Elementary Substitution in Solving Equations - Why it works

Question

To solve a system of linear and certain non-linear equations, the substitution method is widely used by elementary and high school students.

As explained here, to solve this simple system of linear equations,

$$ 2x – 3y = –2 $$ $$ 4x + y = 24$$

we use the first equation to isolate $y$: $y = –4x + 24$

The second equation is then substituted into the first, to solve for x (x=5), then x=5 is used in either equation to solve for y, which equals 4.

However, this method is not explicitly one of the three row operations. How would I prove the solution set obtained using this method is exactly the same as the solution set of the original system of equations?

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EDIT: It has been pointed out that substitution in a linear system is in fact a row operation. But what about the use of this method for a non-linear system? For example, this method surely works for this simple non-linear system:

$$ y=x^3$$ $$ y=2x$$

(Of course, this method may not work with certain non-linear functions that are not one-to-one, say $\sin(x)$.)

Answer 1

I will show you what I meant with, $$y=-4x+24$$ we will get in the first equation, $$2x-3(-4x+24)=-2$$ $$2x+12x-72=-2$$ thus we get, $$x=5$$ and, $$y=-20+24=4$$ thus we can plug our solution set, $(x,y)=(5,4)$ in the original system: $$2\cdot 5-3\cdot 4=-2$$ $$4\cdot 5+4=24$$ and this is true.

Answer 2

This is exactly same as applying row operations. Given system of equations can be represented as:

\begin{bmatrix} 2 & 3 & :-2 \\ 4 & 1 & : 24 \end{bmatrix} Now to find y and substitute in first equation is same as applying row operation change $R_1$ in $R_1+R_2$. \begin{bmatrix} 14 & 0 & :70 \\ 4 & 1 & : 24 \end{bmatrix} apply row operation $R_1$ goes to $1/5 R_1$ \begin{bmatrix} 1 & 0 & :5 \\ 4 & 1 & : 24 \end{bmatrix} apply row operation $R_2$ goes to $R_2-5R_1$ (equivalent to putting value of $x$ in second and then solving for $y$) \begin{bmatrix} 1 & 0 & :5 \\ 0 & 1 & : 4\end{bmatrix} So the solution is $x=5,y=4$