Question

. Five times a number added to three times another gives 41. Three times the second number subtracted from 8 times the first number is 11. Find the numbers

Answer 1

We have :-

• 5 times a number added to three times another gives 41.
• Three times the second number subtracted from 8 times the first number is 11.

To find :-

• Find the numbers .

Solution :-

• let first number be x .
• let second number be y .

A .T.Q

• when five times a number added to three times another ,it gives 41 as a result . i.e, 5x + 3y = 41 ...............eq.(1)
• when 3 times the second number subtracted from 8 times the first numbe, it gives 11 . i.e, 8x – 3y = 11 ...............eq.(2)

$\\ \sf \: 5x + 3y = 41 \: ................. \: eq.(1)$

$\\ \sf \: 8x - 3y = 11 \: ................. \: eq.(2)$

Add eq.(1) and eq.(2) , we get :--

$\sf \: 5x + 3y = 41 \\ \sf \: 8x - 3y = 11 \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ : \implies \sf13x = 52$

$: \implies\sf \: x = \dfrac{52}{13}$

$: \implies\bf \fbox{\bf x= 4}$

Put the value of x = 4 in eq.(1)

$: \implies \sf \: 5 \times 4 + 3y = 41$

$: \implies\sf \: 20 + 3y = 41$

$: \implies \sf \: 3y = 21$

$: \implies\sf \: y = \dfrac{21}{3}$

$: \implies\bf \fbox{\bf y = 7}$

Therefore, first number = 4 .

second number = 7 .

Answer 2

Answer:

5 times a number added to three times another gives 41.

Three times the second number subtracted from 8 times the first number is 11.

To find :-

Find the numbers .

Solution :-

let first number be x .

let second number be y .

A .T.Q

when five times a number added to three times another ,it gives 41 as a result . i.e, 5x + 3y = 41 ...............eq.(1)

when 3 times the second number subtracted from 8 times the first numbe, it gives 11 . i.e, 8x – 3y = 11 ...............eq.(2)

Add eq.(1) and eq.(2) , we get :--

Put the value of x = 4 in eq.(1)

Therefore, first number = 4 .

second number = 7 .

Step-by-step explanation: