Leah is 6 years older, so X + 6

John is 5 older than that, so X + 11

X + (X + 6) + (X + 11) = 41

3X + 17 = 41 (add all the X’s and all the numbers)

3X = 24 (subtract 17 from both sides)

X = 8 (divide both sides by 3)

Sue is 8 years old. To test your answer:

8+(8+6)+(8+6+5) = 41

S= Sue

J= John

Since Leah is 6 years older than Sue,

L= 6+S

and John is 5 years older than Leah’s age (6+S)

J=5+(6+S)

=11+S

S+J+L=41

Sub “J=11+S” and “L= 6+S” into J and L:

S+(11+S)+(6+S)=41

Simplify:

3S+17=41

Subtract 17 from both sides:

3S=24

Divide both sides by 3:

S=8

Sue is eight!

Sue + Leah + John = 41

Y + (Y+6) + (y + 6 + 5) = 41

sub from both sides ( – 6 – 11)

y + y + y = 24

3y = 24

3y/3 = 24/3 Divide 3 from both sides

y = 8

Sue is 8 years old.

Leah is 14 years old.

John is 19 years old.

L = S+6

J = S+6+5

S+L+J=41

S+S+6+S+6+5 = 41

3S + 17 = 41

3S = 41 – 17

3 S = 24 => S = 8

S = 8

L = 14

J = 19

8+14+19=41

Sue = x; Leah = 6 + x; John = 11+ x

x + (6 + x) + (11 + x) = 41

3x + 17 = 41

3x = 24

x = 8

Sue is 8

Leah = x +6

John = x + 6 + 5

the sum their ages,

x + ( x + 6) +( x+ 6 + 5) = 41

Simplify:

3x + 17 = 41

3x = 24

So, dear little Susie is 8.