Variables on both sides

Boxes on both pans

Now there are unknown boxes on both sides. Before reading on, try to get one box alone.

$5x + 2 = 3x + 10$

Take 1 off each side Take an x off each sideSplit into equal groups

Boxes are weights too, so you can take a box off each pan just like a $1$. Removing $3$ boxes from each side leaves $2x + 2 = 10$, which you already know how to solve.

A bigger example

Simplify each side first, then collect.

A scale that can't balance

Here's $2x + 5 = 2x + 1$. Take off what you can.

$2x + 5 = 2x + 1$

Take 1 off each side Take an x off each sideSplit into equal groups

The scale is tipped from the start. Both pans hold the same boxes, so whatever a box weighs, the left pan always has $4$ extra. Removing the boxes leaves $5 = 1$, which is false. This equation has no solution.

A scale that always balances

Now try $3(x + 2) = 3x + 6$. Slide $x$ anywhere. Can you make it tip?

$3(x + 2) = 3x + 6$

−+

$x = -5$$3((-5) + 2) = -9$$3(-5) + 6 = -9$

It never tips, because distributing shows both sides are the same expression. When you collect terms, everything cancels and leaves something always true:

Every value of $x$ is a solution. An equation like this is called an identity.

Telling them apart

After simplifying both sides, compare them:

Key ideas

• Collect the variable terms on one side and the numbers on the other.
• If the variables cancel and leave something false, there is no solution.
• If the variables cancel and leave something true, every number is a solution.