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# Simultaneous Equations by Graphing

Module by: Kenny M. Felder. E-mail the author

Summary: This module introduces how to graph simultaneous equations.

Consider the equation y=2xy=2x size 12{y=2 sqrt {x} } {}. How many (x,y)(x,y) size 12{ $$x,y$$ } {} pairs are there that satisfy this equation? Answer: (0,0)(0,0) size 12{ $$0,0$$ } {}, (1,2)(1,2) size 12{ $$1,2$$ } {}, (4,4)(4,4) size 12{ $$4,4$$ } {}, and (9,6)(9,6) size 12{ $$9,6$$ } {} are all solutions; and there is an infinite number of other solutions. (And don’t forget non-integer solutions, such as (,1)(,1) size 12{ $${ size 8{1} } wideslash { size 8{4} } ,1$$ } {}!)

Now, consider the equation y=x+12y=x+12 size 12{y=x+ { { size 8{1} } over { size 8{2} } } } {}. How many pairs satisfy this equation? Once again, an infinite number. Most equations that relate two variables have an infinite number of solutions.

To consider these two equations “simultaneously” is to ask the question: what (x,y)(x,y) size 12{ $$x,y$$ } {} pairs make both equations true? To express the same question in terms of functions: what values can you hand the functions 2x2x size 12{2 sqrt {x} } {} and x+12x+12 size 12{x+ { { size 8{1} } over { size 8{2} } } } {} that will make these two functions produce the same answer?

At first glance, it is not obvious how to approach such a question-- it is not even obvious how many answers there will be.

One way to answer such a question is by graphing. Remember, the graph of y=2xy=2x size 12{y=2 sqrt {x} } {} is the set of all points that satisfy that relationship; and the graph of y=x+12y=x+12 size 12{y=x+ { { size 8{1} } over { size 8{2} } } } {} is the set of all points that satisfy that relationship. So the intersection(s) of these two graphs is the set of all points that satisfy both relationships.

How can we graph these two? The second one is easy: it is a line, already in y=mx+by=mx+b size 12{y= ital "mx"+b} {} format. The yy size 12{y} {}-intercept is 1212 and the slope is 1. We can graph the first equation by plotting points; or, if you happen to know what the graph of y=xy=x size 12{y= sqrt {x} } {} looks like, you can stretch the graph vertically to get y=2xy=2x size 12{y=2 sqrt {x} } {}, since all the yy size 12{y} {}-values will double. Either way, you wind up with something like this:

We can see that there are two points of intersection. One occurs when xx size 12{x} {} is barely greater than 0 (say, x=0.1x=0.1 size 12{x=0 "." 1} {}), and the other occurs at approximately x=3x=3 size 12{x=3} {}. There will be no more points of intersection after this, because the line will rise faster than the curve.

## Exercise 1

y = 2 x y = 2 x size 12{y=2 sqrt {x} } {}

y = x + 1 2 y = x + 1 2 size 12{y=x+ { { size 8{1} } over { size 8{2} } } } {}

### Solution

From graphing...

x=0.1x=0.1 size 12{x=0 "." 1} {}, x=3x=3 size 12{x=3} {}

Graphing has three distinct advantages as a method for solving simultaneous equations.

1. It works on any type of equations.
2. It tells you how many solutions there are, as well as what the solutions are.
3. It can help give you an intuitive feel for why the solutions came out the way they did.

However, graphing also has two disadvantages.

1. It is time-consuming.
2. It often yields solutions that are approximate, not exact—because you find the solutions by simply “eyeballing” the graph to see where the two curves meet.

For instance, if you plug the number 3 into both of these functions, will you get the same answer?

3 2 x 2 3 3 . 46 3 2 x 2 3 3 . 46 size 12{3 rightarrow 2 sqrt {x} rightarrow 2 sqrt {3} approx 3 "." "46"} {}

3 x + 1 2 3 . 5 3 x + 1 2 3 . 5 size 12{3 rightarrow x+ { { size 8{1} } over { size 8{2} } } rightarrow 3 "." 5} {}

Pretty close! Similarly, 2.10.6322.10.632 size 12{2 sqrt { "." 1} approx 0 "." "632"} {}, which is quite close to 0.6. But if we want more exact answers, we will need to draw a much more exact graph, which becomes very time-consuming. (Rounded to three decimal places, the actual answers are 0.086 and 2.914.)

For more exact answers, we use analytic methods. Two such methods will be discussed in this chapter: substitution and elimination. A third method will be discussed in the section on Matrices.

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