# Homework Help: Algebra help

1. Jan 23, 2013

### 939

1. The problem statement, all variables and given/known data
7((45y + 30)/14) + 9y = 225

45y + 30 + 18y = 450

Where did the 7 go? What was divided by 14?

I'm not sure how this part was done, I undertand how to get the rest though...

2. Relevant equations

7((45y + 30)/14) + 9y = 225

3. The attempt at a solution

I can get the rest, but I simply have no idea how this part was done (it was example in book). It appers to inverse 7/14 and multiply it by 9y, but I don't know why...

Last edited: Jan 23, 2013
2. Jan 23, 2013

### Simon Bridge

7((45y + 30)/14) + 9y = 225

Lets make that more readable: $$7\left ( \frac{45y+30}{14}\right )+9y=225$$ ... multiply everything by 14, divide everything by 7. i.e. multiply through by 2.

3. Jan 23, 2013

### 939

Thanks, the only thing I don't get is...

1) why it becomes 2. If you took 14 out of the denominator wouldn't it become (7)(1/14) = 7/14?

2) Why do you multiply everything, including 225, not simply (45y + 30)?

4. Jan 23, 2013

### Staff: Mentor

7/14 = 1/2. One way to get rid of that 1/2 is to multiply both sides of the equation by 2.
Because whatever you do to one side of an equation, you have to do also to the other side.

5. Jan 23, 2013

### Simon Bridge

I have to do the same thing to both sides in order to keep the expression true.

i.e. if y+2=4 is true
then it is also true that 2y+4=8 (x2 all through)
and it is also true that y+4=6 (+2 to both sides)

But it is not true that 2y+4=4 (x2 to the LHS only).

We can do anything we want to the equation: so long as we do it to both sides, the expression remains true. Some things we can do are more useful than others - i.e. none of the above tells you what value of y makes the expression true.

If we subtracted 2 from both sides, though...

For your problem - you can find the solution by multiplying through by 14, then dividing through by 7. x14/7=x2 !

OR, you could expand out the brackets, then put the LHS over a common denominator...

6. Jan 24, 2013

### 939

Highly appreciated, Mark.

Thanks Simon Bridge, got it!

7. Jan 24, 2013

### Simon Bridge

No worries.

If you are seeing expressions like above then you've probably been doing algebra for some time and being puzzled by it suggests that you were taught by the "transferring numbers from one side to the other" approach - rules like "swap sides swap signs" that sort of thing. Not helpful when you get to more complicated expressions.