Solve by Substitution y=|x^2-3x+1| , y=x-1

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

Eliminate the equal sides of each equation and combine.

$| x^{2} - 3 x + 1 | = x - 1$

Solve $| x^{2} - 3 x + 1 | = x - 1$ for $x$ .

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Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x^{2} - 3 x + 1 = \pm (x - 1)$

Set up the positive portion of the $\pm$ solution.

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

Solve the first equation for $x$ .

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Move all terms containing $x$ to the left side of the equation.

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Subtract $x$ from both sides of the equation.

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

Subtract $x$ from $- 3 x$ .

$x^{2} - 4 x + 1 = - 1$

Move all terms to the left side of the equation and simplify.

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Add $1$ to both sides of the equation.

$x^{2} - 4 x + 1 + 1 = 0$

Add $1$ and $1$ .

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a} + y = x - 1$

Substitute the values $a = 1$ , $b = - 4$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot 2)}}{2 \cdot 1} + y = x - 1$

Simplify.

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Simplify the numerator.

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Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $2$ .

$x = \frac{4 \pm \sqrt{16 - 8}}{2 \cdot 1}$

Subtract $8$ from $16$ .

$x = \frac{4 \pm \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

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Factor $4$ out of $8$ .

$x = \frac{4 \pm \sqrt{4 (2)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{4 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{4 \pm 2 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{4 \pm 2 \sqrt{2}}{2}$

Simplify $\frac{4 \pm 2 \sqrt{2}}{2}$ .

$x = 2 \pm \sqrt{2}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Simplify the numerator.

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Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $2$ .

$x = \frac{4 \pm \sqrt{16 - 8}}{2 \cdot 1}$

Subtract $8$ from $16$ .

$x = \frac{4 \pm \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

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Factor $4$ out of $8$ .

$x = \frac{4 \pm \sqrt{4 (2)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{4 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{4 \pm 2 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{4 \pm 2 \sqrt{2}}{2}$

Simplify $\frac{4 \pm 2 \sqrt{2}}{2}$ .

$x = 2 \pm \sqrt{2}$

Change the $\pm$ to $+$ .

$x = 2 + \sqrt{2}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Simplify the numerator.

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Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $2$ .

$x = \frac{4 \pm \sqrt{16 - 8}}{2 \cdot 1}$

Subtract $8$ from $16$ .

$x = \frac{4 \pm \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

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Factor $4$ out of $8$ .

$x = \frac{4 \pm \sqrt{4 (2)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{4 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{4 \pm 2 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{4 \pm 2 \sqrt{2}}{2}$

Simplify $\frac{4 \pm 2 \sqrt{2}}{2}$ .

$x = 2 \pm \sqrt{2}$

Change the $\pm$ to $-$ .

$x = 2 - \sqrt{2}$

The final answer is the combination of both solutions.

$x = 2 + \sqrt{2}, 2 - \sqrt{2}$

Set up the negative portion of the $\pm$ solution.

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

Solve the second equation for $x$ .

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Simplify $- (x - 1)$ .

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Rewrite.

$x^{2} - 3 x + 1 = 0 + 0 - (x - 1)$

Simplify by adding zeros.

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

Apply the distributive property.

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

Multiply $- 1$ by $- 1$ .

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

Move all terms containing $x$ to the left side of the equation.

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Add $x$ to both sides of the equation.

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

Add $- 3 x$ and $x$ .

$x^{2} - 2 x + 1 = 1$

Subtract $1$ from both sides of the equation.

$x^{2} - 2 x + 1 - 1 = 0$

Combine the opposite terms in $x^{2} - 2 x + 1 - 1$ .

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Subtract $1$ from $1$ .

$x^{2} - 2 x + 0 = 0$

Add $x^{2} - 2 x$ and $0$ .

$x^{2} - 2 x = 0$

Factor $x$ out of $x^{2} - 2 x$ .

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Factor $x$ out of $x^{2}$ .

$x \cdot x - 2 x = 0$

Factor $x$ out of $- 2 x$ .

$x \cdot x + x \cdot - 2 = 0$

Factor $x$ out of $x \cdot x + x \cdot - 2$ .

$x (x - 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 2 = 0 + y = x - 1$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

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Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $x (x - 2) = 0$ true.

$x = 0, 2$

The solution to the equation includes both the positive and negative portions of the solution.

$x = 2 + \sqrt{2}, 2 - \sqrt{2}, 0, 2$

Exclude the solutions that do not make $| x^{2} - 3 x + 1 | = x - 1$ true.

$x = 2 + \sqrt{2}, 2$

Evaluate $y$ when $x = 2 + \sqrt{2}$ .

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Substitute $2 + \sqrt{2}$ for $x$ .

$y = (2 + \sqrt{2}) - 1$

Substitute $2 + \sqrt{2}$ for $x$ in $y = (2 + \sqrt{2}) - 1$ and solve for $y$ .

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Remove parentheses.

$y = 2 + \sqrt{2} - 1$

Remove parentheses.

$y = (2 + \sqrt{2}) - 1$

Subtract $1$ from $2$ .

$y = 1 + \sqrt{2}$

Evaluate $y$ when $x = 2$ .

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Substitute $2$ for $x$ .

$y = (2) - 1$

Substitute $2$ for $x$ in $y = (2) - 1$ and solve for $y$ .

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Remove parentheses.

$y = 2 - 1$

Remove parentheses.

$y = (2) - 1$

Subtract $1$ from $2$ .

$y = 1$

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(2 + \sqrt{2}, 1 + \sqrt{2})$

$(2, 1)$

The result can be shown in multiple forms.

Point Form:

$(2 + \sqrt{2}, 1 + \sqrt{2}), (2, 1)$

Equation Form:

$x = 2 + \sqrt{2}, y = 1 + \sqrt{2}$

$x = 2, y = 1$

Solve by Substitution y=|x^2-3x+1| , y=x-1

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