Solving Linear-Quadratic Systems by Substitution

Linear-Quadratic Systems

Let's say we want to solve a system of equations that is linear AND quadratic. Usually this means one of the equations is linear and another equation is quadratic. In particular, many of these systems take on the form

$$\begin{align} y &= Ax^2 + Bx + C \\ y &= mx + b \end{align}$$

How does one solve these systems? There are multiple ways, including graphing and elimination, but right now we're going to focus on the substitution method. Usually it is the same procedure as solving a system by substitution that is entirely linear. Moreover, if $A = 0$ above, then the system is just linear.

We will do an example to show how this method works on such systems of equations. Our example is

$$\begin{align} y &= x^2 - 12x + 35 \\ y &= x - 1. \end{align}$$

Since this is the method of substitution, you should be inclined to look for ways to write one variable in terms of the other. Luckily, both equations are in this form. They are already solved for $y$ in terms of $x$. We want to substitute $y$ into one of the equations. In this case, substitute the value of $y$ in the second equation into the first equation:

$$x - 1 = x^2 - 12x + 35$$

This is now a one-variable quadratic equation. Move all terms to one side by adding $-x + 1$ to both sides:

$$\begin{align} x - 1 + (-x+1) &= x^2 - 12x + 35 +(-x+1)\\ 0 &= x^2 - 13x + 36 \end{align}$$

This, fortunately, is a quadratic we can factor:

$$(x - 9)(x - 4) = 0$$

The solutions are $x = 9$ and $x = 4$ based on the factors. Now we have two solutions for $x$! This is one of the complications not normally present in a system that is just linear. How we handle it is as follows: we substitute both values of $x$ into an equation to find $y$. This can give us multiple solutions. In this case, we substitute $x = 9$ and $x = 4$ into the equation

$$y = x - 1.$$

If $x = 9$, then $y = 9 - 1 = 8$; on the other hand, if $x = 4$, then $y = 4 - 1 = 3$. This means that our solutions are $(x, y) = (9, 8)$ and $(x, y) = (4, 3)$. I recommend checking that both of these ordered pairs are solutions to the original system of equations.

There are other, fancier tricks that can be used to solve SOME linear-quadratic systems. In particular, while some problems may require the Quadratic Formula, they can all be solved using the method I just demonstrated.

Linear-Quadratic System: Substitution

There are many ways to solve linear-quadratic systems of equations. This lesson will explore substitution.

When solving a linear-quadratic system of equations using substitution, you want to plug in the second equation into the first one. Then, combine like terms, simplify, and factor!

Let's do a practice problem.

Solution

$$y=2x^2-9x+4$$ $$y=4x-2$$ $$4x-2=2x^2-9x+4$$ $$-2=2x^2-13x+4$$ $$0=2x^2-13x+6$$

To solve this quadratic equation, we can factor! $$x=\frac12, x=6$$

(Video) Solving Linear Quadratic Systems Using Substitution

by Aaron Tyson

In this video, Aaron Tyson goes over how to solve a system of equation using substitution.

When looking at the solution for a linear function and a quadratic function, we are looking for the points of intersection between the two functions.

Example 1: Solve the system of equations using substitution.

$$y=(x-3)^2$$ $$x+y=5$$

Step 1: Convert the quadratic equation into standard from.

$$y=(x-3)^2$$ $$y= x^2 -6x +9$$

Step 2: Because the quadratic equation is now equal to $y$, we can plug it into the linear equation.

$$x + (x^2 -6x+9) =5$$

Step 3: Combine like terms, simplify, and factor.

$$x^2-5x+0=5$$ $$x^2 -5x +4=0$$ $$(x-1)(x-4)=0$$

According to the zero product property, we know that: $$x=1, x=4$$

Step 4: Now that we know the $x$ coordinates, we need to know the $y$ coordinates. We can plug in our $x$ values back into the original equations.

We get two coordinates: $(1,4)$ and $(4, 1)$. These two points are where the linear and quadratic functions intersect.

Now try this next example by yourself, and check out the video to see if you got it correct!

Example 2: Solve the following system of equations using substitution.

$$y=x^2+11x-12$$ $$y=12x +30$$