Question

Which of the following is not a possible number of solutions to a system of linear equations?
A. no solution
B. 1 solution
C. 2 solutions
D. infinitely-many solutions

Answer 1

Answer:

A system of linear equations cannot have 2 solutions.

Step-by-step explanation:

If we were to be given a system of linear equations, we could graph them to see if they had intersection points or not.

Let's look at three circumstances.

System One - No Solution

$\displaystyle{\left \{ {{-4x +10y=6} \atop {2x-5y=3}} \right.}$

We can solve this system by multiply the second equation by 2.

$2\times(2x-5y=3)\\\\4x - 10y = 6$

Now, our system becomes:

$\displaystyle\left \{ {{-4x+10y=6} \atop {4x-10y=6}} \right.$

Now, we can add our equations together.

$(-4x + 10y = 6)+(4x - 10y = 6)\\\\0 + 0 = 12\\\\0 \neq 12$

We get a false statement. When you get a false statement with systems, this means that there is no solution for the equations.

System Two - One Solution

$\displaystyle{\left \{ {{4x+3y=-2} \atop {8x-2y=12}} \right.}$

We are given two systems that can be solved by elimination.

$-2\times(4x + 3y = -2)\\\\-8x - 6y = 4$

Our new system is:

$\displaystyle\left \{ {{-8x-6y=4} \atop {8x-2y=12}} \right.$

Now, we can add the equations together.

$\displaystyle(-8x-6y)+(8x-2y) = 0 -8y\\\\4 + 12 = 16\\\\-\frac{8y}{8} = -\frac{16}{8}\\\\y = -2$

Now, we substitute this value of y into either equation and solve for x.

$8x-2(-2)=12\\\\8x + 4 = 12\\\\8x = 8\\\\x = 1$

Therefore, there is one solution at (1, -2).

System Three - Infinitely-Many Solutions

$\displaystyle\left \{ {{y=-2x-4} \atop {y+4=-2x}} \right.$

This system of equations needs simplified first before any action can be taken. We need to place the second equation in slope-intercept form.

$y+4 =-2x\\\\y = -2x - 4$

Now, our system becomes:

$\displaystyle\left \{ {{y=-2x-4} \atop {y=-2x-4}} \right.$

We can already see that there is no way to add or subtract without getting 0 = 0. Therefore, for any value of y or x, this system of equations has a solution. This means that the solution has infinitely-many solutions.

Because there is a way to have no solution, 1 solution, and infinitely-many solutions, it has become apparent that there is no way for a linear system of equations to have 2 solutions.

A quadratic system of equations could have 2 solutions. However, because we are dealing with linear equations, we cannot have 2 solutions.