# Elementary Substitution in Solving Equations - Why it works

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?

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 $\sin(x)$.)

• set the solution set in the original system and you will see it – Dr. Sonnhard Graubner Jul 26 '15 at 6:45
• Could you explain what is meant by "set the solution set"? – FreshAir Jul 26 '15 at 6:51
• The logic goes if $x$ and $y$ satisfy the system of equations, then $y--4x+24$ and then $\dots$. In principle one should then verify that the solution so obtained works, though here since every step is reversible, one often (and not quite correctly) skips that step. – André Nicolas Jul 26 '15 at 6:52
• It may not be explicitly among the "row operations", but inserting the expression for $\ y \$ into the first equation does just the same thing as multiplying the second row by 3, then subtracting that from the first row. (The difference is largely in appearance, since "substitution" does not use arrays or matrices of coefficients.) – colormegone Jul 26 '15 at 6:54
• When we square, or take the sine, or make some other not one to one transformation, verification that we have not introduced an extraneous root becomes necessary. – André Nicolas Jul 28 '15 at 4:08

\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 $x=5,y=4$
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.
• Dr. Sonnhard Graubner, thanks for this detailed answer. However, I think I am missing something. Your answer shows the the solution $(x,y)=(5,4)$ holds. But what I asked was how the solution sets to the original system, and the system after substitution, are the same. Would you point out how this can be generalized? – FreshAir Jul 28 '15 at 3:24