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Find all the real solutions to the system of equations 𝑦 equals 𝑥 squared minus three 𝑥 minus four, and 𝑦 equals two 𝑥 squared minus four 𝑥 minus six.
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Well, the first thing I’m gonna do is I’m gonna label our equations.
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So we’ve got equation one and equation two.
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I do this so it’s easy to see what we’re going to do them.
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Now, we can see at first glance that both of our equations are 𝑦 equals.
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So therefore, what we can do is make them equal to each other.
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So, as we’ve said, because they both have 𝑦 is equal to, we can set them equal to each other.
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So when we do that, we get two 𝑥 squared minus four 𝑥 minus six equals 𝑥 squared minus three 𝑥 minus four, with the second equation on the left-hand side.
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And we’ve just done that because I prefer to have the greater number of 𝑥 squared terms on the left-hand side.
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So now, what we want to do is set our quadratic equal to zero because that’s what we like to do when we want to solve it.
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And to do that, what we’re gonna do is we’re gonna subtract 𝑥 squared add three 𝑥 and add four to each side of our equation.
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And when we do that, we’re gonna be left with 𝑥 squared minus 𝑥 minus two is equal to zero.
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And that’s because if you have two 𝑥 squared minus 𝑥 squared, you get 𝑥 squared.
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And if you have negative four 𝑥 add three 𝑥, you get negative 𝑥.
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And if you have negative six add four, you get negative two.
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Okay, great.
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So now, what we need to do is solve this to find our two 𝑥-values.
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And the way to do that is by using factoring.
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So in order to factor, what we need to do is find a pair of factors whose products is negative two and whose sum is negative one.
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And that’s because we’ve got negative 𝑥.
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So the coefficient will be negative one.
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So now, there are only two pairs of factors that’s gonna give us the product negative two.
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And that’s negative one and two or negative two and one.
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Well, if we look at the sum, the sum needs to be negative one.
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So therefore, this tells us that the greatest value is going to be the negative.
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And the other one is gonna be positive.
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So we’re gonna have negative two and positive one.
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So therefore, we know that 𝑥 minus two multiplied by 𝑥 plus one is gonna be equal to zero.
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So now, what we need to do is set each of our parentheses equal to zero because one of these needs to be zero if the answer to be zero.
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So therefore, we can do that to find out our two 𝑥 values.
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And when we do that, we get 𝑥 is equal to two or 𝑥 is equal to negative one.
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So now, what we’ve done is found the 𝑥-values or 𝑥-coordinates of our solutions.
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So to wrap this up, we need to now find the corresponding 𝑦-values.
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And to enable us to do that, what we need to do is substitute our values for 𝑥 into one of our equations.
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I’ve chosen equation one.
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So I’m gonna just first substitute 𝑥 equals two into equation one.
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So if I substitute in 𝑥 equals two, I’m gonna get 𝑦 is equal to two squared minus three multiplied by two minus four, which is gonna give me four minus six minus four.
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So therefore, I’m gonna get 𝑦 is equal to negative six.
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So that’s great.
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We now need to find the other 𝑦-value.
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So now, to find the other 𝑦-value, what I’m gonna do is substitute 𝑥 equals negative one into equation one again.
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So when I substitute in 𝑥 equals negative one, I’m gonna get 𝑦 is equal to negative one squared minus three multiplied by negative one minus four.
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Which is gonna give us 𝑦 is equal to one plus three minus four.
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And that’s because we had negative one squared.
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And negative multiplied by negative is positive.
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And then, we had negative three multiplied by negative one, which gives us positive three.
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So this is gonna give us a 𝑦-value of zero.
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So therefore, we can say that the solutions to the system equations are 𝑥 equals two and 𝑦 equals negative six and 𝑥 equals negative one and 𝑦 equals zero.