Question

solve for m. picture inside! please help

Answer 1

\[\frac{ 1 }{ m } = \frac{ m - 34 }{2m ^{2} }\]

Answer 2

You could cross-multiply to eliminate the fractions. OR multiply both sides by the LCD - either method will work. Then you won't have any fractions

Then you collect all the terms on one side, set =0. It will be a (really simple) quadratic that you can solve by factoring.

Answer 3

okay that didn't help me at all because I don't know how to cross multiply those kinds of numbers or even find the LCD.

Answer 4

Do you know how to multiply m*m?

Answer 5

m^2

Answer 6

Good. Do you know how to multiply m(m-34) ?

Answer 7

m^2 - 34m

Answer 8

Excellent. Do you know how to multiply 2m^2 * (1) ? :)

Answer 9

so now I have 2m^2 = m^2 - 34m. now what.

Answer 10

then I can get m^2 = -34m

Answer 11

Collect al the terms on one side. I'd suggest moving the terms to the left side, just to make sure that the leading coefficient is positive.

Answer 12

Yes, exactly! Now bring that -34m on over to the LHS.... and you get?

Answer 13

34m + m^2 = 0 ?

Answer 14

Sure,t hat will work (although it's "standard" to put the x^2 term first, then the x term... but it won't make any real difference here).

Answer 15

but isn't there another step

Answer 16

OK, so you have now:

\(\Large 34m+m^2=0\)

That's a lovely little quadratic equation. Do you have any ideas on how to solve that equation? What do you do next?

Answer 17

Oh yes, we aren't done yet. :)

Answer 18

factor out an "m" from each term

Answer 19

m(m + 33m) = 0?

Answer 20

Why did you change the 34 to 33?

m * 33m = 33m^2, not 34m^2.

Answer 21

factor is the opposite of distribute. notice m(m+33m) will not give you the original

Answer 22

okay then what am I supposed to do.

Answer 23

Wait, did you fix your factoring step?

Answer 24

im stuck at m^2 + 34m = 0. don't know how to factor it out correctly.

Answer 25

write m^2 +34 m as m*m + 34*m
notice both terms are multiplied by m, so you can factor one "m" out

Answer 26

Here is an example with different numbers:

\[\Large 25x+x^2=x(25 + x)\]

Answer 27

okay so m(m + 34) = 0

Answer 28

if m=0 what is m(m + 34)= to ?

if (m+34) were 0 what is m(m + 34)= to ?

Answer 29

Good. so there is a rule called the Zero Factor Property. It says that:

If A*B=0 then A=0 or B=0.

Answer 30

s... m = 0 or 34 = 0 ?

Answer 31

m = 0 or m + 34 = 0 ?

Answer 32

Right!

Answer 33

so either m=0, or m = ??

Answer 34

34

Answer 35

or is it -34

Answer 36

You tell me? :) It is one and not the other.

Answer 37

I don't know!

Answer 38

m+34 = 0

What value of m makes that true?

Answer 39

-34

Answer 40

Well, does m=34 make it true?

Answer 41

no only -34

Answer 42

RIGHT! See, stop selling yourself short. You DO know! :) Just take a deep breath and think about it! :)

Answer 43

thank you for your help.

Answer 44

NOW, you have 2 solutions:

m=0 or m=-34

But there is just one little wrinkle here.....

Answer 45

THOSE are solutions to the equation:
\[\Large 34m+m^2=0\]
Which we obtained by cross-multiplying the ORIGINAL equation.

The original equation involves rational expressions. And with rational expressions, you always have to make sure that they are WELL DEFINED, which basically just means that you don't have a 0 in a denominator

So ALWAYS check your solutions to make sure that none of them give you a zero denominator. If so, that is an EXTRANEOUS SOLUTION... meaning that it ISNT a solution to the original equation. So just throw it out, and keep any that aren't extraneous.