Question

HELP PLEASE. FINDING LINEAR EQUATION.

Answer 1

1. \[\frac{(x+2) }{(x+3)}=\frac{ (x+4) }{(x^2+x-6) }+\frac{(x-3) }{ (x-2) } \]
2. \[\frac{ 5 }{ 7x }-3=\frac{ 7x }{ 2 }+\frac{ 3 }{ 14 }\]
3. \[\frac{ 1 }{ 3 }(2x-\frac{ 6 }{ 5 })=\frac{ 1 }{ 2 }(x -\frac{ 4 }{ 5 })\]
4. \[2\left[ 2-\left( 4+2x \right) \right]=6\left[ x-4\left( x+1 \right) \right]\]

Answer 2

anything with an x in the denominator will not be linear

Answer 3

that narrows our options to the last 2

Answer 4

the last one looks simpler to operate thru, so see if that simplifies to a constant is my thought

Answer 5

oh, can you give me the factor of no.1 ill try to solve it and show it to you.

Answer 6

6 reduces to 3 and 2 soo

(x+3)(x-2)

Answer 7

x+2(x-2)=x+4+x-3(x+3)

Answer 8

a curve is not a line; any x with a power greater then 1 will curve ...

Answer 9

\[\frac{(x+2) }{(x+3)}=\frac{ (x+4) }{(x^2+x-6) }+\frac{(x-3) }{ (x-2) }\]
\[\frac{(x+2)(x-2) }{(x+3)(x-2)}=\frac{ (x+4) }{(x^2+x-6) }+\frac{(x+3)(x-3) }{ (x+3)(x-2) }\]

\[\frac{x^2-4}{(x+3)(x-2)}=\frac{ (x+4) }{(x^2+x-6) }+\frac{x^2-9}{ (x+3)(x-2) }\]

Answer 10

\[\frac{ 5 }{x+3 }-\frac{ 3 }{ 2x-5 }=\frac{ 15 }{ 2x^2+x-15 }\]
\[5(2x-5)-3(x+3)=15\]
\[10x-25-3x-9=15\]
\[10x-3x=15+25+9\]
\[\frac{ 7x }{ 7 }=\frac{ 49 }{ 7 }\]
\[x=7\]

can this be applied to 1?

Answer 11

i simple test is to determine if, for all values x,y ... that f(x+y) = f(x) + f(y)

Answer 12

i dont understand it yet, still can you help me out with the other 3 numbers?

ill try to study it later.

Answer 13

one method is:
\[\frac{ 5 }{ 7x }-3=\frac{ 7x }{ 2 }+\frac{ 3 }{ 14 }\]
\[\frac{ 5 }{ 7x }-3-\frac{ 7x }{ 2 }-\frac{ 3 }{ 14 }=0\]
\[\frac{ 5(7x) }{ 7x }-3(7x)-\frac{ 7x(7x) }{ 2 }-\frac{ 3(7x) }{ 14 }=0\]
\[5-21x-\frac{ 49 }{ 2 }x^2-\frac{ 3x }{ 2 }=0\]
notice that we have an x^2 in here, therefore this wont be linear

Answer 14

ok, in my understanding if theres an x in a denominator it will be automatically not be a linear

Answer 15

lets test out this f(x+y) = f(x) + f(y); let x = 1, and y = 2
\[\frac{ 5 }{ 7x }-3-\frac{ 7x }{ 2 }-\frac{ 3 }{ 14 }=0\]
\[\frac{ 5 }{ 7(3) }-3-\frac{ 7(3) }{ 2 }-\frac{ 3 }{ 14 }\\~~~~~~~=\frac{ 5 }{ 7(1) }-3-\frac{ 7(1) }{ 2 }-\frac{ 3 }{ 14 }+\frac{ 5 }{ 7(2) }-3-\frac{ 7(2) }{ 2 }-\frac{ 3 }{ 14 }\]

\[\frac{ 5 }{ 21}-3-\frac{ 21}{ 2 }-\frac{ 3 }{ 14 }\\~~~~~~~=\frac{ 5 }{ 7 }-3-\frac{ 7 }{ 2 }-\frac{ 3 }{ 14 }+\frac{ 5 }{ 14 }-3-\frac{ 14 }{ 2 }-\frac{ 3 }{ 14 }\]
my calculator says:
-13.476... = -15.857...

Answer 16

clearly thats false

Answer 17

x in the denominator becomes: x^(-1) which of course is not x^1

so an x in the denominator is a good indication that its not linear

Answer 18

\[\frac{ 1 }{ 3 }(2x-\frac{ 6 }{ 5 })=\frac{ 1 }{ 2 }(x -\frac{ 4 }{ 5 })\]
\[\frac{ 2 }{ 3 }x-\frac{ 2 }{ 5 }=\frac{ 1 }{ 2 }x -\frac{ 2 }{ 5 }\]
\[\frac{ 2 }{ 3 }x=\frac{ 1 }{ 2 }x \]
\[\frac{ 4 }{ 3 }x=x \]
\[\frac{ 4 }{ 3 }x-x=0 \]
\[\frac{ 1 }{ 3 }x=0 \]

Answer 19

in this case
\[\frac{1}{3}(x)+\frac13(y)=\frac{1}{3}(x+y)\]
so its linear

Answer 20

ok, i get it, but im lost at one and two.

Answer 21

\[2\left[ 2-\left( 4+2x \right) \right]=6\left[ x-4\left( x+1 \right) \right]\]
\[4-2( 4+2x )= 6x-24( x+1 )\]
\[4-8-4x = 6x-24x-24\]
\[-4x-4 = -18x-24\]
\[14x+20=0\]this of course is linear as well
\[14(x+y)+20=14(x)+14(y)+20\]
\[14(x+y)=14(x)+14(y)\]

Answer 22

well, we simplified number 2 as

\[5-21x-\frac{ 49 }{ 2 }x^2-\frac{ 3x }{ 2 }=0\]
\[10-42x-49x^2-3x=0\]
\[49x^2+45x-10=0\]the x^2 is a dead give away that this is not linear

\[49(x+y)^2+45(x+y)-10=49(x)^2+45(x)-10~+~49(y)^2+45(y)-10\]
\[49(x^2+2xy+y^2)+45(x)+45(y)-10=49(x)^2+45(x)-10~+~49(y)^2+45(y)-10\]
\[49(x^2+2xy+y^2)-10=49(x)^2-10~+~49(y)^2-10\]
\[49(x)^2+96xy+49(y)^2-10=49(x)^2-10~+~49(y)^2-10\]
\[96xy-10=-10~-10\]
\[96xy=-10\]

Answer 23

wow, this thing is going to give me much headache. now im going to review it.
thank you very much for your effort and patience mr amistre, appreciated it.

see you soon.

Answer 24

good luck :)

Answer 25

thank you, btw how did you get no. 3? i cant seem to follow it.

is it always x=0?

Answer 26

if we forget the more advanced: f(x+y) = f(x) + f(y)

the setup simplified to a linear form, \(\frac13x\)