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Solving Linear-Quadratic Systems

Solving Linear-Quadratic Systems 1.1 - Solution

arrow_back Return to Solving Linear-Quadratic Systems
We want to solve the given system of equations using the substitution method. The -variable is isolated in Equation (II). This allows us to substitute its value 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
We found that when One solution of the system, which is a point of intersection of the parabola and the line, is To find the other solution, we will substitute for in Equation (II) again.
We found that when Therefore, our second solution, which is the other point of intersection of the parabola and the line, is