A few weeks ago I told the story of the exams I passed to enter the “Bachillerato Bilingüe” class in my high school. Among all the things we had to learn for the exam there was the elimination algorithm to solve systems of linear equations, which I’m going to quickly show here.
$$ \begin{cases} (1) \quad x + y + z = 6 \\ (2) \quad 2x - y + z = 3 \\ (3) \quad 3x + 2y - z = 4 \end{cases} $$We can solve this step by step as follows. First, eliminate $y$ adding $(1)$ and $(2)$ together:
$$ (x + y + z) + (2x - y + z) = 6 + 3 $$$$ 3x + 2z = 9 \quad \mathbf{(4)} $$Eliminate $y$ multiplying equation $(2)$ by $2$ and adding it to equation $(3)$:
$$ 2(2x - y + z) = 2(3) \implies 4x - 2y + 2z = 6 $$Adding this to equation $(3)$ we get:
$$ (4x - 2y + 2z) + (3x + 2y - z) = 6 + 4 $$$$ 7x + z = 10 \quad \mathbf{(5)} $$Solve the $2 \times 2$ system for $x$ and $z$: from equation $(5)$, we can see that $z = 10 - 7x$. Substitute this into equation $(4)$:
$$ 3x + 2(10 - 7x) = 9 $$$$ 3x + 20 - 14x = 9 $$$$ -11x = -11 \implies \mathbf{x = 1} $$Now, find $z$:
$$ z = 10 - 7(1) \implies \mathbf{z = 3} $$Finally, find $y$ using equation 1:
$$ 1 + y + 3 = 6 $$$$ y + 4 = 6 \implies \mathbf{y = 2} $$And the final solution is:
$$ \mathbf{(x, y, z) = (1, 2, 3)} $$Et voilà! This simple algorithm was one of my discoveries of 1989. Then came other methods, including substitution and matrices (aka Cramer’s rule), but traditional calculus (that is, derivatives and integrals) wasn’t a part of the curriculum for the exam.