2–3 Minute Worked Example (Video)
Watch how a short video demonstrates solving systems of linear equations and a brief narration. Notice: minimal text on slides, large equations, and slow cursor movements.
As you watch, notice how the movement within the equations works together with the narration—this is where video adds value beyond static notes.
Transcript
In this lesson, we're going to talk about solving systems of linear equations by using the elimination method.
The first thing we want to do is make sure each of our equations is in the same format. If you take a look at Equation 1 and Equation 2, you'll notice they’re both in what we call standard form: Ax + By = C.
The goal of the elimination method is to eliminate one of the variables by combining the equations. Normally, you can’t solve an algebraic equation with two variables unless one is removed. So, our first step is to make either the x-coefficients or the y-coefficients match and be opposite in sign.
Looking at the two equations:
- 6x − 3y = −3
- 4x + 5y = 9
We have x-coefficients of 6 and 4, and y-coefficients of −3 and 5. Neither pair matches yet, but both have least common multiples we can work with.
Let’s eliminate the x-values. To do that, we’ll multiply the first equation by 2 (to get 12x) and the second equation by −3 (to get −12x). That gives us:
- 12x − 6y = −6
- −12x − 15y = 27
Now, when we add the two equations together, the x terms cancel out:
(12x − 6y) + (−12x − 15y) = (−6 + 27)
That leaves us with: −21y = 21. Divide both sides by −21 and we get y = −1.
Once we know y, we substitute it back into one of the original equations to solve for x. That’s it!
Production Notes
- Keep duration to 2–3 minutes.
- Script first; then record screen + mic.
- Add captions (.vtt) and verify timing.