Richard is 6 years older than Lisa, and in 5 years the sum of their ages will be twice as much as the sum of their ages 4 years ago. Find their present ages.
R = L + 6
R + 5 + L + 5 = 2 ( R - 4 + L - 4 )
R + L + 10 = 2 ( R + L - 8 )
R + L + 10 = 2 * R + 2 * L - 2 * 8
R + L + 10 = 2 R + 2 L - 16
Replace R = L + 6 in this equation
L + 6 + L + 10 = 2 ( L + 6 ) + 2 L - 16
2 L + 16 = 2 * L + 2 * 6 + 2 L - 16
2 L + 16 = 2 L + 12 + 2 L - 16
2 L + 16 = 4 L - 4 Add 4 to both sides
2 L + 16 + 4= 4 L - 4 + 4
2 L + 20 = 4 L Subtract 2 L to both sides
2 L + 20 - 2 L = 4 L - 2 L
20 = 2 L Divide both sides by 2
10 = L
L = 10 yrs
R = L + 6 = 10 + 6 = 16 yrs
To solve this problem, we can first set up equations using the given information.
Let's assume Lisa's age is L, and Richard's age is R.
According to the problem, Richard is 6 years older than Lisa. So we can write the equation:
R = L + 6 ----- (equation 1)
We also know that in 5 years, the sum of their ages will be twice as much as the sum of their ages 4 years ago. Let's calculate their ages in 5 years.
Lisa's age in 5 years = L + 5
Richard's age in 5 years = R + 5
The sum of their ages in 5 years:
(L + 5) + (R + 5) = 2 × [(L - 4) + (R - 4)]
L + 5 + R + 5 = 2(L - 4 + R - 4)
Simplifying this equation:
L + R + 10 = 2L - 8 + 2R - 8
L + R + 10 = 2L + 2R - 16
Now, let's rearrange equation 1 and substitute it into this equation:
L + R + 10 = 2(L + 6) + 2R - 16
L + R + 10 = 2L + 12 + 2R - 16
L + R + 10 = 2L + 2R - 4
Subtracting L and R from both sides:
10 = L - 4
Moving -4 to the left side:
L = 10 + 4 = 14
Now, substitute the value of L into equation 1:
R = L + 6 = 14 + 6 = 20
So, Lisa's present age is 14, and Richard's present age is 20.