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

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Answer 1 · 2017-07-17T08:28:56

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

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 · 2017-07-17T08:30:34

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

$3x + 11y = 4$
$-3x + 6y = -9$ (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 $y=-5/17$

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