Question Video: Solving a System of Linear and Quadratic Equations to Find the Set of Points of Intersection of Two Given Graphs | Nagwa Question Video: Solving a System of Linear and Quadratic Equations to Find the Set of Points of Intersection of Two Given Graphs | Nagwa

Question Video: Solving a System of Linear and Quadratic Equations to Find the Set of Points of Intersection of Two Given Graphs Mathematics • Third Year of Preparatory School

Find the set of points of intersection of the graphs of 𝑥 − 𝑦 = 0 and 6𝑥² − 𝑦² = 45.

02:24

Video Transcript

Find the set of points of intersection of the graphs of 𝑥 minus 𝑦 equals zero and six 𝑥 squared minus 𝑦 squared equals 45.

So we’ve been asked to find the points of intersection of 𝑥 minus 𝑦 equals zero, which is a straight line, and six 𝑥 squared minus 𝑦 squared equals 45, which is a curve. This is equivalent to solving the linear-quadratic system of equations 𝑥 minus 𝑦 equals zero, six 𝑥 squared minus 𝑦 squared equals 45. We’re going to do this using the method of substitution. We begin by rearranging the linear equation to give one variable in terms of the other. We find that 𝑥 is equal to 𝑦.

We’re now going to substitute our expression for 𝑥 into the second equation. Doing so gives six 𝑦 squared minus 𝑦 squared is equal to 45. We could equally have substituted 𝑦 equals 𝑥 into our second equation, which would give six 𝑥 squared minus 𝑥 squared equals 45. Both approaches will lead us to the same solution. We can now solve this quadratic equation for 𝑦.

Simplifying the left-hand side, six 𝑦 squared minus 𝑦 squared, gives five 𝑦 squared. We can then divide through by five, giving 𝑦 squared equals nine and solve by square rooting. Remembering, we must take plus or minus the square root. So we have that 𝑦 is equal to plus or minus the square root of nine, which is positive or negative three.

Having found our 𝑦-values, we now need to find the corresponding 𝑥-values by substituting into the linear of equation. And it’s very straightforward. As our linear equation can be expressed as 𝑥 equals 𝑦, then each 𝑥-value is just the same as the corresponding 𝑦-value. We find then that there are two points of intersection between these graphs, the point three, three and the point negative three, negative three, which we can express as the set containing these two coordinates.

Now, you may not immediately recognize what the graph of six 𝑥 squared minus 𝑦 squared equals 45 looks like. But if you have access to a graphics calculator or some graphical plotting software, then you can plot these two graphs. 𝑥 minus 𝑦 equals zero is a straight line and six 𝑥 squared minus 𝑦 squared equals 45 is what’s known as a hyperbola. And by considering these two graphs, you can confirm that the two points of intersection are indeed the points we’ve given here.

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