Question

sum of 5 times of an number and 3 times of another number is63. sum of 7 times of the first number and 3 times of second number is 81. which are the numbers?​

Answer 1

Answer:

first number = 9 and second number = 6

Step-by-step explanation:

let the first number be x and the second number be y.

then by the given conditions we get

5x + 3y = 63 ( first equation )

7x + 3y = 81 ( second equation)

subtracting the second equation with the first equation we get:

2x = 18

x = 18/2 = 9

x = 9

substitute x = 9 in first equation :

(5 × 9) + 3y = 63

45 + 3y = 63

3y = 63 - 45 = 18

3y = 18

y = 18/3 = 6

y= 6

therefore the first number(x) = 9

the second number(y) = 6

Answer 2

$\sf{\underline{\boxed{\large{\blue{\mathsf{Solution}}}}}}$

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

• Sum of 5 times of a number and 3 times of a number is 63 .
• Sum of 7 times a number and 3 times of a number is 81 .

_______________________

$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

• Find the original no. = ??

______________________

$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

Let the first no. be x and second no. be y

• Sum of 5 times of a number and 3 times of a number is 63 .

5x + 3y = 63 ---- ( i )

• Sum of 7 times of a number and 3 times of a number is 81

7x + 3y = 81 -----( ii )

• Subtracting eq ( i ) from ( ii )

7x + 3y - 5x - 3y = 81 - 63

2x = 18

x = 9

X = 9

• putting value of x in eq i

5 x 9 + 3y = 63

45 + 3y = 63

3y = 63 - 45

3y = 18

y = 6

Y = 6

_______________

$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

• First no = 9
• Second no = 6