# Systems Using Substitution

## Key Questions

• Well, I would say that it is easier when you have few equations and variables. If you have 2 equations and 2 variables it is ok; when you get to 3 equations and 3 variables it becomes more complicated, it is still possible, but you have more work to do. The number of substitutions increases together with the possibility to make mistakes.
More than 3 equations and 3 variables and it gets almost impossible and other methods would be better.

• For an answer to have an infinite solution, the two equations when you solve will equal $0 = 0$.

Here is a problem that has an infinite number of solutions.

$3 x + 2 y = 12$
$- 6 x - 4 y = 24$

If you solve this your answer would be $0 = 0$ this means the problem has an infinite number of solutions.

For an answer to have no solution both answers would not equal each other.

Here is a problem that has no solution.

$4 x - 8 y = 5$
$- 3 x + 6 y = 11$

Again, if you solve this your answer would be $0 = 59$, this is obviously not true, 0 does not equal 59 so this problem would have no solution.

I assume you are interested in linear equations. In general you need $n$ equations if you have $n$ variables.
Let us have $3$ equations and $3$ variables $x , y$ and $z$. Now pick up an equation with $x$ and segregate it say $x$ in terms of $y , z$. When we put this value of $x$ in two other equations we get two equations in $y$ and $z$.
We can now find $y$ in terms of $z$ say using second equation and when we put in third equation we get value of $z$.
Once $z$ is known, it is easy to find $y$ and then $x$.