FunctionsMAT 102

Subsection2.B.1Simultaneous Equations by Graphing

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 \(y = 2\sqrt{x}\) is the set of all points that satisfy that relationship; and the graph of \(y = x + \frac{1}{2}\) 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 \(y = mx + b\) format. The \(y\)-intercept is \(\frac{1}{2}\) and the slope is \(1\text{.}\) We can graph the first equation by plotting points; or, if you happen to know what the graph of \(y = \sqrt{x}\) looks like, you can stretch the graph vertically to get \(y = 2\sqrt{x}\text{,}\) since all the \(y\)-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 \(x\) is barely greater than \(0\) (say, \(x = 0.1\)), and the other occurs at approximately \(x = 3\text{.}\)

There will be no more points of intersection after this, because the line will rise faster than the curve.

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.

Let's consider the two equations from the graph above:

Taking our estimated solution of \(3\text{,}\) if you plug the number \(3\) into both of these functions, will you get the same answer?

Pretty close! Similarly, \(2\sqrt{.1} \approx 0.632\text{,}\) which is quite close to \(.1+\frac{1}{2} = 0.6\text{.}\) But if we want more exact answers, we will need to draw a much more exact graph, which becomes very time-consuming.

For more exact answers, we use analytic methods. Two such methods will be discussed in this chapter: substitution and elimination.

Subsection2.B.2Solving by Substitution

Here is the algorithm for substitution:

1. Solve one of the equations for one variable.
2. Substitute the expression for the variable in the other equation.
3. Solve the second equation, which now has only one variable.
4. Finally, use the equation you found in step (1) to find the other variable.

Why does substitution work? We found in the first step that \(y = 2x - 8\text{.}\) This means that \(y\) and \(2x - 8\) are equal in the sense that we discussed in the first chapter on functions-they will always be the same number, in these equations-they are the same. This gives us permission to simply replace one with the other, which is what we do in the second (“substitution”) step.

Subsection2.B.3Solving by Elimination

Here is the algorithm for elimination:

1. Multiply one equation (or in some cases both) by some number, so that the two equations have the same coefficient for one of the variables.
2. Add or subtract the two equations to make that variable go away.
3. Solve the resulting equation, which now has only one variable.
4. Finally, plug back in to find the other variable.

Why does elimination work? As you know, you are always allowed to do the same thing to both sides of an equation. If an equation is true, it will still be true if you add \(4\) to both sides, multiply both sides by \(6\text{,}\) or take the square root of both sides.

Now-consider, in the second step above, what we did to the equation \(3x + 4y = 1\text{.}\) We added something to both sides of this equation. What did we add? On the left, we added \(8x - 4y\text{;}\) on the right, we added \(32\text{.}\) It seems that we have done something different to the two sides.

However, the second equation gives us a guarantee that these two quantities, \(8x-4y\) and \(32\text{,}\) are in fact the same. So by adding \(8x - 4y\) to the left, and \(32\) to the right, we really have done exactly the same thing to both sides of the equation \(3x + 4y = 1\text{.}\)

Subsection2.B.4Special Cases

What happens in this situation if we had tried to solve by the method of substitution?

All of this is much easier to understand graphically! Remember that one way to solve simultaneous equations is by graphing them and looking for the intersection.

Example2.B.6Solving by Graphing: No Solution

Recall Example 2.B.3and Example 2.B.4. We found no solution to the system

\begin{align*} 2x+3y &= 8\\ 4x+6y &= -3 \end{align*}

In this case, we see the original equations represented as two parallel lines.

There is no point of intersection, so there is no simultaneous solution.\(\checkmark\)

Example2.B.7Solving by Graphing: Infinite Solutions

Recall Example 2.B.5. We found many solutions to the system

\begin{align*} 2x+3y &= 8\\ 4x+6y &= 16 \end{align*}

In this case, we see that the original equations represented the same line, in two different forms.

Any point on the line is a solution to both equations.\(\checkmark\)

GENERAL RULE FOR THE SPECIAL CASES: If you solve an equation and get a mathematical impossibility such as \(0 = 13\text{,}\) there is no solution. If you get a mathematical tautology such as \(0 = 0\text{,}\) there are an infinite number of solutions and you should find an expression for the solutions in terms of \(x\text{.}\) For example, \(\left(x, \frac{2}{3}x+\frac{8}{3} \right)\text{,}\) where the \(y\) coordinate is replaced by \(mx+b\text{,}\) the equation for the line. Remember \(mx+b\) equals \(y\text{.}\) When you express \(y\) in terms of \(x\text{,}\) someone can plug in any value for the \(x\)-coordinate and determine the corresponding \(y\)-coordinate on the line.