Step by step
The question
Question: if the exterior angle is (3x + 10)° and the far inside angles are x° and 50°, what is x?
The exterior angle is outside the triangle, on the pushed-out side.
1 Use the rule
The exterior angle is the two far inside angles added together.
So x + 50 = 3x + 10.
Careful!
The outside angle is not inside the triangle.
It does not go in the 180° sum.
2 Get x on one side
There is x on both sides.
We want x on one side only.
So we remove the x from the left side.
To remove x, we take away x.
We do it on both sides.
Why both sides?
The = sign means both sides are the same amount.
If we change only one side, they are not the same any more.
x and − x make zero.
Cross them out.
3x − x = 2x.
Now we have 50 = 2x + 10.
Let's swap the two sides, so x is on the left.
We want x by itself.
Two things are stuck to x: times 2 and + 10.
The + 10 is on the outside, so it comes off first.
Then the times 2.
3 Remove the + 10
To remove + 10, we take away 10.
We do it on both sides, so they stay equal.
+ 10 and − 10 make zero.
Cross them out.
50 − 10 = 40.
Now we have 2x = 40.
4 Remove the 2
2x means 2 times x.
We want just x.
To remove times 2, we divide by 2.
We do it on both sides, so they stay equal.
2 divided by 2 is 1.
So just x is left.
40 divided by 2 is 20.
So x = 20!
5 Check
Let's check: with x = 20, the inside angles are 20° and 50°.
Together 70°, the outside angle.
So the answer is right!
Let's look back.
First: use the rule.
Then: get x on one side.
Then: remove the + 10.
Then: remove the 2.
Last: check.