Question

Solve the following for all real values of x.
2/x+1=x-2/2

5/e^x +1=1

Answer 1

\[\frac{ 2 }{ x+1 }=\frac{ x- 2}{ 2 }\]

or

\[\frac{ 2 }{ x }+1=\frac{ x-2 }{ 2 }\]

Answer 2

Which one?

Answer 3

reply @peanut1995

Answer 4

first one

Answer 5

ok

Answer 6

\[\frac{ 2 }{ x+1 }=\frac{ x-2 }{ 2 }\]

Cross Multiply
\[4=(x+1)(x-2)\]

\[4=x^2-2x+x-2\]

\[4=x^2-x-2\]

\[x^2-x-6=0\]

Factorize

\[(x+2)(x-3)=0\]

\[x=-2,3\]

Answer 7

\[\frac{ 5 }{ e^x+1 }=1\]

or

\[\frac{ 5 }{ e^x }+1=1\]

Answer 8

Which one?

Answer 9

first one

Answer 10

ok

Answer 11

\[\frac{ 5 }{ e^x+1 }=1\]

Cross Multiply

\[5=e^x+1\]

\[e^x=5-1\]

\[e^x=4\]

Take the natural log of both sides: Note: ln[e] = 1

e^x = 4

ln[e^x] = ln[4]

x * ln[e] = ln[4]

x = ln[4]

x ≈ 1.386

Answer 12

okay would you mind helping me on one more?

Answer 13

i will try.

Answer 14

\[\sqrt{x-1} -5/\sqrt{x-1}=0\]

Answer 15

\[\sqrt{x-1}-\frac{ 5 }{\sqrt{x-1} }=0\]

Taking LCM

\[\frac{ \sqrt{x-1}\sqrt{x-1}-5 }{ \sqrt{x-1} }=0\]

Cross Multiply

\[x-1-5=0\]

\[x-6=0\]

\[x=6\]

Answer 16

Thank you so so much:) I really appreciate it

Answer 17

My Pleasure.