Question

Four years ago peter was three times as old as Sylvia, and in 5 years the sum of their ages will be 38 years. Find their present ages

Answer 1

Answer:

Let P = Peter's age and S = Sylvia's age.

We will first translate "four years ago Peter was three times as old as Sylvia." This can be written P = 4 + 3*(S-4).

Next, we translate "in 5 years the sum of their ages will be 38 years" as (P + 5) + (S + 5) = 38.

Now we have two unknowns (P & S) and two equations, so we are ready to solve. We substitute our first equation for P into our second equation:

(4+3*(S-4)+5) + (S+5) = 38

9 + 3S - 12 + S + 5 = 38

2 + 4S = 38

4S = 36

S = 9.

Substitute this back into the first equation to find P = 4+3*(S-4) = 4+3*(9-4) = 4+15 = 19.

Answer 2

$\bold\green{\underline{Let:}}$

The present age of Sylvia be Y

The present of age of Peter be X

$\bold\green{\underline{Given:}}$

Four years ago Peter was 3 times as old as Sylvia,

The age of Sylvia=3(y-4)

The age of Peter=X-4

⟹X-4=3y-12

⟹X-3Y=-8

$\bold\purple{\boxed{x-3y=-8-----(1)}}$

Than,

In 5 years the sum of their ages will be 38 years

The age of Sylvia=Y+5

The age of Peter=X+5

$\bold\red{\underline{A.T.Q}}$

⟹X+5+Y+5=38

⟹X+Y=38-10

⟹X+Y=28

$\bold\purple{\boxed{X+y=28-----(2)}}$

Now , solving equation 1 and 2

⟹X-3Y=-8

⟹X+Y=28

⟹-4Y=-36

⟹Y=9

Than, putting the value of y in EQ 1

⟹X-3Y=-8

⟹X-3*9=-8

⟹X=-8+27

⟹X=19

Hence,

$\bold\purple{\boxed{The\: present\:age\:of\: Sylvia=9\: years}}$

$\bold\orange{\boxed{The\:present\:age\:of\:Peter=19\: years}}$