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Find the set of points of intersection of the graphs of 𝑥 plus five 𝑦 equals zero and 𝑦 squared equals negative 𝑥.
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To solve this problem and actually find the set of points of intersection, we need to actually make the graphs equal to each other because when they’re equal to each other, this is where they’re going to intersect.
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So therefore, we’re actually gonna solve this problem as a simultaneous equation problem.
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And as you can see, I’ve actually labeled each equation: equation one and equation two.
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And it’s actually equation one that we’re gonna look at first because we’re actually gonna rearrange equation one.
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We’re gonna do this because we actually want to make 𝑥 the subject.
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And we do that so that we can then substitute it in to equation two.
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So if we have 𝑥 plus five 𝑦 equals zero, if we subtract five 𝑦 from each side, we’re gonna be left with 𝑥 is equal to negative five 𝑦.
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So we’ve achieved our goal and we’ve actually made 𝑥 the subject.
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So now, we’re gonna substitute 𝑥 equals negative five 𝑦 into equation two.
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We’re gonna do that so that there’s only one unknown in our equation because we’ll then be able to find 𝑦.
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So this gives us 𝑦 squared is equal to negative negative five 𝑦.
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So therefore, as we have negative negative, then we actually get positive.
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So we can say that 𝑦 squared is equal to five 𝑦.
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And now because we actually have a quadratic term because we got 𝑦 squared, we want to make it equal to zero.
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So we subtract five 𝑦 from each side.
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So we get 𝑦 squared minus five 𝑦 is equal to zero.
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So now, we can solve this using factoring.
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So therefore, we start with 𝑦 outside our parentheses and we get that because it’s a factor of both terms.
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And then inside the parentheses, we have 𝑦 minus five because 𝑦 multiplied by 𝑦 gives us 𝑦 squared and 𝑦 multiplied by negative five gives us negative five 𝑦.
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Okay, great, so therefore, we get that 𝑦 is equal to zero and five.
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And that’s because if we had 𝑦 equal to zero, then the parentheses would be multiplied by zero.
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So it’d give us zero.
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And if we had 𝑦 is equal five, then five minus five is zero.
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So again, we’d get zero.
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And these values are the 𝑦-coordinates of our points of intersection.
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So now, we need to find the 𝑥-coordinates.
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So now to find 𝑥, what we’re gonna do is substitute 𝑦 equals zero and 𝑦 equals five into equation one.
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You can also substitute it into equation one or equation two.
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But we’re just gonna do it into equation one.
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So we’re gonna start with 𝑦 equals zero.
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So if we can substitute that in, we get 𝑥 plus five multiplied by zero equals zero, which gives us 𝑥 plus zero equals zero.
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So therefore, 𝑥 is equal to zero and our first point of intersection is at zero, zero.
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So now, we can move on to 𝑦 equals five.
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So we get 𝑥 plus five multiplied by five equals zero.
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So 𝑥 plus 25 equals zero.
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And therefore, 𝑥 is equal to negative 25, which means that our second point of intersection is negative 25, five.
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And if we look back to our question, it says find the set of points of intersection.
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So therefore, we can write it as the set of points that actually are the points of intersection of the graphs 𝑥 plus five 𝑦 equals zero and 𝑦 squared equals negative 𝑥 are zero, zero and negative 25, five.