Solving Linear Relations

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. 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 cost. We want to know the cost when \(x=15\). Plug in 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 \(y = $32.50\).

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 \(y=19\). Plug in and isolate for x:

\(y=1.5x+10\)

\(19=1.5x+10\)

\(19-10=1.5x\)

\(9=1.5x\)

\(9/1.5=x\)

\(6=x\)

Therefore, there were \(6\) toppings if the pizza cost \(y = $19.00\).


Given the table of values for x and y, solve for when x = 40.
X 1 2 3 4 5 6 7
Y 2 5 8 11 14 17 20


Using the same table as the question above, what would be the x value when y = 40?

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