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Consider the equations [tex]\( y = |x-1| \)[/tex] and [tex]\( y = 3x + 2 \)[/tex].

The approximate solution of this system of equations is [tex]\(\square\)[/tex].


Sagot :

To find the solution to the system of equations [tex]\( y = |x - 1| \)[/tex] and [tex]\( y = 3x + 2 \)[/tex], we need to determine the points where the two graphs intersect.

### Step-by-Step Solution:

1. Analyze the Piecewise Function [tex]\( y = |x - 1| \)[/tex]:
The absolute value function [tex]\( |x - 1| \)[/tex] can be written as a piecewise function:
[tex]\[ y = \begin{cases} x - 1 & \text{if } x \geq 1 \\ 1 - x & \text{if } x < 1 \end{cases} \][/tex]

2. Solve [tex]\( 3x + 2 = x - 1 \)[/tex] for [tex]\( x \geq 1 \)[/tex]:
[tex]\[ 3x + 2 = x - 1 \][/tex]
Subtract [tex]\( x \)[/tex] from both sides:
[tex]\[ 2x + 2 = -1 \][/tex]
Subtract 2 from both sides:
[tex]\[ 2x = -3 \][/tex]
Divide both sides by 2:
[tex]\[ x = -\frac{3}{2} \][/tex]
Since [tex]\( x = -\frac{3}{2} \)[/tex] does not satisfy [tex]\( x \geq 1 \)[/tex], there is no valid solution in this interval.

3. Solve [tex]\( 3x + 2 = 1 - x \)[/tex] for [tex]\( x < 1 \)[/tex]:
[tex]\[ 3x + 2 = 1 - x \][/tex]
Add [tex]\( x \)[/tex] to both sides:
[tex]\[ 4x + 2 = 1 \][/tex]
Subtract 2 from both sides:
[tex]\[ 4x = -1 \][/tex]
Divide both sides by 4:
[tex]\[ x = -\frac{1}{4} \][/tex]
Since [tex]\( x = -\frac{1}{4} \)[/tex] satisfies [tex]\( x < 1 \)[/tex], it is a valid solution.

4. Calculate the [tex]\( y \)[/tex]-value for [tex]\( x = -\frac{1}{4} \)[/tex]:
Substitute [tex]\( x = -\frac{1}{4} \)[/tex] into either original equation [tex]\( y = 3x + 2 \)[/tex]:
[tex]\[ y = 3\left(-\frac{1}{4}\right) + 2 = -\frac{3}{4} + 2 = \frac{8}{4} - \frac{3}{4} = \frac{5}{4} \][/tex]

Therefore, the approximate solution to the system of equations is:
[tex]\[ x \approx -\frac{1}{4}, \, y \approx \frac{5}{4} \][/tex]

So, in the drop-down menu, the correct answer would be:
[tex]\[ \left( -\frac{1}{4}, \frac{5}{4} \right) \][/tex]
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