Question

How do you check your solutions to a system of equations using the substitution method?

Answer 1

Hint: If you find out the solution for any system of equations, that means that the solution should satisfy all the equations used to find the result. If they satisfy then your result is correct, if not then it is wrong.

Complete step by step solution:
Whenever we solve a system of equations, we get a set of solutions. Now one needs to understand that the result is found using those equations only so the result when put in those equations, the full equation must satisfy.
Let us consider the simplest set of equations and verify the same.
$\
  x + y = 2 \\
  x - y = 0 \\
\ $
When we solve this set, the result we get will be $x = 1$ and $y = 1$. So, if we substitute the values of x and y in both the equations, let’s see what happens.
$\
  x + y = 2 \\
  1 + 1 = 2 \\
  2 = 2 \\
  x - y = 0 \\
  1 - 1 = 0 \\
  0 = 0 \\
\ $
When substituting the value of the result, we get a satisfying equation. Now, suppose we found a wrong result, say $x = 2$ and $y = 3$. Now according to the substitution method, the result must dissatisfy the equations.
$\
  x + y = 2 \\
  2 + 3 = 2 \\
  5 = 2..............\left( i \right) \\
  x - y = 0 \\
  2 - 3 = 0 \\
   - 1 = 0..............\left( {ii} \right) \\
\ $
In both cases (i) and (ii), we get illogical results which signify that the solutions were wrong. As we know, the result of a pair of linear equations has only one result and as such, put any other value of x and y in the equation and you will end up with some of the other illogical equations.
This method is effective on all types of equations, be it quadratic or equations of any other higher degree. In fact, this is the method that is most effective in checking if the results are correct or not.

Note:
If one of the equations satisfies with the result and others don’t, then the result is still wrong and as such, only when all the equations satisfy with the result, can we say that we have the correct result. In the case of the above equation, put any value of x and y where $x = y$, the second equation will satisfy but the first will not and so it is wrong except when $x = 1,y = 1$.