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Video: Determining the Number of Solutions of a System of Three Equations

Bethani Gasparine

Determine the number of solutions to the system of equations −6𝑥 − 2𝑦 + 9𝑧 = −4, −24𝑥 − 8𝑦 + 36𝑧 = −22, −12𝑥 − 4𝑦 + 18𝑧 = −10.

03:54

Video Transcript

Determine the number of solutions to the system of equations:

negative six 𝑥 minus two 𝑦 plus nine 𝑧 equals negative four, negative 24𝑥 minus eight 𝑦 plus 36𝑧 equals negative 22, and negative 12𝑥 minus four 𝑦 plus 18𝑧 equals negative 10.

In order to determine the number of solutions to the system of equations, we need to pick two equations and eliminate either 𝑥, 𝑦, or 𝑧. That will be the first step. So deciding on which variable to eliminate 𝑥, 𝑦, or 𝑧. When you look at their coefficients, 𝑥 has coefficients of negative six, negative 24, and negative 12. These are all divisible by six. 𝑦 has coefficients of negative two, negative eight, and negative four. Those are all divisible by two. And lastly, 𝑧 has coefficients of nine, 36 and 18. All divisible by nine. So any of these variables 𝑥, 𝑦, or 𝑧 wouldn’t be too difficult to eliminate because if they’re all divisible by a common number, we can multiply each number by something and make them all the same. And then they can wipe out.

The way that we’ll do this is we’ll take two equations, and we’ll stack them using the elimination method. And we will get the coefficients to be the same but one positive and one negative. That way when we add them, they cancel each other out. So since 𝑥, 𝑦, and 𝑧 are all pretty equal, out of the two equations to pick, let’s go ahead and pick the first equation and the last equation. Their numbers are the smallest.

So here we have them stacked so that way we can use the elimination method where we’ll add them together. We need to eliminate either 𝑥, 𝑦, or 𝑧. But 𝑦s have the smallest coefficients. So let’s eliminators those. So we have negative two and negative four. We can multiply the top equation by negative two and that would make negative two 𝑦 turn into four 𝑦. And then four 𝑦 plus negative four 𝑦 would make the 𝑦s cancel. So let’s go ahead and multiply that first equation by negative two.

So we have 12𝑥 plus four 𝑦 minus 18𝑧 equals eight. And now we’ll rewrite that next equation. And now we’re ready because four 𝑦 minus four 𝑦 makes the 𝑦s cancel just like we want it. So now we can work with the 𝑥 and the 𝑧. However, 12𝑥 plus negative 12𝑥 completely cancels. And negative 18𝑧 plus 18𝑧 cancels. So there’s nothing on the left, and on the right-hand side of the equation, eight plus negative 10 is negative two. And zero is not equal to negative two. Therefore, our answer is: no solution.

If we would have gotten something like zero equals zero or negative two equals negative two, something true, our answer would’ve been all real numbers. Also it could’ve been that the 𝑥s and the 𝑧s wouldn’t have cancelled. And you would’ve had an equation with 𝑥 and 𝑧 equal to some constant. Then we have to go back to our original three equations and pick a different set of two. And then use that set and eliminate 𝑦. And have another equation of 𝑥 and 𝑧, and put the 𝑥 and 𝑧 equations together. And solve for either 𝑥 or 𝑧, and keep going from there. However, with this particular problem, zero did not equal negative two.

So our answer again is: no solution.