 = \frac{1}{2} \][/tex]
[tex]\[ 2 - 1.5 = \frac{1}{2} \][/tex]
[tex]\[ 0.5 = 0.5 \][/tex] (Correct!)
Both equations are satisfied with [tex]\( x = 1 \)[/tex] and [tex]\( y = 1.5 \)[/tex]. Therefore, the solution to the system is:
[tex]\[ (x, y) = (1.0, 1.5) \][/tex]
This set of values corresponds to the point (1.0, 1.5) on the graph where the two lines intersect.
