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 \(x\):
\(y=1.5x+10\)
\(19=1.5x+10\)
\(19-10=1.5x\)
\(9=1.5x\)
\(\cfrac{\cancel{1.5}x}{\cancel{1.5}} = \cfrac{9}{1.5}\)
\(x = 6\)
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 |
Show Answer
i. First, we can determine the slope. The value of \(y\) increases by \(3\) each time while the value of \(x\) increase by \(1\). We can plug these values into the slope equation to solve:
\(m = \cfrac{\Delta y}{\Delta x} = \cfrac{3}{1} = 3\)
Therefore, the slope is \(3\).
Next, we can determine the \(y\)-intercept (or the point where \(x=0\)). In this case, it would be \(2 - 3 = -1\), working backwards from \(x=1\).
Then, we can write the equation of the line as such:
\( y = 3x - 1\)
Finally, we can substitute \(40\) for \(x\) and solve for \(y\):
\(y = 3 (40) - 1\)
\(y = 120 - 1\)
\(y = 119\)
Therefore, we can determine that \(y = 119\) when \(x = 40\).
ii. All we need to do is substitute \(40\) for \(y\) and isolate \(x\):
\(y = 3x - 1\)
\(40 = 3x - 1\)
\(3x = 40 + 1\)
\(3x = 41\)
\(\cfrac{\cancel{3}x}{\cancel{3}} = \cfrac{41}{3}\)
\(x = \cfrac{41}{3}\)
Therefore, we can determine that \(x = \cfrac{41}{3}\) when \(y = 40\).
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.
Show Answer
The dependent variable is the internet bill cost, \(C\) which changes based on the amount of data, \(d\), over the limit. The amount of data (GB) over the limit would be the independent variable.
Since the problem states the base payment is \($30/\text{month}\), that would be the starting value or the \(y\)-intercept. This is the price of the bill when Gabriel is \(0 \; [\text{GB}]\) over the limit.
Next, the question states that for every GB over the limit, an additional \($10\) would be charged. That means the bill will increase \($10\) for each GB over the limit. This value represents the slope.
Therefore, we can write the equation as such:
\(C = 10d + 30\)