Question

different of two numbers is 3 and the differens of their squres is 54 what is the numbers ? (pairs of equation)​

Answer 1

Answer :

x = 21/2 , y = 15/2

Solution :

Let the required numbers be x and y such that x > y .

According to the question , the difference between two numbers is 3 .

Thus ,

x - y = 3 -----(1)

Also ,

The difference of their squares is 54 .

Thus ,

x² - y² = 54 -----(2)

Eq-(2) can be rewritten as ;

=> (x - y)(x + y) = 54

=> 3(x + y) = 54

=> x + y = 54/3

=> x + y = 18 ------(3)

Now ,

Adding eq-(1) and (3) , we have ;

=> x - y + x + y = 3 + 18

=> 2x = 21

=> x = 21/2

Now ,

Putting x = 21/2 in eq-(1) , we have ;

=> x - y = 3

=> y = x - 3

=> y = 21/2 - 3

=> y = (21 - 6)/2

=> y = 15/2

Hence ,

x = 21/2 , y = 15/2

Answer 2

$\huge\bold{\underline{Question:}}$

different of two numbers is 3 and the differens of their squres is 54 what is the numbers ? (pairs of equation)

$\huge\bold{\underline{Answer:}}$

Given:

• Difference of two numbers is 3
• The difference of their square is 54

To find:

what is the numbers ?

Solution:

Let the two numbers be ' x ' and ' y '

According to the question:

x - y = 3 ............ (1)

and

x² - y² = 54...........(2)

From equation (2) we get,

$\sf{\implies x² - y² = 54}$

$\sf{\implies (x+y)(x-y) = 54}$

$\sf{\implies (x+y)×3 = 54}$ [from equation (I)]

$\sf{\implies x+y =\dfrac{54}{3}}$

$\sf{\implies x+y = 18}$..........(3)

Adding equation (1) and (3) ,we get

$\sf{\implies 2x = 21}$

$\sf{\implies x =\dfrac{21}{2}}$

Now, putting the value of X in equation (1)

$\sf{\implies \dfrac{21}{2} - y = 3}$

$\sf{\implies y = \dfrac{21}{2}-3}$

$\sf{\implies y = \dfrac{21-6}{2}}$

$\sf{\implies y = \dfrac{15}{2}}$

Therefore,

the given numbers are = 21/2 and 15/2

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