Solution not the same as answer for variable.

[bobisaka]

Equation:
(y+5)/(5-3) + -3/15 = 1/-3

y = -3

However, when i use -3 to solve the equations i get..

(-3+5)/(5-3) + -3/15 = 1/-3

(2(15))/(2(15)) + (-3(2))/(15(2)) = 1/-3

(30 + -6)/30 = 1/-3

24/30 = 1/-3

4/5 = 1/-3

What am i doing wrong?

 

[Cubist]

You are correct that y=-3 is not a solution to the above equation.

Can you post the full question, because it seems that either the equation is wrong OR the value y=-3 is wrong.

 

[Dr.Peterson]

I observe that y = -3 would be a solution if the equation were not (y+5)/(5-3) + -3/15 = 1/-3 , but (y+5)/(5*-3) + -3/15 = 1/-3 .

It also is written oddly, so I wonder it you had already replaced some of the y's with -3, and the actual equation was (y+5)/(5y) + y/15 = 1/y .

Is that what happened here? If so, then your main error was in replacing y with -3 without parentheses, so that a multiplication changed into a subtraction.

 

[bobisaka]

Hi guys, this is the original question.. I will give it another crack.

 

[Dr.Peterson]

In other words, I guessed right. The solution is y = -3, and substituting as I showed, replacing y with "(-3)", will make the check work.

Note that we ask you to show your entire original question, for exactly this reason: If you show something you arrived at halfway through, there may be a mistake before that point that we could recognize if we saw it all:

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[pka]

this is the original question.. I will give it another crack.
View attachment 18968

\(15y\left(\dfrac{y+5}{5y}+\dfrac{y}{15}\right)=15y\left(\dfrac{1}{y}\right)\)
\(3(y+5)+y^2=15\)