Question

How do you solve the following system: #11y + 3x = 4, x-2y=3 #?

Answer 1

`x = 41/17`
`y = -5/17`

Explanation:

Let,
`3x + 11y = 4` be eq 1
`x - 2y = 3` be eq 2

now, multiply eq 2 by -3 so as it will be cancelled out in the next step.
we get,
`-3x + 6y = -9`

`rArr 3x + 11y =4`
`-3x + 6y = -9`

`rArr 17y = -5`
`therefore y =-5/17`
now, substitute value of y in any equation,
u get `x = 41/17`

ENJOY MATHS !!!!!

Answer 2

By arranging equation, `x=41/17` and `y=-5/17`

Explanation:

`3x + 11y = 4`
`-3x + 6y = -9` (when you enlarge the second equation with -3)
Combine these two equations and get
`17y = -5`

`y = -5/17`

Put this value in any original equation (for instance the first)

`3x - 55/17 = 4`

`3x = (68+55)/17`

`x = 123/51`

`x= 41/17`

This is your answer `x=41/17` and `y=-5/17`