Find Three Ordered Pair Solutions y=8/3x+4

$y = \frac{8}{3} x + 4$

Step 1

Combine $\frac{8}{3}$ and $x$ .

$y = \frac{8 x}{3} + 4$

Step 2

Choose any value for $x$ that is in the domain to plug into the equation.

Step 3

Choose $0$ to substitute in for $x$ to find the ordered pair.

Remove parentheses.

$y = \frac{8 (0)}{3} + 4$

Simplify $\frac{8 (0)}{3} + 4$ .

Multiply $8$ by $0$ .

$y = \frac{0}{3} + 4$

Divide $0$ by $3$ .

$y = 0 + 4$

Add $0$ and $4$ .

$y = 4$

Use the $x$ and $y$ values to form the ordered pair.

$(0, 4)$

Step 4

Choose $1$ to substitute in for $x$ to find the ordered pair.

Remove parentheses.

$y = \frac{8 (1)}{3} + 4$

Simplify $\frac{8 (1)}{3} + 4$ .

Multiply $8$ by $1$ .

$y = \frac{8}{3} + 4$

To write $4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = \frac{8}{3} + 4 \cdot \frac{3}{3}$

Combine $4$ and $\frac{3}{3}$ .

$y = \frac{8}{3} + \frac{4 \cdot 3}{3}$

Combine the numerators over the common denominator.

$y = \frac{8 + 4 \cdot 3}{3}$

Simplify the numerator.

Multiply $4$ by $3$ .

$y = \frac{8 + 12}{3}$

Add $8$ and $12$ .

$y = \frac{20}{3}$

Use the $x$ and $y$ values to form the ordered pair.

$(1, \frac{20}{3})$

Step 5

Choose $2$ to substitute in for $x$ to find the ordered pair.

Remove parentheses.

$y = \frac{8 (2)}{3} + 4$

Simplify $\frac{8 (2)}{3} + 4$ .

Multiply $8$ by $2$ .

$y = \frac{16}{3} + 4$

To write $4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = \frac{16}{3} + 4 \cdot \frac{3}{3}$

Combine $4$ and $\frac{3}{3}$ .

$y = \frac{16}{3} + \frac{4 \cdot 3}{3}$

Combine the numerators over the common denominator.

$y = \frac{16 + 4 \cdot 3}{3}$

Simplify the numerator.

Multiply $4$ by $3$ .

$y = \frac{16 + 12}{3}$

Add $16$ and $12$ .

$y = \frac{28}{3}$

Use the $x$ and $y$ values to form the ordered pair.

$(2, \frac{28}{3})$

Step 6

These are three possible solutions to the equation.

$(0, 4), (1, \frac{20}{3}), (2, \frac{28}{3})$

Step 7

Find Three Ordered Pair Solutions y=8/3x+4

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