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,
Penny

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.