 # [SOLVED] Compound Assignment With Augmented Addition (Help needed)

Tell us what’s happening:
I’m so stuck on this, I really don’t know what to do

``````var a = 3;
var b = 17;
var c = 12;

// Only modify code below this line

a = a + 12;
b = 9 + b;
c = c + 7;

You’re gonna use the compound assignment operator `+=` to add a number to a variable and assign it at the same time.

If you have

``````a = a + 42;
``````

you can use `+=` as a shorter version:

``````a += 42;
``````

Thank You so much, do you also know how to do the subtraction bit? I’m confused
var a = 11;
var b = 9;
var c = 3;

// Only modify code below this line

a -= a - 6;
b -= b - 15;
c -= c - 1;
that’s what I have so far

It’s also the same idea/pattern with subtraction (and in general, operations that require two operands).

Is it the same for Multiplication and Division or is it another concept?

Yep. Totally the same I’m really confused Can you tell what exactly confuses you?

This is what I came up with, for the multiplication

var a = 5;
var b = 12;
var c = 4.6;

// Only modify code below this line

a *= 25;
b *= 36;
c *= 46;

What am I doing wrong?

Why are you multiplying a by 25 and assigning it back to a? The original code multiplied a by 5 and assigned it back to a.

I’m very confused, I thought what I was doing was making it right

I will just use the code involving variable a and then you can apply the logic to the other variables (b and c).

Original code for variable a is below:

``````var a = 5;
a = a * 5;
``````

You are just going to replace the 2nd line with code that uses the *= operator. So you will have:

``````a *= ???;
``````

You just have figure out what ??? needs to be in the above line. Remember, the original code multiplied a by 5 and then assigned it back to a.

Thanks, I got how to do it. I really appreciate your help

The pattern of operating on a variable and assigning it back to that variable is so common that there’s a short-hand way of writing it.

Instead of writing `a = a + 3` you can just write `a += 3`. Same for all the other arithmetic operators. It’s just a different way of writing the same thing, but you only need to refer to the named variable once.

Thanks, I got the hang of it now 