If X/a-b+c = Y/b-c+a = Z/c-a+b, Prove that : X+Y/a = Y+Z/b

jejo1

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If X/a-b+c = Y/b-c+a = Z/c-a+b
Prove that : X+Y/a = Y+Z/b


Can I have a detailed proof please :)
 
Can I have a detailed proof please
Sorry, but the policy you saw in the "Read Before Posting" announcement remains in effect; namely, we don't "do" students' work for them, nor do we provide complete solutions. Students still need to show some effort of their own. :oops:

If X/a-b+c = Y/b-c+a = Z/c-a+b
Prove that : X+Y/a = Y+Z/b
What you have posted means the following:

. . . . .\(\displaystyle \mbox{If }\, \dfrac{X}{a}\, -\, b\, +\, c\, =\, \dfrac{Y}{b}\, -\, c\, +\, a\, =\, \dfrac{Z}{c}\, -\, a\, +\, b,\)

. . . . .\(\displaystyle \mbox{then prove that }\, X\, +\, \dfrac{Y}{a}\, =\, Y\, +\, \dfrac{Z}{b}\)

Was this what you meant?

When you reply, please include a clear listing of your thoughts and working so far. Thank you! ;)
 
Can I have a detailed proof please
First the answer is absolutely not.


If X/a-b+c = Y/b-c+a = Z/c-a+b
Prove that : X+Y/a = Y+Z/b

Second what you posted is an absolute unreadable mess.
Where are any grouping symbols? We cannot tell what numerator and/or denominator you mean.
It is also required that you show some work!
 
Sorry, but the policy you saw in the "Read Before Posting" announcement remains in effect; namely, we don't "do" students' work for them, nor do we provide complete solutions. Students still need to show some effort of their own. :oops:


What you have posted means the following:

. . . . .\(\displaystyle \mbox{If }\, \dfrac{X}{a}\, -\, b\, +\, c\, =\, \dfrac{Y}{b}\, -\, c\, +\, a\, =\, \dfrac{Z}{c}\, -\, a\, +\, b,\)

. . . . .\(\displaystyle \mbox{then prove that }\, X\, +\, \dfrac{Y}{a}\, =\, Y\, +\, \dfrac{Z}{b}\)

Was this what you meant?

When you reply, please include a clear listing of your thoughts and working so far. Thank you! ;)
Oh, I didn't mean I need a complete work.. I wanted a complete proof to help me understand how this is made.
Besides, I don't really know how you wrote the equation this way but that isn't what I meant.
I meant that the whole part after the (/) sign was the denominator. for example (X/a-b+c) X is the numerator and a-b+c is the denominator. after the equal sign the same applies. the part after prove that is alright.
 
Oh, I didn't mean I need a complete work.. I wanted a complete proof to help me understand how this is made.
I'm sorry, but what is the difference between "a complete work" and "a complete proof" of the work, especially since "the work" is exactly "the proof"? :shock:

Besides, I don't really know how you wrote the equation this way but that isn't what I meant.
I meant that the whole part after the (/) sign was the denominator. for example (X/a-b+c) X is the numerator and a-b+c is the denominator.

To learn how to type math as text (especially about how to use grouping symbols), try here. Your exercise, I think, is meant to be as follows:

. . . . .If X/(a-b+c) = Y/(b-c+a) = Z/(c-a+b), then prove that X + Y/a = Y + Z/b.

This typesets as:

. . . . .\(\displaystyle \mbox{If }\, \dfrac{X}{a\, -\, b\, +\, c}\, =\, \dfrac{Y}{b\, -\, c\, +\, a}\, =\, \dfrac{Z}{c\, -\, a\, +\, b},\)

. . . . .\(\displaystyle \mbox{then prove that }\, X\, +\, \dfrac{Y}{a}\, =\, Y\, +\, \dfrac{Z}{b}\)

Please reply showing your thoughts and efforts so far. For instance, you started by picking a pair of equal "fractions" from the original equation (say, the first two), cross-multiplied, and... then what? Please be complete. Thank you! ;)
 
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