Find the approximate solution of this system of equations.
y = |x² - 3x + 1|, y = x - 1

Solution:

We will solve the equation using the absolute values of the equation.

Let y = |x² - 3x + 1| ……………… be equation (1)

y = x - 1 ……………...be equation (2)

Step 1: Put both the equations equal to each other.

|x² - 3x + 1| = x - 1

Since we know that |x| = - x or x

Step 2: Put the absolute equation equal to both positive and negative value.

|x² - 3x + 1| = ± (x - 1)

|x² - 3x + 1| = +1 (x - 1) or |x² - 3x + 1| = -1(x - 1)

Step 3: Solve the equation for both the values.

x² - 3x + 1 = x - 1 or x²- 3x + 1 = - x + 1

x² - 3x + 1 - x + 1 = 0 or x² - 3x + 1 + x - 1 = 0

x² - 4x + 2 = 0 or x² - 2x = 0

Step 4: Use the quadratic formula to find the values of x which satisfy both equations.

x² - 4x + 2 = 0 or x² - 2x = 0

Quadratic formula x = [- b ± √ b² - 4ac ]/ 2a where a is the coefficient of x², b is the coefficient of x and c is the constant term.

x = [ 4 ± √ (-4)² - 4(1)(2) ] / 2(1)or x = [ 2 ± √ (-2)² - 4(1)(0) ] / 2(1)

x = [ 4 ± √16 - 8 ] / 2 or x = [ 2 ± √4 ] / 2

x = [ 4 ± √8 ] / 2 or x = [ 2 ± √4 ] / 2

x = [ 4 ± 2√2 ] / 2 or x = [ 2 ± 2 ] / 2

x = [ 2 ± √2 ]or x = [ 1 ± 1 ]

x = (2 + √2), (2 - √2)or x = (1 + 1), (1 - 1)

x = (2 + √2), (2 - √2)or x = 2, 0

Step 5: Use the substitution method to find the value of y for the equation.

Substitute the values of x = (2 + √2), (2 - √2) or x = 2, 0 in equation (2).

y = x - 1 or y = x - 1

y = (2 + √2) - 1, (2 - √2) - 1or y = 2 - 1, 0 - 1

y = (1 + √2), (1 - √2) or y = 1, - 1

The solution made by equation x - 1 that can make the solution real is (2, 1).

---

Find the approximate solution of this system of equations.
y = |x² - 3x + 1|, y = x - 1

Summary:

The approximate solution of this system of equations y = |x² - 3x + 1| and y = x - 1 that make the solution real (x, y) is (2, 1).