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Intersections of the Graphs of Two Equations as Solutions

Instructor: Laura Pennington

Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.

This lesson will do a quick review of equations and their graphs and solutions. We will then look at intersections of graphs of two equations and see how these intersections actually represent solutions to the equations.

Graphs of Equations

Suppose you just found a deal while grocery shopping where you get to take $3 off the total cost of trail mix, where each pound costs $7. In other words, the total cost can be represented by the following formula:

  • f(x) = 7x - 3, where x is the number of pounds of trail mix you buy

This formula is called an equation in mathematics, and it is a statement saying that two algebraic expressions are equal. A solution to an equation is a point (a, b), such that if we plug a in for x and b in for y (or f(x)) in the equation, we get a true statement.

For instance, the point (2, 11) is a solution to our cost equation, because if we plug x = 2 and y = 11 into the equation, we get a true statement:

  • 11 = 7(2) - 3 = 14 - 3 = 11

This solution tells us that if we buy two pounds of trail mix, our total cost will be $11.

Because solutions of equations are points, it makes sense that we can represent our equation graphically. Basically, the graph of an equation is the set of all the points that make the equation true, so it is the set of all the solutions to the equation.

Consider the graph of our cost equation.


intsol1


The graph is a line and every point on that line is a solution to the equation f(x) = 7x - 3. We see that our point (2,11) is on there, which we already knew was a solution. We also see that the point (4,25) is on there as well, so this should be a solution to the equation.

  • 25 = 7(4) - 3 = 28 - 3 = 25

Sure enough! If we plug x = 4 and y = 25 into our equation, we get a true statement.

Intersections of Graphs of Two Equations as Solutions

Now suppose you find another deal on dried apples, such that you get $1 off the total cost of dried apples, where one pound costs $6. In this case, we can represent the total cost of dried apples with the following equation:

  • g(x) = 6x - 1

You decide that you want to get trail mix and dried apples, but you want to spend the same amount of money on each of them, so you want to find the number of pounds of each that would result in the same cost. In other words, we want to find the value of x that would make f(x) = g(x).

Your first thought may be to solve this algebraically by setting f(x) = g(x) and solving for x.

7x - 3 = 6x - 1 Add 3 to both sides
7x = 6x + 2 Subtract 6x from both sides
x = 2 This is our solution

This certainly works, but did you know that we could also solve this problem graphically? You see, since we are trying to find a value of x that makes f(x) = g(x), we are looking for the x-value of a point, (a,b), where (a,b) is a solution to both f(x) and g(x).

As we said, if a point is on the graph of an equation, it is a solution to that equation, so if we can find a point that is on both f(x) and g(x), then that point would be a solution to both equations, and that would happen where the graphs of the two equations intersect.

Ah-ha! This tells us that the intersection point of two graphs, f(x) and g(x), is equal to a solution to both graph's equations, so the x-value of this intersection point would be a solution to f(x) = g(x). Let's look at the intersection point of f(x) and g(x) in our example.


intsol2


We see that the intersection point is (2,11), so when x = 2, the two functions have the same value of 11. Therefore, if we buy two pounds of trail mix and 2 pounds of dried apples, we will pay the same amount of $11 for each. This is exactly what we got when we solved the problem algebraically.

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