To find the solution to the system of equations [tex]\( y = |x - 1| \)[/tex] and [tex]\( y = 3x + 2 \)[/tex], we need to solve these equations algebraically. We will consider the two cases for [tex]\( |x - 1| \)[/tex], since the absolute value function depends on whether the expression inside is positive or negative.
Case 1: [tex]\( x - 1 \geq 0 \)[/tex] (which means [tex]\( x \geq 1 \)[/tex])
In this case, [tex]\( |x - 1| = x - 1 \)[/tex]. So our equations become:
[tex]\[ y = x - 1 \][/tex]
[tex]\[ y = 3x + 2 \][/tex]
Setting these two expressions for [tex]\( y \)[/tex] equal to each other:
[tex]\[ x - 1 = 3x + 2 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x - 3x = 2 + 1 \][/tex]
[tex]\[ -2x = 3 \][/tex]
[tex]\[ x = -\frac{3}{2} \][/tex]
Since [tex]\(-\frac{3}{2} < 1 \)[/tex], this solution does not satisfy the condition [tex]\( x \geq 1 \)[/tex], so we discard this solution for Case 1.
Case 2: [tex]\( x - 1 < 0 \)[/tex] (which means [tex]\( x < 1 \)[/tex])
In this case, [tex]\( |x - 1| = -(x - 1) = -x + 1 \)[/tex]. So our equations become:
[tex]\[ y = -x + 1 \][/tex]
[tex]\[ y = 3x + 2 \][/tex]
Setting these two expressions for [tex]\( y \)[/tex] equal to each other:
[tex]\[ -x + 1 = 3x + 2 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 1 - 2 = 3x + x \][/tex]
[tex]\[ -1 = 4x \][/tex]
[tex]\[ x = -\frac{1}{4} \][/tex]
Since [tex]\(-\frac{1}{4} < 1 \)[/tex], this solution satisfies the condition [tex]\( x < 1 \)[/tex]. Now, we find the corresponding [tex]\( y \)[/tex] value:
[tex]\[ y = 3x + 2 \][/tex]
[tex]\[ y = 3(-\frac{1}{4}) + 2 \][/tex]
[tex]\[ y = -\frac{3}{4} + 2 \][/tex]
[tex]\[ y = \frac{8}{4} - \frac{3}{4} \][/tex]
[tex]\[ y = \frac{5}{4} \][/tex]
Thus, the solution to the system of equations is:
[tex]\[ x = -\frac{1}{4}, \quad y = \frac{5}{4} \][/tex]
The approximate solution of this system of equations is [tex]\( \left( -\frac{1}{4}, \frac{5}{4} \right) \)[/tex].
