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Solving Quadratic Systems

Solving Quadratic Systems 1.10 - Solution

arrow_back Return to Solving Quadratic Systems
We want to solve the given system of equations using the substitution method. Fortunately the variable is isolated in Equation (II). Let's substitute for in Equation (I).
Simplify
Notice that in Equation (I), we have a quadratic equation in terms of only the variable. Now, recall the Quadratic Formula. We can substitute and into this formula to solve the quadratic equation.
Solve for
This result tells us that we have two solutions for One of them will use the positive sign and the other one will use the negative sign.
Now, consider Equation (II). We can substitute and into the above equation to find the values for Let's start with
Simplify
We found that when One solution of the system, which is a point of intersection of the two parabolas, is We can also use decimal notation to express this point as To find the other solution, we will substitute for in Equation (II) again.
Solve for
We found that when Therefore, our second solution, which is the other point of intersection of the parabola and the line, is Using decimal notation we can write this point as Let's see how the graph looks like.

As we can see, the parabolas intersect at the points we have found.