Question

sum of 5 times of an number and 3 times of another number is 63. sum of 7 times of the first number and 3 times of second number is 81. which are the numbers ?​

Answer 1

Answer:

x= 9 and y =6

Step-by-step explanation:

let the first no be x and second be y

so according to question, 5x+3y=63 and 7x +3y=81 by solving the above equation you will get x=9 and y = 6

Answer 2

GivEn:

• Sum of 5 times of a number and 3 times of another number is 63.
• Sum of 7 times of the first number and 3 times of second number is 81.

To find:

• Original numbers?

Solution:

☯ Let the first and second number be x and y respectively.

★ According to the Question:

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

➯ 5x + 3y = 63 ⠀⠀⠀⠀⠀⠀⠀❬ eq (❶) ❭

• Sum of 7 times of the first number and 3 times of second number is 81.

➯ 7x + 3y = 81⠀⠀⠀⠀⠀⠀⠀ ❬ eq (❷) ❭

⠀━━━━━━━━━━━━━━━━━━━━━

Using Elimination method:

• Subtracting eq (1) from eq (2),

➯ (7x + 3y) - (5x + 3y) = 81 - 63

➯ 2x = 18

➯ x = 18/2

➯ x = 9

Now, Substituting value of "x" in eq (1),

➯ 5(9) + 3y = 63

➯ 45 + 3y = 63

➯ 3y = 63 - 45

➯ 3y = 18

➯ y = 18/3

➯ y = 6

Hence,

• The First number, x = 9
• The second number, y = 6