How do you solve the system $3 x + 2 y = 19$ $3x+2y=19$ and $2 x - 3 y = - 9$ $2x-3y=-9$ by substitution?

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Answer 1 · 2015-05-24T22:48:05

$3 x + 2 y = 19 \to 3 x = 19 - 2 y \to x = \frac{19 - 2 y}{3}$ $3x + 2y = 19 -> 3x = 19 - 2y -> x = (19 - 2y)/3$

$2 x - 3 y = - 9 \to (\frac{2}{3}) . (19 - 2 y) - 3 y = - 9$ $2x - 3y = -9 -> (2/3).(19 - 2y) - 3y = -9$

$38 - 4 y - 9 y = - 27 \to - 13 y = - 27 - 38 = - 65$ $38 - 4y - 9y = -27 -> -13y = -27 - 38 = - 65$

$y = \frac{65}{13}$ $y = 65/13$ = 5

$x = \frac{19 - 10}{3} = \frac{9}{3} = 3$ $x = (19 - 10)/3 = 9/3 = 3$

Check: 3(3) + 2(5) = 19. OK

How do you solve the system 3x+2y=193x+2y=193x+2y=19 and 2x-3y=-92x−3y=−92x-3y=-9 by substitution?