Question

plzzz help

Answer 1

@pooja195

Answer 2

So you are solving for b, yes?

Answer 3

yes

Answer 4

Ok, well to solve for b, what we want to do is get b alone. This means dividing the fraction away. When we divide a fraction, what we really do is multiply by its inverse. What would the inverse of \[\large \sf -\frac{1}{3}\] be?

Answer 5

inverse?

Answer 6

Yes, the inverse. If a normal fraction is \[\large \sf \frac{numerator}{denominator}\] the inverse is \[\large \sf \frac{denominator}{numerator}\] For example, the inverse of \(\large \sf \frac{1}{2}\) would be \(\large \sf \frac{2}{1}\)

Answer 7

ohhhh ok

Answer 8

-3/1

Answer 9

So now you would multiply \(\large \sf -\frac{3}{1}\) on both sides of the equation.

Answer 10

ok

Answer 11

hmmm im kinda stuck on this one

Answer 12

Remember that -3/1 is the same as -3. Also, remember than inverses cancel out with their original fraction.

Answer 13

ohhh so now its -3b=9

Answer 14

No no, we'd get to this point of \[\large \sf -\frac{3}{1} \times -\frac{1}{3}b~=~9 \times -3\] Now since negatives cancel out, the left side would be more like this \[\large \sf \frac{3}{1} \times \frac{1}{3}b\] and the left side would be \[\large \sf 9 \times -3\]

Answer 15

so b=27?

Answer 16

Close, but you are multiplying 9 by a NEGATIVE 3, so your end product of multiplication will be negative.

Answer 17

ojhhhh -27 lolthanks

Answer 18

No problem :)