WEBVTT
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Determine the number of solutions to the system of equations:
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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.
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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.
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That will be the first step.
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So deciding on which variable to eliminate 𝑥, 𝑦, or 𝑧.
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When you look at their coefficients, 𝑥 has coefficients of negative six, negative 24, and negative 12.
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These are all divisible by six.
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𝑦 has coefficients of negative two, negative eight, and negative four.
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Those are all divisible by two.
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And lastly, 𝑧 has coefficients of nine, 36 and 18.
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All divisible by nine.
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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.
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And then they can wipe out.
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The way that we’ll do this is we’ll take two equations, and we’ll stack them using the elimination method.
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And we will get the coefficients to be the same but one positive and one negative.
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That way when we add them, they cancel each other out.
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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.
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Their numbers are the smallest.
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So here we have them stacked so that way we can use the elimination method where we’ll add them together.
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We need to eliminate either 𝑥, 𝑦, or 𝑧.
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But 𝑦s have the smallest coefficients.
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So let’s eliminators those.
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So we have negative two and negative four.
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We can multiply the top equation by negative two and that would make negative two 𝑦 turn into four 𝑦.
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And then four 𝑦 plus negative four 𝑦 would make the 𝑦s cancel.
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So let’s go ahead and multiply that first equation by negative two.
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So we have 12𝑥 plus four 𝑦 minus 18𝑧 equals eight.
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And now we’ll rewrite that next equation.
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And now we’re ready because four 𝑦 minus four 𝑦 makes the 𝑦s cancel just like we want it.
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So now we can work with the 𝑥 and the 𝑧.
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However, 12𝑥 plus negative 12𝑥 completely cancels.
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And negative 18𝑧 plus 18𝑧 cancels.
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So there’s nothing on the left, and on the right-hand side of the equation, eight plus negative 10 is negative two.
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And zero is not equal to negative two.
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Therefore, our answer is: no solution.
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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.
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Also it could’ve been that the 𝑥s and the 𝑧s wouldn’t have cancelled.
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And you would’ve had an equation with 𝑥 and 𝑧 equal to some constant.
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Then we have to go back to our original three equations and pick a different set of two.
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And then use that set and eliminate 𝑦.
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And have another equation of 𝑥 and 𝑧, and put the 𝑥 and 𝑧 equations together.
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And solve for either 𝑥 or 𝑧, and keep going from there.
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However, with this particular problem, zero did not equal negative two.
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So our answer again is: no solution.