Question

Solve each of the following system of equations by elimination method.

Answer 1

SOLUTION :
Given :
65x -33y = 97 ……….(1)
33x -65y = 1…………..(2)

On Adding eq 1 &2
98x - 98y = 98
x - y = 1 ………….,.(3)
[ On dividing both sides by 98]

On Subtracting eq 2 & from eq 1,
32x + 32y = 96
x + y = 3 ………………(4)
[ On dividing both sides by 32]

On Adding eq 3 & 4,
x - y = 1
x + y = 3
-------------
2x = 4
x = 4/2= 2
x = 2

On Substituting x = 2 in eq 3,
x - y = 1
2 - y = 1
2 -1 = y
y = 1

Hence, x = 2 & y = 1 is the required solution.

HOPE THIS WILL HELP YOU….

Answer 2

Solution:-

given by:'
$65x - 33y = 97.........(1) \\ 33x - 65y = 1............(1) \\ multipy \: by \: in \: eq \: (1)with \: 33 \: and \: in \: eq(2) \: with \: 65 \: we \: get \\ 2145x - 1089y = 3201.........(3) \\ \\ 2145x - 4225y = 65..........(4) \\ substrac \: eq(1) \: in \: (2)... \\ - 3136 y= - 3136 \\ y = 1 \\ put \: y = 1 \: in \: eq(1). \\ 65x = 97 + 33 \\ x = \frac{130}{65} \\ x = 2$
here (x , y) = (2 ,1)

■I HOPE ITS HELP■