Quadratic Systems of Equations

In an earlier chapter, you learned how to solve a system of two equations that were both linear, such as: $$\{\,\cl"tight"{\table x,+,2y,=,-7; 2x,-,3y,=,0}$$

In this lesson, you will learn a new method for solving such a system, by using substitution to eliminate a variable in one of the equations. This method can also be used when one of the equations is quadratic (involves $x^2$ or $y^2$).

Solving linear systems by substitution

We’ll start by looking for a solution to the system of equations

$$ \{\,\cl"tight red"{\table y,=,2x+1; x+3y,=,10} $$

Notice that the first equation is in slope-intercept form and the second equation is in standard form.

Think about a solution $(x, y)$ that makes both equations true. The first equation tells us that for this solution, $y$ is the same as $2x+1$. That means we can substitute $2x+1$ for $y$ in the second equation, and the resulting equation will still be true at this solution. When we do this, we get the new equation:

$$ x+3(2x+1)=10 $$

What do you get if you simplify this new equation?

What is the solution to this equation? Use scratch paper if necessary.

Now that you’ve found $x$, use the first equation $y=2x+1$ to find $y$.

What is the point $(x, y)$ that solves this system of equations?

The system is graphed on the grid to the left. Is the solution you see on the graph the same as the solution you just found using algebra?

We say a variable is isolated if it appears alone on one side of an equation, and does not occur in the other side of the equation. If a system of two equations has an isolated variable (say, $y$) in one of the equations, you can solve the system using the method of subExample:

First, substitute the formula for $y$ this gives you into the other equation, eliminating $y$ from that equation;
then solve the resulting equation for $x$;
and finally use the value of $x$ you just found, to now find $y$.

Solve the following systems of equations, each of which includes an isolated variable.

This method of solving systems is known as solving by substitution.

If you have a system where both equations are in standard form, you can still solve it by substitution. You just need to start by rewriting one of those equations in slope-intercept form (that is, by isolating the variable $y$).

We would like to solve the system

$$ \{\,\cl"tight red"{\table 4x, +, 2y, =, 8; x, +, 2y, =, -1} $$

Begin by writing the equation $4x+2y=8$ in slope-intercept form.

Now, use that rewritten equation, along with the other original equation $x+2y=-1$, to find $x$ and $y$.

Sometimes, isolating $x$ can be easier than isolating $y$. We’ll now look at a system where that is the case. Click to see the system

$$ \{\,\cl"tight red"{\table 2x, +, 6y, =, -14; -3x, +, 2y, =, 10} $$

graphed on the grid to the left. Rewrite the first equation ($2x+6y=-14$) in a form that isolates $x$.

Now, use that rewritten equation, along with the other original equation $-3x+2y=10$, to find $y$ and $x$.

Quadratic systems of equations

We’ll now look at systems of equations where one equation is quadratic (involves $x^2$ or $y^2$) and the other linear (involves only $x$ and $y$). You can solve these systems by substitution, just as you did the systems of linear equations in the previous section.

We’ll look at the following system of equations:

$$ \{\,\cl"tight red"{\table 2x+y, =, 1; y, =, x^2-3x-1 } $$

This system is graphed on the grid to the left. How many solutions does it appear to have?

Using scratch paper, isolate $y$ in $2x+y=1$.

Now, substitute that expression for $y$ into the quadratic equation $y=x^2-3x-1$.

Using scratch paper, rewrite this equation in the form $ax^2+bx+c=0$. (Remember to write $x^2$ as “x^2”.)

Now, solve the resulting equation for $x$, by either factoring or using the quadratic formula.

If $x$ is the first value you typed in (), what is $y$?

If $x$ is the second value you typed in (), what is $y$?

What two points $(x,y)$ are solutions to the system?

Are the solutions you see graphed to the left the same as the solutions you just found using algebra?

The system of equations

$$ \{\,\cl"tight red"{\table 3x-3y, =, 3; x, =, y^2-3y+2 } $$

is graphed on the grid to the left. This is also a quadratic system of equations because of the $y^2$ in the second equation. Notice that the graph of $x=y^2-3y+2$ is a parabola that opens to the right.

How many solutions does this system appear to have?

Using scratch paper, isolate $y$ in the first equation ($3x-3y=3$).

Now, we want to substitute that expression for $y$ into the equation $x=y^2-3y+2$. As a first step toward doing that, use the expression you just found for $y$ to find a simplified expression for $y^2$ involving only $x$.

Now, substitute the expressions you found for $y$ and $y^2$ into the equation $x=y^2-3y+2$, and simplify the resulting equation.

Using scratch paper if necessary, rewrite this equation in the form $ax^2+bx+c=0$.

Now, use the quadratic formula to solve the resulting equation for $x$.

If we use the “+” sign in the above expression for $x$, what is $y$?

If we use the “$-$” sign in the above expression for $x$, what is $y$?

Quadratics with other shapes

The equation $x^2+y^2=9$ is graphed on the grid to the left. Notice that this graph is no longer a parabola, because it involves both $x^2$ and $y^2$. (In fact, it is a circle.)

Click to see the graph of the system of equations

$$ \{\,\cl"tight red"{\table 2x,+,y, =, 8; x^2,+,y^2, =, 9 } $$

Based on its graph, does this system of equations appear to have any solutions?

Using scratch paper, isolate $y$ in the first equation.

From this, find a simplified expression for $y^2$ in terms of $x$.

Now, substitute that expression for $y^2$ into the equation $x^2+y^2=9$ and simplify the result.

Rewrite this equation in the form $ax^2+bx+c=0$.

What is the discriminant of this quadratic equation? (Remember that the discriminant of $ax^2+bx+c=0$ is the quantity $b^2-4ac$ that appears under the square root sign in the quadratic formula.)

Is the discriminant positive, negative, or zero?

Does this system of equations have any real solutions?

We’ll now look at the system

$$ \{\,\cl"tight red"{\table y, =, 2x+2; xy, =, 4 } $$

which is graphed on the grid to the left. (The two curves in the graph give the solutions to the second equation.)