To find the solution to the system of equations given by [tex]\( y = |x - 1| \)[/tex] and [tex]\( y = 3x + 2 \)[/tex], we need to analyze each equation and see where they intersect.
### Step 1: Analyze the Equations
Equation 1: [tex]\( y = |x - 1| \)[/tex]
This represents a V-shaped graph with the vertex at [tex]\( (1, 0) \)[/tex]. It breaks into two linear parts:
1. For [tex]\( x \ge 1 \)[/tex]: [tex]\( y = x - 1 \)[/tex]
2. For [tex]\( x < 1 \)[/tex]: [tex]\( y = -(x - 1) = -x + 1 \)[/tex]
Equation 2: [tex]\( y = 3x + 2 \)[/tex]
This represents a straight line with a slope of 3 and a y-intercept of 2.
### Step 2: Solve for Intersections
Case 1: [tex]\( x \ge 1 \)[/tex]
For [tex]\( x \ge 1 \)[/tex], [tex]\( y = x - 1 \)[/tex]:
Set [tex]\( x - 1 = 3x + 2 \)[/tex]:
[tex]\[
x - 1 = 3x + 2
\][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[
x - 3x = 2 + 1 \\
-2x = 3 \\
x = -\frac{3}{2}
\][/tex]
Since [tex]\( x \ge 1 \)[/tex], [tex]\( x = -\frac{3}{2} \)[/tex] is not valid in this interval.
Case 2: [tex]\( x < 1 \)[/tex]
For [tex]\( x < 1 \)[/tex], [tex]\( y = -x + 1 \)[/tex]:
Set [tex]\( -x + 1 = 3x + 2 \)[/tex]:
[tex]\[
-x + 1 = 3x + 2
\][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[
1 - 2 = 3x + x \\
-1 = 4x \\
x = -\frac{1}{4}
\][/tex]
Since [tex]\( x < 1 \)[/tex], [tex]\( x = -\frac{1}{4} \)[/tex] is valid here.
### Step 3: Find the corresponding [tex]\( y \)[/tex]-value
Substituting [tex]\( x = -\frac{1}{4} \)[/tex] into [tex]\( y = 3x + 2 \)[/tex]:
[tex]\[
y = 3\left(-\frac{1}{4}\right) + 2 \\
y = -\frac{3}{4} + 2 \\
y = \frac{5}{4}
\][/tex]
### Conclusion
The approximate solution to the system of equations [tex]\( y = |x - 1| \)[/tex] and [tex]\( y = 3x + 2 \)[/tex] is: [tex]\( \left( -\frac{1}{4}, \frac{5}{4} \right) \)[/tex].
So, you should select:
[tex]\[
\boxed{\left( -\frac{1}{4}, \frac{5}{4} \right)}
\][/tex]
