Algebra Examples

$f (x) = x^{2} - 2 x + 3$ , $f (x) = - 6 x$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} - 2 x + 3 = - 6 x$

Step 2

Solve $x^{2} - 2 x + 3 = - 6 x$ for $x$ .

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Step 2.1

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

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Step 2.1.1

Add $6 x$ to both sides of the equation.

$x^{2} - 2 x + 3 + 6 x = 0$

Step 2.1.2

Add $- 2 x$ and $6 x$ .

$x^{2} + 4 x + 3 = 0$

Step 2.2

Factor $x^{2} + 4 x + 3$ using the AC method.

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Step 2.2.1

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 $3$ and whose sum is $4$ .

$1, 3 + f (x) = - 6 x$

Step 2.2.2

Write the factored form using these integers.

$(x + 1) (x + 3) = 0$

Step 2.3

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

$x + 1 = 0$

$x + 3 = 0 + f (x) = - 6 x$

Step 2.4

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

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Step 2.4.1

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

$x + 1 = 0$

Step 2.4.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.5

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 2.5.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.6

The final solution is all the values that make $(x + 1) (x + 3) = 0$ true.

$x = - 1, - 3$

Step 3

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

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Step 3.1

Substitute $- 1$ for $x$ .

$f (- 1) = - 6 \cdot - 1$

Step 3.2

Divide each term in $f (- 1) = - 6 \cdot - 1$ by $- 1$ and simplify.

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Step 3.2.1

Divide each term in $f (- 1) = - 6 \cdot - 1$ by $- 1$ .

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

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

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

$- 1 \cdot (f \cdot - 1) = \frac{- 6 \cdot - 1}{- 1}$

Step 3.2.2.2

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

$- (f \cdot - 1) = \frac{- 6 \cdot - 1}{- 1}$

Step 3.2.2.3

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

$- (- 1 \cdot f) = \frac{- 6 \cdot - 1}{- 1}$

Step 3.2.2.4

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

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

Step 3.2.2.5

Multiply $- - f$ .

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Step 3.2.2.5.1

Multiply $- 1$ by $- 1$ .

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

Step 3.2.2.5.2

Multiply $f$ by $1$ .

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

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Move the negative one from the denominator of $\frac{- 6 \cdot - 1}{- 1}$ .

$f = - 1 \cdot (- 6 \cdot - 1)$

Step 3.2.3.2

Rewrite $- 1 \cdot (- 6 \cdot - 1)$ as $- (- 6 \cdot - 1)$ .

$f = - (- 6 \cdot - 1)$

Step 3.2.3.3

Multiply $- (- 6 \cdot - 1)$ .

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Step 3.2.3.3.1

Multiply $- 6$ by $- 1$ .

$f = - 1 \cdot 6$

Step 3.2.3.3.2

Multiply $- 1$ by $6$ .

$f = - 6$

Step 4

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

$f = - 6, x = - 1 f = - 6, x = - 3$

Step 5

List all of the solutions.

$f = - 6, x = - 1$

$f = - 6, x = - 3$