Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis; Jay Abramson; Sharon North; Amy Baldwin; Alyssa Howell

2.7.3 Sorting Systems of Equations Based on Number of Solutions

2.7.3 • Sorting Systems of Equations Based on Number of Solutions

Activity

Your teacher will give you a set of cards. Each card contains a system of equations.

Sort the systems into three groups based on the number of solutions each system has. Be prepared to explain how you know where each system belongs.

Compare your answer:

No solution: Cards 2 and 5
One solution: Cards 1, 4, 6, and 9
Infinitely many solutions: Cards 3, 7, and 8

Are you ready for more?

Extending Your Thinking

1.

In the cards, for each system with no solution, change a single constant term so that there are infinitely many solutions to the system.

Compare your answer:

Card 2: change 3 to 13; Card 5: change $- 12$ to 12.

2.

For each system with infinitely many solutions, change a single constant term so that there are no solutions to the system.

Compare your answer:

Any change to a constant term would work.

3.

Explain why in these situations it is impossible to change a single constant term so that there is exactly one solution to the system.

Compare your answer:

Changing a constant term won’t change the slope of the lines, so the graphs will be either distinct parallel lines or the same line.

Self Check

Which of the following is true of the following system of equations?

$\begin{array}{rcl}&&\left\{\begin{array}{l}7x\;+\;6y\;=\;2\\–28x\;–\;24y\;=\;–8\end{array}\right.\end{array}$

If you write both equations in slope-intercept form, they have the same slope.
equations in slope-intercept form, they have different $y$ -intercepts.
The system of equations has exactly one solution.
The system of equations has no solutions.

Additional Resources

Classifying and Rewriting Systems of Equations

Find a solution to the system of equations using the addition method.

$x + 3 y = 2$

$3 x + 9 y = 6$

Solution

With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating $x$ . If we multiply both sides of the first equation by $- 3$ , then we will be able to eliminate the $x$ -variable.

Now add the equations:

$\begin{matrix} x + 3 y & = & 2 \\ (- 3) (x + 3 y) & = & (- 3) (2) \\ - 3 x - 9 y & = & - 6 \end{matrix}$

$\begin{matrix} - 3 x - 9 y & = & - 6 \\ + 3 x + 9 y & = & 6 \\ 0 & = & 0 \end{matrix}$

We can see that there will be an infinite number of situations that satisfy both equations.

Try it

Classifying and Rewriting Systems of Equations

Change a single constant term in the system from above so that there are no solutions.

$\begin{matrix} x + 3 y & = & 2 \\ 3 x + 9 y & = & 6 \end{matrix}$

Show that you were successful in changing the system to have no solution.

Compare your answer: Here is how to rewrite the system so there is no solution.

We can simply change the 2 in the first equation to any other number, such as 5.

$\begin{matrix} x + 3 y & = & 5 \\ 3 x + 9 y & = & 6 \end{matrix}$

Let’s try to prove there is no solution. Using substitution:

$\begin{matrix} x + 3 y & = & 5 \\ x & = & 5 - 3 y \end{matrix}$

Substitute this expression into the second equation.

$\begin{matrix} 3 x + 9 y & = & 6 \\ 3 (5 - 3 y) + 9 y & = & 6 \\ 15 - 9 y + 9 y & = & 6 \\ 15 & \neq & 6 \end{matrix}$

Since the equation is not true, the system has no solution.