Question Video: Solving Systems of Quadratic Equations Graphically Mathematics • 9th Grade

Use the graph to solve the system of equations 𝑦 = π‘₯Β² βˆ’ 10π‘₯ + 32, 𝑦 = βˆ’2π‘₯Β² + 20π‘₯ βˆ’ 43.

02:46

Video Transcript

Use the graph to solve the system of equations 𝑦 equals π‘₯ squared minus 10π‘₯ plus 32 and 𝑦 equals negative two π‘₯ squared plus 20π‘₯ minus 43.

Well, the first thing we can do easily is identify which of the graphs is in fact which equation. Because we know that if we have a positive π‘₯ squared term, so we have a positive coefficient of π‘₯ squared, then we have a u-shaped parabola. However, if we have a negative π‘₯ squared coefficient, then what we have is an n-shaped parabola or inverted u-shaped parabola. So therefore, what we can say is the top half of the graph is gonna be 𝑦 equals π‘₯ squared minus 10π‘₯ plus 32. And that’s cause it’s a u-shaped parabola. And then, for the bottom half, we’ve got an inverted u. So it’s 𝑦 equals negative two π‘₯ squared plus 20π‘₯ minus 43.

So now what we want to do is use the graph to solve the system of equations. So to do this, what we need to do is see where the two parts of the graph in fact meet. Well, we can see straightaway there’s gonna be only one real solution. And that’s because our two parts of the graph only meet in one point. And this is the point where they touch. And in fact, it’s at the minimum point of 𝑦 equals π‘₯ squared minus 10π‘₯ plus 32 and at the maximum point of 𝑦 equals negative two π‘₯ squared plus 20π‘₯ minus 43.

So therefore, to find a solution to our system of equations, what we do is read across and read up from the graph. And we can see that we have an π‘₯-value of five and a 𝑦-value of seven. So therefore, we can say that the solution to the system of equations is π‘₯ equals five and 𝑦 equals seven.

What we can do is a quick check just to make sure that these are definitely correct. And we’ll do that by substituting them back into the original equations. Well, if we substitute in π‘₯ equals five and 𝑦 equals seven to the first equation, we’ll get seven equals five squared minus 10 multiplied by five plus 32, which is gonna give us seven equals 25 minus 18. Well, this is correct because seven does in fact equal 25 minus 18. So we can know that the first equation is correct with our solutions.

So then we can move on to the second equation. So when we do that, what we’re gonna get is seven is equal to negative two multiplied by five squared plus 20 multiplied by five minus 43, which is gonna be seven is equal to negative 50 plus 100 minus 43, which is gonna give us seven is equal to 100 minus 93. Well, again, this is correct cause 100 minus 93 is seven. So therefore, we can definitely say that the solution to our system of equations is π‘₯ equals five and 𝑦 equals seven.

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