Question

Two numbers differ by 3. The sum of twice the smaller no. & thrice the
greater no. is 28. find the numbers.
​

Answer 1

Given :

• Difference of the two numbers = 3

• Sum of twice of smaller number and thrice of smaller number = 28.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

To Find :

⠀⠀⠀⠀⠀⠀⠀The actual numbers.

Solution :

Let the two numbers be x and y.

So According to the Question, their Difference is 3 , so the equation formed is :

$\underline{\bf{x - y = 3}}$⠀⠀⠀⠀⠀⠀⠀Eq.(i)

Now , it said that the sum of twice of smaller number and thrice of bigger no. is 28.

So let the smaller no. be y and the bigger number be x , so we get the Equation as :

$\underline{\bf{3x + 2y = 28}}$⠀⠀⠀⠀⠀⠀⠀Eq.(ii)

Now, putting the two Equations together and solving them by Elimination method , we get :

⠀⠀$\bf{x - y = 3}$

⠀⠀$\bf{3x + 2y = 28}$

⠀ ￣￣￣￣￣￣

Now , multiplying 2 in the first Equation , we get :

⠀⠀⠀⠀⠀⠀⠀ ⠀$\bf{x - y = 3 \times (2)}$⠀⠀⠀⠀ ⠀

⠀⠀⠀⠀⠀⠀⠀⠀ $\bf{3x + 2y = 28}$

⠀⠀⠀⠀ ⠀⠀⠀⠀￣￣￣￣￣￣

⠀⠀$\bf{2x - 2y = 6}$

⠀⠀$\bf{3x + 2y = 28}$

⠀⠀ ￣￣￣￣￣￣⠀⠀⠀⠀

Now , by adding , we get :

⠀⠀$\bf{2x = 6}$

⠀⠀$\bf{3x = 28}$

⠀⠀￣￣￣￣

$\bf{5x = 34}$

$:\implies \bf{x = \dfrac{34}{5}}$

Hence, the value of x is 34/5.

Now , putting the value of x in the equation (i) , we get :

$:\implies \bf{x - y = 3}$

$:\implies \bf{\dfrac{34}{5} - y = 3}$

$:\implies \bf{ - y = 3 - \dfrac{34}{5}}$

$:\implies \bf{ - y = \dfrac{15 - 34}{5}}$

$:\implies \bf{ - y = \bigg(\dfrac{-19}{5}\bigg)}$

$:\implies \bf{ \not{-} y = \not{-} \dfrac{19}{5}}$

$:\implies \bf{y = \dfrac{19}{5}}$

Hence, the values of x and y are 34/5 and 19/5 respectively.

Answer 2

Given:

• Difference between two numbers = 3

• Sum of twice of the smaller number and thrice the greater number = 28

Find:

• What are the numbers = ?

Solution:

Let, the larger number be p

and the smaller number be q

Now,

$\pink{\bold{ACCORDING \: TO \: QUESTION}}$

➨ $\sf p - q = 3.......(i)$

➨ $\sf 3p + 2q = 28.......(ii)$

Now, in eq(i)

$\sf\to p - q = 3$

$\sf\to p = 3 + q$

Put this value of p in eq(ii)

$\sf \longmapsto 3p + 2q = 28$

$\sf \longmapsto 3(3 + q) + 2q = 28$

$\sf \longmapsto 9+ 3q+ 2q = 28$

collect like terms

$\sf \longmapsto 3q+ 2q = 28 - 9$

$\sf \longmapsto 5q = 19$

$\underline{\boxed{\sf \hookrightarrow q = \dfrac{19}{5}}}$

Now, substitute this value of q in eq(i)

$\sf \leadsto p - q = 3$

$\sf \leadsto p - \dfrac{19}{5} = 3$

$\sf \leadsto p = 3 + \dfrac{19}{5}$

$\sf \leadsto p = \dfrac{15 + 19}{5}$

$\underline{\boxed{\sf \to p = \dfrac{34}{5}}}$

Hence, the value of p will be $\dfrac{34}{5}$ and the value of q will be $\dfrac{19}{5}$