R=5z; 15R=3y. What is R? R=y, but how?

Question

This is from a Georgia Practice Test and my friend cant figure it out. All she knows is that R = y

Ok, from what I know, I stated the question correctly. This is what was emailed to me by my friend. I've tried to do it and I havent been able to come up with it either. All I know is that this is what she got from the practice book for a Georgia State Test she has to take in order to attend a college there. I appreciate anyone's help.

Answer 1

R does not = y, (unless they are both 0).

15R=3y divide both sides by 15 to get R in terms of y
R = (3/15)y
R = (1/5)y

Now using R=5z
(1/5)y = 5z multiply both sides by 5 to get y in terms of z
y = 25z

And if y=0, then z has to = 0 too.

Answer 2

is r=y part of the problem or the answer, because there is always the possiblity of a typeo

if it is given that r=y
then no matter how you work it out r=y=0
because if you work it out normally, r=25, z = 5 and y = 125

look at the ratios they are all off, no matter how you slice the numbers y will always be 5 times as big as r so y = 5r and r = 5z so y = 5(5Z). Assuming i did my math right...

15R = 3Y
: 5R = Y

R=5Z
: Y = 5(5Z)
: Y = 25Z

15(5Z) = 3Y
5Z = 3/15(Y)
25Z = Y

Sorry wish there was a more straight forward answer.

no matter which way you do the math, y will always be 25 times bigger than Z

if r has to = y then they both have to be zero and the math above illustrates that.

Answer 3

R=5z; 15R=3y. What is R? R=y, but how?

The only way for R=y and 15R=3y to be true is for both R and y to equal 0

The thing confusing me is why has the z variable been introduced?
To solve 2 unknowns you need 2 equations
To solve 3 unknowns you need 3 equations
But the answer does not involve z.
z would also have to be 0

If the question is to show that R=y from the other two equation it is impossible as you have 3 unknowns and 2 equations.

Answer 4

You have left something out. It is not possible to solve three unknowns with just two equations.... And r cannot equal y because it violates your 2nd equation. If r = y, then 15 r = 15 y It is not possible for 15r = 3y if r = y... Restate the question properly.

Ron.

Answer 5

R cannot = y

r MUST be 5 *greater than y if 15R=y ,
unless loads of the variables = 0, which is pointless!

Answer 6

R = y and also = z

If y = R then substitute R for Z (as per R=5z)
and since R=5z then 15R=3(5z)
therefore...15R=15z
therefore...R=z

...and also y (as R=y)

Answer 7

As equations, you can't solve it.
Must be wrong question or trick question, so some lateral thinking could be useful...
Why is "R" upper case, and the others lower?
Big R could be radius?
Solve to remove R and you get 25z=y. The alphabet?
Puzzling. I'll look forward to the answer...

Answer 8

R can't equal y
If R = y, then:

15R = 3y
15y = 3y
so, 5 = 1
Since that isn't true, then R can't equal y

Answer 9

15R would be 75 z. 75z = 3y Divide both sides by 3. You get 25z=y.

25z would be 5R. 5R=y.

R can't equal y.

Answer 10

Just substitute. If R=y, you can replace R with y.

y=5*z; 15*y=3*y

15*y=3*y is impossible, so I don't see how this works. . . Did you type it wrong?