Math CentralQuandaries & Queries


Question from Lily, a student:

On the substitution method what happens when you have zero's on both sides of the equation? Is that considered no solution or infinitely many?

We have two solutions for you

Hi Lily,

Let's consider a hypothetical example. Suppose that you have two linear equations in the variables x and y and you solve the first equation for x and obtain

x = 5 + 3y

You then substitute this expression for x into the second equation to obtain an equation in the variable y. You then proceed to solve this equation for y and you end with 0 = 0. What this means is that if you have the pair (x, y) that satisfy x = 5 + 3y then (x, y) satisfies the second equation also because it yields the true statement 0 = 0. Thus you can choose any value for y, find x using x = 5 + 3y and the pair (x, y) is a solution to both equations. Thus you have a system with infinitely many solutions.

You might however have solved the first equation for x yielding

x = 5 + 3y

and ten substituted into the second equation to obtain 0 = 5. What this means is that no pair (x, y) that satisfies x = 5 + 3y can also satisfy the second equation so the system has no solutions.

I hope this helps,


Hi Lily.

Here's a simple example of two parallel lines (I know they are
parallel because they have the same slope):

y = 5x
y = 5x + 1

5x = 5x + 1
0 = 1
No solution for the intersection of these lines. So no solution gives
an invalid result as you see.

What if the two lines are identical?
y = 5x
y = 5x

5x = 5x
0 = 0
Infinite intersection points.

Hope this helps,
Stephen La Rocque.

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