How do you solve the following system: $5 x + 2 y = 12, 4 x - y = 6$ $5x+2y=12, 4x-y=6$ ?

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Answer 1 · 2016-01-02T07:18:22

$x = \frac{24}{23}, y = \frac{18}{13}$ $x=24/23, y=18/13$

By substitution

$5 x + 2 y = 12$ $5x+2y=12$ --- $(1)$ $(1)$

$4 x - y = 6$ $4x-y=6$

$y = 4 x - 6$ $y=4x-6$ --- $(2)$ $(2)$

Substitute equation $(2)$ $(2)$ into equation $(1)$ $(1)$

$5 x + 2 (4 x - 6) = 12$ $5x+2(4x-6)=12$

$5 x + 8 x - 12 = 12$ $5x+8x-12=12$

$13 x = 24$ $13x=24$

$x = \frac{24}{13}$ $x=24/13$

Substitute $x = \frac{24}{13}$ $x=24/13$ into equation $(2)$ $(2)$

$y = 4 (\frac{24}{13}) - 6$ $y=4(24/13)-6$

$y = \frac{96}{13} - 6$ $y=96/13-6$

$y = \frac{18}{13}$ $y=18/13$

thus, $x = \frac{24}{23}, y = \frac{18}{13}$ $x=24/23, y=18/13$

How do you solve the following system: 5x+2y=12, 4x-y=6 5x+2y=12,4x−y=65x+2y=12, 4x-y=6 ?