How do you solve the following system: $3 x + 2 y = 1, 2 x - 3 y = 10$ $3x + 2y = 1, 2x - 3y = 10$ ?

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Answer 1 · 2018-04-16T03:19:27

$x = \frac{23}{13}, y = - \frac{28}{13}$ $x=23/13, y=-28/13$

$3 x + 2 y = 1$ $3x+2y=1$ equation (1)
$2 x - 3 y = 10$ $2x-3y=10$ equation (2)

From (2): $2 x = 3 y + 10 \Rightarrow x = \frac{3 y + 10}{2}$ $2x=3y+10 => x=(3y+10)/2$

Substitute $x = \frac{3 y + 10}{2}$ $x=(3y+10)/2$ into (1);

$3 (\frac{3 y + 10}{2}) + 2 y = 1$ $3((3y+10)/2)+2y=1$

$3 (3 y + 10) + 4 y = 2$ $3(3y+10)+4y=2$

$9 y + 30 + 4 y = 2$ $9y+30+4y=2$

$13 y = - 28$ $13y=-28$

$y = - \frac{28}{13}$ $y=-28/13$

Substitute $y = - \frac{28}{13}$ $y=-28/13$ into $x = \frac{3 y + 10}{2}$ $x=(3y+10)/2$ :

$x = \frac{3 (- \frac{28}{13}) + 10}{2} = \frac{23}{13}$ $x=(3(-28/13)+10)/2=23/13$

How do you solve the following system: 3x + 2y = 1, 2x - 3y = 10 3x+2y=1,2x−3y=103x + 2y = 1, 2x - 3y = 10 ?