-3b+ac=4
Move c to the left and - 3b to the right
ac-c=-4+3b
Now factor by c at the left
c(a-1)=-4+3b
Divide the right by a-1
C=(-4+3b)/(a-1)
Expand
x^2+3x-10=18
minus 18 both sides
x^2+3x-28=0
find what 2 numbers mutiply to -28 and add to 3
7 and -3
(x+7)(x-3)=0
set each to zero
x+7=0
x=-7
x-3=0
x=3
x=-7 and 3
The answer to your question is 120
So if John gives a pen and pencil to each and all of them are over we know that the no. of pencils and pens should be the same
So we will find the LCM of 15 and 40 that is equal to 120
So John brought 120/15 that is 8 packets of pen and 120/40 that is three packets of pencils
1 is L
2 is N
3 is M
Look back at your notes if you have any.
