addition: 5/x + 3/x+2 = -6/x(x+2)

[BB]

What am i missing in this problem?

5/x + 3/x+2 = -6/x(x+2)

x(x+2)[5/x + 3/x+2] = x(x+2)(-6/x+2)

5 + 3 (x+2) = -6x

5 + 3x + 6 = -6x

3x + 11 = -6x

 

[tkhunny]

Re: addition

x(x+2)[5/x + 3/x+2] = x(x+2)(-6/x+2)

5 + 3 (x+2) = -6x

You should have

5(x+2) + 3x = -6

There are several errors. One that might not be noticed is the magical disappearance of the 'x' from the denominator on the RHS.

Give it another go.

 

[BB]

addition

Should the problem read

x(x+2) [5/x + 3/x+2] = x(x+2) (-6/x+2)

5x +3(x+2) = -6x(x+2)

5(x + 2) + 3x = -6

 

[soroban]

Re: addition

Hello, BB!

Be careful! . . . You're making all kinds of errors . . .

\(\displaystyle \L\frac{5}{x}\, + \,\frac{3}{x\,+\,2}\;=\;\frac{-6}{x(x\,+\,2)}\)

Note that: \(\displaystyle \,x\,\neq\,0\) and \(\displaystyle x\,\neq\,-2\)


Multiply both sides by the common denominator: \(\displaystyle \:x(x\,+\,2)\) . . . Write it out!

. . \(\displaystyle \L x(x\,+\,2)\cdot\frac{5}{x}\,+\,x(x\,+\,2)\cdot\frac{3}{x\,+\,2} \;=\;x(x\,+\,2)\cdot\frac{-6}{x(x\,+\,2)\)


Reduce carefully:

. . \(\displaystyle \L \not{x}(x\,+\,2)\cdot\frac{5}{\not{x}}\,+\,x(\sout{x\,+\,2})\cdot\frac{3}{\sout{x\,+\,2}} \;=\;\not{x}(\sout{x\,+\,2})\cdot\frac{-6}{\not{x}(\sout{x\,+\,2})\)

And we get: \(\displaystyle \L\:5(x\,+\,2)\,+\,3x\;=\;-6\)

This simplifies to: \(\displaystyle \L\:8x\:=\:-16\;\;\Rightarrow\;\;x\:=\:-2\)


But \(\displaystyle x\,=\,-2\) cannot be a solution.

. . Therefore, this equation has no solution.

 

[BB]

addition

soroban

Thank you very much for the help this is very hard for me but i will not give up.