Algebra Examples

y = 2 f (x)

$y = 2 f (x)$

Step 1

Multiply

2 f

$2 f$ by

x

$x$ .

y = 2 f x

$y = 2 f x$

Step 2

Choose any values for

x

$x$ and

y

$y$ that are in the domain to plug into the equation.

Step 3

Choose

0

$0$ to substitute in for

x

$x$ and

1

$1$ to substitute for

z

$z$ .

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Step 3.1

Multiply

2 f

$2 f$ by

0

$0$ .

y = 2 f \cdot 0

$y = 2 f \cdot 0$

Step 3.2

Remove parentheses.

y = 2 f (0)

$y = 2 f (0)$

Step 3.3

Multiply

2 f (0)

$2 f (0)$ .

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Step 3.3.1

Multiply

0

$0$ by

2

$2$ .

y = 0 f

$y = 0 f$

Step 3.3.2

Multiply

0

$0$ by

f

$f$ .

y = 0

$y = 0$

y = 0

$y = 0$

Step 3.4

Use the

x

$x$ ,

y

$y$ , and

z

$z$ values to form the ordered pair.

(0, 0, 1)

$(0, 0, 1)$

(0, 0, 1)

$(0, 0, 1)$

Step 4

Choose

1

$1$ to substitute in for

x

$x$ and

2

$2$ to substitute for

z

$z$ .

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Step 4.1

Multiply

2 f

$2 f$ by

1

$1$ .

y = 2 f \cdot 1

$y = 2 f \cdot 1$

Step 4.2

Remove parentheses.

y = 2 f (1)

$y = 2 f (1)$

Step 4.3

Multiply

2

$2$ by

1

$1$ .

y = 2 f

$y = 2 f$

Step 4.4

Use the

x

$x$ ,

y

$y$ , and

z

$z$ values to form the ordered pair.

(1, 2 f, 2)

$(1, 2 f, 2)$

(1, 2 f, 2)

$(1, 2 f, 2)$

Step 5

Choose

2

$2$ to substitute in for

x

$x$ and

3

$3$ to substitute for

z

$z$ .

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Step 5.1

Multiply

2 f

$2 f$ by

2

$2$ .

y = 2 f \cdot 2

$y = 2 f \cdot 2$

Step 5.2

Remove parentheses.

y = 2 f (2)

$y = 2 f (2)$

Step 5.3

Multiply

2

$2$ by

2

$2$ .

y = 4 f

$y = 4 f$

Step 5.4

Use the

x

$x$ ,

y

$y$ , and

z

$z$ values to form the ordered pair.

(2, 4 f, 3)

$(2, 4 f, 3)$

(2, 4 f, 3)

$(2, 4 f, 3)$

Step 6

These are three possible solutions to the equation.

(0, 0, 1), (1, 2 f, 2), (2, 4 f, 3)

$(0, 0, 1), (1, 2 f, 2), (2, 4 f, 3)$

⎡ ⎢ ⎣ \begin{matrix} x^{2} & \frac{1}{2} \sqrt{π} & \int x d x \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$