Question Video: Solving Systems of Linear and Quadratic Equations to Find the Set of Points of Intersection of Two Given Graphs Mathematics

Find the set of points of intersection of the graphs of π‘₯ + 7 = 8 and π‘₯Β² + 𝑦² = 65.

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Video Transcript

Find the set of points of intersection of the graphs of π‘₯ plus seven equals eight and π‘₯ squared plus 𝑦 squared equals 65.

Whenever we’re given the equations of two graphs and asked to find their points of intersection, that’s an indication to us that we’re going to solve a system of equations. In this case, we have a linear equation, π‘₯ plus seven equals eight, and an equation involving some quadratic terms. In fact, the graph of the equation π‘₯ squared plus 𝑦 squared equals 65 is a circle. This circle is centered at the origin, and it has a radius of root 65. Now it’s actually outside the scope of this video to discuss where that comes from, but it does give us an indication as to how many potential solutions we’ll get.

Subtracting seven from both sides of our other equation, and we get the equation π‘₯ equals one. The line π‘₯ equals one, in fact, is a vertical line that passes through the π‘₯-axis at one. And so this tells us there are three possible outcomes for the number of solutions, the number of points of intersection. There could be no points of intersection. That is, the line π‘₯ equals one does not pass through the circle. There could be one point of intersection. In this case, the line π‘₯ equals one would be a tangent to the circle; it touches the circumference exactly once. And finally, our line could intersect the circle twice, giving us two sets of solutions, two points of intersection of our graphs.

Now that we have an idea of what our solutions might look like, let’s solve these equations simultaneously. When we’re working with two equations whose degrees are different β€” in this case, we have a linear, degree one, and a quadratic, degree two β€” we tend to use the method of substitution. To make our lives easier, let’s define the equation π‘₯ equals one to be equation one and the quadratic, the equation of our circle, to be equation two.

When we solved the equation, we found that π‘₯ equals one. This means that if there are any points of intersection of our graphs, their π‘₯-coordinates will always be one. So we can substitute π‘₯ equals one into our second equation and then solve for 𝑦. The left-hand side, π‘₯ squared plus 𝑦 squared, is one squared plus 𝑦 squared in this case. Or equivalently, one plus 𝑦 squared equals 65. To begin to solve for 𝑦 squared, let’s subtract one from both sides. And that gives us 𝑦 squared equals 64.

So what do we do next? Well, the inverse operation to squaring is to find the square root. But of course, when we’re solving an equation by finding the square root, we need to consider two possible outcomes. We have to take both the positive and negative square root of 64. In fact, the square root of 64 is eight. So 𝑦 is either positive eight or negative eight. So there are two sets of solutions here. When π‘₯ is equal to one, 𝑦 is equal to eight. And when π‘₯ is equal to one, 𝑦 is equal to negative eight. This means, in turn, that there are two points of intersection of our graphs. They have coordinates one, eight and one, negative eight.

And wherever possible, it’s sensible to try and check our answers. We can do so by substituting each pair of values into the equation involving π‘₯ and 𝑦. When we substitute π‘₯ equals one and 𝑦 equals eight, we get one squared plus eight squared, which is equal to 65 as expected. Similarly, when π‘₯ is one and 𝑦 is negative eight, we also get 65. So the set of solutions and, in fact, the set of points of intersection of our graphs is the set containing the elements one, eight and one, negative eight.

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