Question

answer with full explanation pls and fast​

Answer 1

Solution

Given :-

• Sonu is twice of age Monu
• Five year ago, Sonu's age was three times Monu's age .

Find :-

• Present age of Both .

Explanation

Let,

• Age of Sonu = x years
• Age of Monu = y years

According to question,

Case 1.

==> x = 2y

==> x - 2y = 0___________(1)

Case 2.

==> (x - 5) = 3 × (y - 5)

==> x - 3y = -15 + 5

==> x - 3y = -10__________(2)

Subtract equ(1) & equ(2)

==> -2y + 3y = 10

==> y = 10

Keep value of y in equ(1)

==> x - 2 × 10 = 0

==> x = 20

Hence

• Present age of Sonu (x) = 20 years
• Present age of Monu(y) = 10 years

_____________________

Answer Verification

Case 1.

(Sonu is twice of age Monu )

==> x = 2y

keep value of x & y .

==> 20 = 2 × 10

==> 20 = 20

L.H.S. = R.H.S.

Case 2.

==> x - 3y = -10

keep value of x & y .

==> 20 - 3 × 10 = -10

==> 20 - 30 = -10

==> -10 = -10

L.H.S. = R.H.S.

That's Proved

____________________

Answer 2

Step-by-step explanation:

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