Once you have an equation to describe a linear relation, you can use it to solve for other terms. To solve for the \(y\)-value, substitute the value for \(x\) and solve. To solve for the \(x\)-value, substitute the value for \(y\) and isolate to get \(x\).

Consider the following relationship between the number of toppings on a pizza and the cost:

# Toppings | 0 | 1 | 2 | 3 |
---|---|---|---|---|

Cost ($) | 10.00 | 11.50 | 13.00 | 14.50 |

Since I love toppings, I want to know how much a pizza would cost if I had \(15\) toppings. One option is to continue the pattern. In this instance, the pattern starts at \(10\) and increases by \(1.50\) each time.

This could take a long time. Instead, we can write an equation and solve the equation for \(15\) toppings:

\(y=mx+b\)

\(y=1.5x+10\)

Here, \(x\) represents the number of toppings and \(y\) represents the total cost. We want to know the cost when \(x=15\). Substitute this value for \(x\) and solve:

\(y=1.5x+10\)

\(y=1.5(15)+10\)

\(y=22.50+10\)

\(y=32.50\)

Therefore, the cost of a pizza with \(15\) toppings is \($32.50\).

We can also use the equation to solve for the number of toppings given the price. How many toppings are on the pizza if the price of the pizza was \($19\).

We want to know the number of toppings when \(y=19\). Substitute this value for \(y\) and isolate for \(x\):

\(y=1.5x+10\)

\(19=1.5x+10\)

\(19-10=1.5x\)

\(9=1.5x\)

\(9/1.5=x\)

\(\cfrac{\cancel{1.5}x}{\cancel{1.5}} = \cfrac{9}{1.5}\)

\(x = 6\)

Therefore, we can determine that there are \(6\) toppings if the pizza costs \($19.00\).

Given the following table of values:

- Solve for \(y\) when \(x = 40\)
- Solve for \(x\) when \(y = 40\)

x Values | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
---|---|---|---|---|---|---|---|

y Values | 2 | 5 | 8 | 11 | 14 | 17 | 20 |

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Gabriel's internet bill is \($30/\text{month}\) plus \($10/\text{GB}\) for every data over the limit (\(25 \; [\text{GB}]\)). Write a linear equation to represent Gabriel's bill.

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