Question

help again D: lol
y/(y + 3) = 1/(y + 3)

Answer 1

i'm not good at fractions

Answer 2

The denominators are the same

So the numerators must be equal

Answer 3

okay

Answer 4

Or you can cross multiply

y/(y + 3) = 1/(y + 3)

y(y + 3) = 1(y + 3)

y^2 + 3y = y + 3

y^2 + 3y - y - 3 = 0

y^2 +2y - 3 = 0

(y+3)(y-1) = 0

y+3=0 or y-1=0

y=-3 or y=1

So the *possible* solutions are y = -3 or y = 1, but when y = -3, we get a division by zero error. So we throw out y = -3

Therefore the only solution is y = 1

Answer 5

oh I forgot all about cross multiplying hah
thank you :)

Answer 6

This backs up what I said at the top

"The denominators are the same So the numerators must be equal"

So if the denominators are the same (which they are -- they're both y+3), then the numerators y and 1 must be the same

aka

y = 1

Answer 7

you're welcome

Answer 8

that makes a lot of sense :) good job

Answer 9

thanks

Answer 10

despite the fact that you can use this trick, i'd still go with cross multiplication since that works in a lot more cases

Answer 11

can you help me on one more?

Answer 12

(x^2 + 2x)/x * (3x - 6)/(x^2 - 4)

Answer 13

sure

Answer 14

x^2 + 2x factors to x(x+2) when you factor out the GCF x

3x-6 factors to 3(x-2) when you factor out the GCF 3

and

x^2-4 factors to (x-2)(x+2) by the difference of squares rule

-------------------------------------------------------

So we then get...

\[\Large \frac{x^2+2x}{x} * \frac{3x-6}{x^2-4}\]

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

\[\Large \frac{x(x+2)}{x} * \frac{3(x-2)}{x^2-4}\]

\[\Large \frac{x(x+2)}{x} * \frac{3(x-2)}{(x-2)(x+2) }\]

which combines to

\[\Large \frac{3x(x+2)(x-2)}{x(x-2)(x+2)} \]

See anything cancelling?

Answer 15

hmm possibly everything but the 3 on the end?

Answer 16

exactly

Answer 17

awesome :)

Answer 18

\[\Large \frac{x^2+2x}{x} * \frac{3x-6}{x^2-4}\]

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

\[\Large \frac{x(x+2)}{x} * \frac{3(x-2)}{x^2-4}\]

\[\Large \frac{x(x+2)}{x} * \frac{3(x-2)}{(x-2)(x+2) }\]

\[\Large \frac{3x(x+2)(x-2)}{x(x-2)(x+2)} \]

\[\Large \frac{3x(x+2)(x-2)}{x(x-2)(x+2)} \]

\[\Large \frac{3x\cancel{(x+2)}(x-2)}{x(x-2)\cancel{(x+2)}} \]

\[\Large \frac{3x(x-2)}{x(x-2)} \]

\[\Large \frac{3x\cancel{(x-2)}}{x\cancel{(x-2)}} \]

\[\Large \frac{3x}{x} \]

\[\Large \frac{3\cancel{x}}{\cancel{x}} \]

\[\Large \frac{3}{1} \]

\[\Large 3 \]


So
\[\Large \frac{x^2+2x}{x} * \frac{3x-6}{x^2-4} = 3\] for all allowed values of x.

Answer 19

oops have one duplicated step, but oh well...lol

Answer 20

lol not a problem :) thanks for all the help
you really are awesome

Answer 21

you're welcome

Answer 22

thanks for the compliment

Answer 23

super math dude

Answer 24

lol to the rescue...

Answer 25

hint: solve by cross multiply

Answer 26

y=1 {Denominators are same}