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Lesson 150: Solving Linear-Quadratic Systems of Equations Algebraically and Graphically
We've learned how to graph both linear and quadratic equations. In this lesson we'll learn how to graph a system that is comprised of one of each, and we'll also learn how to solve such a system algebraically.
Before starting this lesson, be sure to review all of the previous lessons on graphing linear and quadratic equations, and solving systems of equations graphically and algebraically. If you do not fully understand all of those lessons, this lesson will not make sense.
Let's look at this system of equations: x + y = 1, and y = -x2 + 4x - 3. We know from experience that the first equation is linear, and the second is quadratic. That means we'll have a line and a parabola. To solve this system of equations graphically, we'll need to plot them both on the same coordinate plane, and see where they intersect. They will intersect in either 0, 1, or 2 places. Think about why.
We can use any method that we have learned to plot these equations graphically. The easiest way is to just graph one equation at a time by picking values of x, and substituting to find the corresponding y values. Pick a variety of x values such as -2, -1, 0, 1, and 2. After doing this, if you see that you need more values of x, just expand your range in the positive and negative directions.
You can also use more sophisticated methods for graphing. For example, you can convert the linear equation into standard y=mx+b format, and plot the line using the slope and y-intercept. You can also use the formulas that we learned to determine the nature of the graph of the parabola. In any case, by plotting both equations on the graph, you will be able to see the points of intersection.
The system can also be solved algebraically. As always, we need to do something to eliminate one of the variables. In this case, the easiest thing to do will be to eliminate y. Eliminating x will be difficult because we have the x2 term. The easiest thing to do is manipulate the linear equation to get it in terms of y. Doing so, we get y = -x + 1. Now that we have each equation in terms of y, we can actually set the two equations equal to one another, and y is eliminated. We have -x + 1 = -x2 + 4x - 3.
What we want to do is get everything over to one side, therefore leaving 0 on the other side. Doing so using algebra, we end up with x2 - 5x + 4 = 0. What I did was move things such that the x2 term would not be negative. This should look familiar to you. We can use reverse-FOIL to factor the polynomial. We get (x - 4)(x -1) = 0. We know that if either binomial equals 0, the left side equals 0. That means that x = 4 and x = 1 will be solutions. Substitute each x value into either of the two original equations to determine the corresponding y values. Doing so, we see that (1,0) and (4, -3) are both solutions to the system of equations. Those will be the points where the line and parabola intersect.
These problems can be tricky, but they all follow this same pattern. Do as many as you can from your textbook for practice, and contact me if you run into difficulty.
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