Solving Quadratic Systems by Substitution or Elimination

How to solve simultaneous equations?

Solving simultaneous equation is a basic skill in solving various kind of problems. One of the ways of doing this is by substitution. For example, you are given the following equations:

$3x=7y+1$, $x^2+2y^2+3xy=13$

The degree of both equations are different. So one can substitute $x=\frac{7y+1} {3}$ into the latter equation to obtain a quadratic equation with one variable and it is much easier to solve it.

(Video) Solving Nonlinear Systems with Substitution

by ThinkwellVids

Here's a video by ThinkwellVids showing you how to solve a system of nonlinear equations by using substitution.

He gives the following system of equations. $$\begin{align}y&=x^2-x-1\\ 3x-y&=4 \end{align}$$ First, we need to put the second equation into a form we can work with, $y=3x-4$. After this quick fix, our system is: $$\begin{align}y&=x^2-x-1\\ y&=3x-4 \end{align}$$ We can set these equal to each other and solve for $x$. $$\begin{align}3x-4&=x^2-x-1\\ 0&=x^2-4x+3\\ 0&=(x-1)(x-3)\\ x&=1, 3 \end{align}$$ We need to plug both of these into one of our equations. We're going to use the first one, $y=x^2-x-1$. $$\begin{align}f(1)&=(1)^2-(1)-1\\ f(1)&=-1\\ \\ f(3)&=(3)^2-(3)-1\\ f(3)&=5 \end{align}$$

Our final answers as points are $(1, -1)$ and $(3, 5)$.

(Video) Solving Nonlinear Systems with Elimination

by ThinkwellVids

Here's a video by ThinkwellVids showing you how to solve a system of nonlinear equations by eliminating a variable.

He gives the following system: $$\begin{align}y&=x^2+8x+19\\ 2x-y&=-10\end{align}$$ We can use elimination to solve this system. We can eliminate the $y$ by adding the two equations together, and we can make this easier by lining up similar terms. $$\begin{align}y&=x^2+8x+19\\ 2x-y&=\phantom{0|00000}-10\\ 2x&=x^2+8x+9\\ 0&=x^2+6x+9 \end{align}$$

Now we have a form that we can factor to solve for $x$. $$\begin{align}x^2+6x+9&=0\\ (x+3)(x+3)&=0\\ x&=-3\end{align}$$

We're going to plug this back into one of our original equations, $2x-y=-10$. $$\begin{align}2(-3)-y&=-10\\ -6-y&=-10\\ -y&=-4\\ y&=4 \end{align}$$

Our answer as a point is $(-3,4)$.

He then gives a second set of equations: $$\begin{align}y&=x^2+x+2\\ 3x-2y&=1 \end{align}$$ He explains that this could be simplified with elimination by multiplying the top equation by $2$ before adding them together. This would give us the equation $0=2x^2-x+5$. Unfortunately this can't be factored, so we need to use the quadratic formula. $$\begin{align}x&=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ x&=\frac{1 \pm \sqrt{1^2-4(2)(5)}}{2(5)}\\ x&=\frac{1 \pm \sqrt{-39}}{10} \end{align}$$

The discriminant ($b^2-4ac$) is negative, so this set of equations has no real solutions. The two functions would not touch on a graph.