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

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Answer 1 · 2017-06-14T14:22:04

By arranging equations. $x=32/13$ and $y=-4/13$

$11y + 3x = 4$
$3y + 2x = 4$

Enlarge the first equation by 2 and the second by -3:

$22y + 6x = 8$
$-9y - 6x = -12$

Now combine these:

$22y - 9y + 6x - 6x =-4$

$13 y = -4$

$y=-4/13$

From here, you can find x using any original equation

$(11times-4/13) + (3x) = 4$

$3x = 4+(44/13)$

$3x = 96/13$

$x = 96/39$

or

$x=32/13$

Your answer is $x=32/13$ and $y=-4/13$

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