How do you solve the following system?: $3 x + 3 y = - 7, 3 x - y = 30$ $3x+3y= -7 , 3x − y = 30$

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Answer 1 · 2017-02-01T09:34:20

$y = - 9 \frac{1}{4}, x = 6 \frac{11}{12}$ $y=-9 1/4, x=6 11/12$

$3 x + 3 y = - 7 - - - - - - - 1$ $3x+3y=-7 -------1$

$3 x - y = 30 - - - - - - - - 2$ $3x-y=30--------2$

$4 y = - 37$ $4y=-37$ ---------------------------------- $1 - 2$ $1-2$

$y = - \frac{37}{4}$ $y=-37/4$

$y = - 9 \frac{1}{4}$ $y=-9 1/4$

substitute $y = - \frac{37}{4}$ $y= -37/4$ in-------2

$3 x - (- \frac{37}{4}) = 30$ $3x-(-37/4)=30$

$3 x + \frac{37}{4} = 30$ $3x+37/4=30$

$3 x = 30 - \frac{37}{4}$ $3x=30-37/4$

$3 x = \frac{120}{4} - \frac{37}{4}$ $3x=120/4-37/4$

$3 x = \frac{83}{4}$ $3x=83/4$

multiply both sides by 4

$12 x = 83$ $12x=83$

$x = \frac{83}{12}$ $x=83/12$

$x = 6 \frac{11}{12}$ $x=6 11/12$

substitute $y = - \frac{37}{4}, x = \frac{83}{12}$ $y=-37/4, x= 83/12$ in-------2

$3 (\frac{83}{12}) - (- \frac{37}{4}) = 30$ $3(83/12)-(-37/4)=30$

$\frac{249}{12} + \frac{37}{4} = \frac{30}{1}$ $249/12+37/4=30/1$

$\frac{249 + 111 = 360}{12}$ $(249+111=360)/12$

$\frac{249}{12} + \frac{111}{12} = \frac{360}{12}$ $249/12+111/12=360/12$

$\frac{360}{12} = \frac{360}{12}$ $360/12=360/12$

How do you solve the following system?: 3x+3y=−7,3x−y=30 3x+3y= -7 , 3x − y = 30