![]() To identify the \(y\)-value for each point, substitute each \(x\)-value back into the original equation, \(x+y=5\), and solve. Isolate the variable to find that \(x=-2\). Isolate the variable to find that \(x=1\). Next, separate \((3x-3)(x+2)=0\) into two separate equations and solve for \(x\). ![]() \(-3\) and \(2\) are the two numbers that work for the factored equation. We can see that our graphs intersect at the points (\(-\frac)=0\). Now we are able to graph both equations, and our graph will look like this: We can do this by adding y and subtracting 1 from both sides. In order to graph our linear equation, we want to first rearrange it so that y is by itself on one side. Solve this system graphically and then algebraically: If we wanted one more way to check our answers, we can plug them back into our original equations and make sure our answers match up. So our points of intersection for this graph are (-4,0) and (7,22), which is what we found by graphing them. I’m going to plug them into the linear equation because it will require fewer steps to get y. To find our y-values for each of these points, we simply want to plug in our x-values to either equation and solve for y. These are our values for x at the two points where our graphs intersect. This tells us that our zeroes for this quadratic equation are -4 and 7. Or, subtract 4 from both sides and get that x is equal to negative 4. Now we can add 7 to both sides, and get that x is equal to 7. Which gives us, when we break it apart into two different equations, 0 equals x minus 7, or 0 equals x plus 4. Zero is equal to, we have our x-terms, we’re wanting to reverse FOIL, so 28, we can have 7 times 4 gives us 28 and we want a negative 3 in the middle, so we subtract 7 and add 4. To do this, we will subtract 8 and 2x from both sides. In order to solve a quadratic equation for x, we need to get all of our terms on one side and zero on the other. ![]() When we write it out, that will look like this: Looking back at our equations from earlier, we can see that since \(y=2x+8\), we can replace the y in \(y=x^2–x–20\) with \(2x+8\), or in other words, set our equations equal to one another. To find at what points our equations are equal to each other, we can substitute our y-values for each other. The other way to solve systems like this is very similar to the substitution method of solving regular systems of linear equations. (-4,0) and (7,22).īut what if we aren’t able to graph our equations? Well, that’s where algebra comes in. We can see from our graph that the two equations intersect at the points negative 4,0) and (7,22). If we graph these two equations next to each other, we get something that looks like this: To do this, we simply graph our equations, either with a calculator or by hand, and see where the graphs intersect. ![]() The first solving method we are going to look at is solving graphically. When we have a linear equation and a quadratic equation, we will have zero, one, or two points of intersection. When we solve systems of equations, we are taking two or more equations and finding the point, or points, where they intersect. ![]() Notice that the y term is still not to a power or a root, but we do have an x term that is squared. Quadratic equations are equations of the form \(y=ax^2+bx+c\), and they graph as a parabola. Notice that neither y nor x is to a power or a root. They are equations that can be written as y equals ax plus b. Linear equations are equations that graph as a line. Let’s get started!īefore we get into how to solve these systems, let’s review the differences between linear and quadratic equations. Hi, and welcome to this video on solving systems of equations with a linear and a quadratic equation! In this video, we will take a look at two different ways to approach these problems: graphically and algebraically. ![]()
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