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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 $y^2$).
We’ll start by looking for a solution to the system of equations
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:
What do you get if you simplify this new equation?
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:
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
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
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 $x$.
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:
If $x$ is the first value you typed in (), what is $y$?
If $x$ is the second value you typed in (), what is $y$?
The system of equations
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.
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.
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 $y$?
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
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.
We’ll now look at the system
which is graphed on the grid to the left. (The two curves in the graph give the solutions to the second equation.)
Notice that this simplified equation is quadratic. Because of this, the original system (involving an $xy$ term) is also a quadratic system, even though there aren’t any explicit $x^2$ or $y^2$ terms in it.