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

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Answer 1 · 2017-06-14T14:22:04

By arranging equations. $x = \frac{32}{13}$ $x=32/13$ and $y = - \frac{4}{13}$ $y=-4/13$

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

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

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

Now combine these:

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

$13 y = - 4$ $13 y = -4$

$y = - \frac{4}{13}$ $y=-4/13$

From here, you can find x using any original equation

$(11 \times - \frac{4}{13}) + (3 x) = 4$ $(11times-4/13) + (3x) = 4$

$3 x = 4 + (\frac{44}{13})$ $3x = 4+(44/13)$

$3 x = \frac{96}{13}$ $3x = 96/13$

$x = \frac{96}{39}$ $x = 96/39$

or

$x = \frac{32}{13}$ $x=32/13$

Your answer is $x = \frac{32}{13}$ $x=32/13$ and $y = - \frac{4}{13}$ $y=-4/13$

How do you solve the following system: 11y + 3x = 4, 2x+3y =4 11y+3x=4,2x+3y=411y + 3x = 4, 2x+3y =4 ?