Question

Three times of a number added to two times of another number gives 33. When three times of the second number is subtracted from four times the other number got 10. taking the numbers as x and y form the equations and find the numbers?​

Answer 1

Let one number be x and other be y

A/q

3x+2y=33____(i)

4x-3y=10_____(ii)

Multiplying (i) by 3 and (ii) by 2, we get:

9x+6y=99____(iii)

8x-6y=20____(iv)

Adding (iii) and (iv)

9x+6y+8x-6y=99+20

17x=119

x=7

Putting, x=7, in (i) we get:

21+2y=33

2y=33-21=12

y=6

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Answer 2

Answer:

Let the no. be x and y

A/Q

3x+2y=33...(i)

4x-3y=10...(ii)

multiplying (i) by 3 and (ii) by 2, we get

9x+6y=99...(iii)

8x-6y=20...(iv)

by adding eq. (iii) and (iv), we have

17x = 119

So, x=7

now, putting value of x in eq. (i)

3×7+2y=33

21+2y=33

2y=33-21

2y= 12

y = 6

hope it's help you amit bro..