Solve the System of Equations f(x)=x^2-2x+3 , f(x)=-x+5

$f (x) = x^{2} - 2 x + 3$ , $f (x) = - x + 5$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} - 2 x + 3 = - x + 5$

Step 2

Solve $x^{2} - 2 x + 3 = - x + 5$ for $x$ .

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

Add $x$ to both sides of the equation.

$x^{2} - 2 x + 3 + x = 5$

Add $- 2 x$ and $x$ .

$x^{2} - x + 3 = 5$

Subtract $5$ from both sides of the equation.

$x^{2} - x + 3 - 5 = 0$

Subtract $5$ from $3$ .

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

Factor $x^{2} - x - 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $- 1$ .

$- 2, 1 + f (x) = - x + 5$

Write the factored form using these integers.

$(x - 2) (x + 1) = 0$

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

$x - 2 = 0$

$x + 1 = 0 + f (x) = - x + 5$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(x - 2) (x + 1) = 0$ true.

$x = 2, - 1$

Step 3

Evaluate $f$ when $x = 2$ .

Substitute $2$ for $x$ .

$f (2) = - (2) + 5$

Substitute $2$ for $x$ in $f (2) = - (2) + 5$ and solve for $f$ .

Simplify.

Multiply $- 1$ by $2$ .

$f (2) = - 2 + 5$

Add $- 2$ and $5$ .

$f (2) = 3$

Divide each term in $f (2) = 3$ by $2$ and simplify.

Divide each term in $f (2) = 3$ by $2$ .

$\frac{f (2)}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{f \cdot 2}{2} = \frac{3}{2}$

Divide $f$ by $1$ .

$f = \frac{3}{2}$

Step 4

Evaluate $f$ when $x = - 1$ .

Substitute $- 1$ for $x$ .

$f (- 1) = - (- 1) + 5$

Divide each term in $f (- 1) = - (- 1) + 5$ by $- 1$ and simplify.

Divide each term in $f (- 1) = - (- 1) + 5$ by $- 1$ .

$\frac{f \cdot - 1}{- 1} = \frac{- (- 1)}{- 1} + \frac{5}{- 1}$

Simplify the left side.

Move the negative one from the denominator of $\frac{f \cdot - 1}{- 1}$ .

$- 1 \cdot (f \cdot - 1) = \frac{- (- 1)}{- 1} + \frac{5}{- 1}$

Rewrite $- 1 \cdot (f \cdot - 1)$ as $- (f \cdot - 1)$ .

$- (f \cdot - 1) = \frac{- (- 1)}{- 1} + \frac{5}{- 1}$

Move $- 1$ to the left of $f$ .

$- (- 1 \cdot f) = \frac{- (- 1)}{- 1} + \frac{5}{- 1}$

Rewrite $- 1 f$ as $- f$ .

$- - f = \frac{- (- 1)}{- 1} + \frac{5}{- 1}$

Multiply $- - f$ .

Multiply $- 1$ by $- 1$ .

$1 f = \frac{- (- 1)}{- 1} + \frac{5}{- 1}$

Multiply $f$ by $1$ .

$f = \frac{- (- 1)}{- 1} + \frac{5}{- 1}$

Simplify the right side.

Combine the numerators over the common denominator.

$f = \frac{- (- 1) + 5}{- 1}$

Simplify the expression.

Multiply $- 1$ by $- 1$ .

$f = \frac{1 + 5}{- 1}$

Add $1$ and $5$ .

$f = \frac{6}{- 1}$

Divide $6$ by $- 1$ .

$f = - 6$

Step 5

The solution of the system of equations is all the values that make the system true.

$f = \frac{3}{2}, x = 2 f = - 6, x = - 1$

Step 6

List all of the solutions.

$f = \frac{3}{2}, x = 2$

$f = - 6, x = - 1$

Solve the System of Equations f(x)=x^2-2x+3 , f(x)=-x+5

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