Solve Algebra Homework: 7((45y+30)/14)+9y=225

[939]

Homework Statement

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

the next line reads

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...

Homework Equations

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

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...

 

[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.

 

[939]

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.

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)?

 

[Mark44]

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?

7/14 = 1/2. One way to get rid of that 1/2 is to multiply both sides of the equation by 2.

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

Because whatever you do to one side of an equation, you have to do also to the other side.

 

[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...

 

[939]

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.

Highly appreciated, Mark.

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...

Thanks Simon Bridge, got it!

 

[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.