# Solving Two Equations in Two Unknowns

Alignments to Content Standards: A-REI.C.5

Lisa is working with the system of equations $x + 2y = 7$ and $2x - 5y = 5$. She multiplies the first equation by 2 and then subtracts the second equation to find $9y = 9$, telling her that $y = 1$. Lisa then finds that $x = 5$. Thinking about this procedure, Lisa wonders

There are lots of ways I could go about solving this problem. I could add 5 times the first equation and twice the second or I could multiply the first equation by -2 and add the second. I seem to find that there is only one solution to the two equations but I wonder if I will get the same solution if I use a different method?
1. What is the answer to Lisa's question? Explain.
2. Does the answer to (a) change if we have a system of two equations in two unknowns with no solutions? What if there are infinitely many solutions?

## IM Commentary

The goal of this task is to help students see the validity of the elimination method for solving systems of two equations in two unknowns. That is, the new system of equations produced by the method has the same solution(s) as the initial system. This is a subtle and vital point, though students should already be familiar with implementing this procedure before working on this task.

It is not difficult to verify that a solution to the initial system of equations is also a solution to the new system. The key to the success of the elimination method, however, is that all steps in the algorithm are reversible. This is why a solution to the simpler system of equations is also a solution to the original system. This can be seen geometrically when taking a multiple of an equation since, for example, $x + 2y = 7$ and $2x + 4y = 14$ define the same line in the plane. The geometric intuition is lost, however, when equations are added or subtracted as this creates a new line, having the same point of intersection with the line defined by $2x - 5y = 5$.

## Solution

1. Fortunately the answer to Lisa's question is yes, she will get the same answer regardless of which (non-zero) multiples of the equations she takes in order to eliminate a variable. The key fact that makes this work is that the process is reversible so she can also manipulate the new, simpler system ($2x-5y=5$ and $9y=9$) to get back the original system. Concretely, suppose $x=a, y=b$ is a solution to $x+2y=7$ and $2x-5y=5$: for these equations, this means that $a = 5$ and $b = 1$ but this is not essential for the reasoning. So $2a-5b=5$ and $a+2b=7$. Lisa's first step is to multiply $x+2y=7$ by 2 to get $2x+4y=14$. Since $a+2b=7$ we also have $2a+4b=14$. The second step is to subtract $2x-5y=5$ from $2x+4y=14$ giving the equation $9y=9$. Since $2a-b=5$ and $2a+b=14$ the same reasoning applies to $a$ and $b$. We have just shown that if $x=a$ and $y=b$ is a solution to $2x-5y=5$ and $x+2y=7$ then it is also a solution to $9y=9$ and $x+2y=7$.

The same reasoning can be applied going back in the other direction. Briefly, if $x=a$ and $y=b$ is a solution to the pair of equations $9y=9$ and $x+2y=7$, then we can multiply the second equation by 2 and subtract the first equation to give $2x-5y=5$ and $x=a$ and $y=b$ will be a solution to this as well as to $x+2y=7$. This means that the original equations have the same solutions as the simpler, modified version. This argument will apply no matter how Lisa manipulates the equations, as long as the steps are reversible, that is as long as she does not multiply by 0 (undoing this would mean dividing by 0 which is not defined).

2. The method of elimination of variables works for any system of two equations in two unknowns.  The reasoning in part (a) applies to any solution $x=a$ and $y=b$ to a system of two linear equations in two unknowns. If there are infinitely many solutions to one set of equations there will be infinitely many solutions to a new set obtained by adding and multiplying the original two equations (as long as the multiple is not 0). Similarly, if there are no solutions to a set of two linear equations in two unknowns there will also be no solutions to a new set obtained by adding and multiplying these (if there were a solution, it would solve the original two equations when the process is reversed as seen in the solution to part (a)).