In the second, third, and fourth quadrants the values of x or y are negative.
A
y=0.7x
B
Graph:
Explanation: See solution.
Practice makes perfect
We are asked to write an equation that will relate the price of the ticket x to the price of the discounted ticket y. We are told that the discounted price of the ticket y is 70 % of the real price of the ticket x.
yis70 %ofx ⇔ y=70 % * x
We can write 70 % as 0.7.
y=70 % * x ⇔ y=0.7x
We wrote an equation that relates x and y.
We will graph the equation that we found in Part A.
y=0.7x
To do that, we can plot the point (0,0) and use the equation to find another point.
To find another point of the graph, (x,y), we can first take any value of x. For example, let's take x=8. Then, to find the corresponding y-value we can use the equation y=0.7x.
We found another point of the graph, (8,5.6). Let's add it to the graph.
Now, we can sketch a line that crosses the points (0,0) and (8,5.6).
We graphed the equation and the whole line is in the first quadrant. This is because in the second, third, and fourth quadrants the values of x or y are negative.
Since both x and y represent prices and negative prices do not make sense to us, we do not want to consider negative values of x or y. This is why the line should be only in the first quadrant where both x and y are positive.
