Question

Find the two nos such that there
times the fuest no added to twice the second no is equal to 36, and
twice the first no added to second
no is equal to 21.​

Answer 1

Given :

• Three times the first number when added to twice the second number gives sum of 36.
• And when 2 times first number is added to second number is 21.

To Find :

• The Two Numbers.

Solution :

Let the first number be x.

Let the second number be y.

Case 1 :

When the first number is multiplied by 3 and second number is multiplied by 2, the sum of product is 36.

Equation :

$\sf{\longrightarrow{3x\:+\:2y\:=\:36}}$

$\sf{\longrightarrow{3x=36-2y}}$

$\sf{\longrightarrow{x=\dfrac{36-2y}{3}}\:\:\:\:\:(1)}$

Case 2 :

Twice the first number added to the second number gives sum of 21.

Equation :

$\sf{\longrightarrow{2x+y=21}}$

$\sf{\longrightarrow{2\:\Big(\dfrac{36-2y}{3}\Big)+y=21}}$

$\bold{\Big[From\:equation\:(1)\:x\:=\:\dfrac{36-2y}{3}\Big]}$

$\sf{\longrightarrow{\Big(\dfrac{72-4y}{3}\Big)+y=21}}$

$\sf{\longrightarrow{\dfrac{72-4y}{3}+y=21}}$

$\sf{\longrightarrow{\dfrac{72-4y+3y}{3}=21}}$

$\sf{\longrightarrow{72-4y+3y=21\:\times\:3}}$

$\sf{\longrightarrow{72-4y+3y=63}}$

$\sf{\longrightarrow{-4y+3y=63-72}}$

$\sf{\longrightarrow{-y=-9}}$

$\sf{\longrightarrow{y=9}}$

Substitute, y = 9 in equation (1),

$\sf{\longrightarrow{x=\dfrac{36-2y}{3}}}$

$\sf{\longrightarrow{x=\dfrac{36-2(9)}{3}}}$

$\sf{\longrightarrow{x=\:\dfrac{36-18}{3}}}$

$\sf{\longrightarrow{x=\:\dfrac{18}{3}}}$

$\sf{\longrightarrow{x=6}}$

$\large{\boxed{\bold{First\:Number\:=\:x\:=\:6}}}$

$\large{\boxed{\bold{Second\:Number\:=\:y\:=\:9}}}$

Answer 2

Correct question: Find the two numbers such that three times the first number added to twice the second number is equal to 36. And, twice the first number added to second number is equal to 21.

Answer:

$\large{\boxed{\red{\bf 6 \: and \: 9}}}$

Step-by-step explanation:

Let the first number be x. And, the second number be y.

When three times the first number added to twice the second number, the sum is equal to 36.

Equation :

$\implies \sf 3x + 2y = 36\\\\\implies \sf 2y = 36 - 3x \\\\\implies \sf y= \frac{36-3x}{2} \: \: \dots (i)$

Again, when twice the first number added to second number, the sum is equal to 21.

Equation :

$\implies \sf 2x + y = 21 \\\\\implies \sf y = 21 - 2x \: \: \dots (ii)$

On equating both the equation (i) and (ii),

$\implies \sf \frac{36-3x}{2} = 21 - 2x \\\\\underline{\bf On \: cross-multiplying :} \\\\\implies \sf 36-3x = 2(21-2x) \\\\\implies \sf 36 - 3x = 42 - 4x \\\\\implies \sf - 3x + 4x = 42 - 36 \\\\\underline{\boxed{\bf \therefore \: x = 6}}$

Substituting the value of x in (i),

$\sf \implies y = \frac{36-3x}{2} \\\\\implies \sf y = \frac{36-3(6)}{2} \\\\\implies \sf y = \frac{36-18}{2}\\\\\implies \sf y = \frac{18}{2} \\\\\underline{\boxed{\bf \therefore \: y = 9} }$

Hence, the required numbers are 6 and 9.