A box contains 42 red counters, 90 yellow counters and no other counters. A group of students share these counters between them so that they each receive the
same number of red counters and the same number of yellow counters.
There are no counters left over,
How many students could be in the group?
Give all possible answers and show your reasoning.

Please help

jeff jeff

step-by-step explanation:

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !

the graph is negative for all real values of x where   -6 < x < -2.

step-by-step explanation:

this refers to the values of f(x) in the area below the x-axis.