# Find the approximate solution of this system of equations.

y = |x^{2} - 3x + 1|, y = x - 1

**Solution:**

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

Let y = |x^{2} - 3x + 1| ……………… be equation (1)

y = x - 1 ……………...be equation (2)

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

|x^{2} - 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.

|x^{2} - 3x + 1| = ± (x - 1)

|x^{2} - 3x + 1| = +1 (x - 1) or |x^{2} - 3x + 1| = -1(x - 1)

**Step 3: **Solve the equation for both the values.

x^{2} - 3x + 1 = x - 1 or x^{2}- 3x + 1 = - x + 1

x^{2} - 3x + 1 - x + 1 = 0 or x^{2} - 3x + 1 + x - 1 = 0

x^{2} - 4x + 2 = 0 or x^{2} - 2x = 0

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

x^{2} - 4x + 2 = 0 or x^{2} - 2x = 0

Quadratic formula x = [- b ± √ b^{2} - 4ac ]/ 2a where a is the coefficient of x^{2}, 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 = |x^{2} - 3x + 1|, y = x - 1

**Summary:**

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

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