How do you solve the following system: $10 y = 42 + 2 x, 3 x - 3 y = 2$ $10y=42+2x , 3x - 3y = 2$ ?

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Answer 1 · 2016-08-31T11:23:08

$y = \frac{65}{12} and x = \frac{73}{12}$ $y=65/12 and x=73/12$

$x - y = \frac{2}{3}$ $x-y =2/3$

or $x = y + \frac{2}{3}$ $x=y + 2/3$

$10 \cdot y = 42 + 2 \cdot (y + (\frac{2}{3}))$ $10*y = 42 + 2*(y+(2/3))$

$10 \cdot y - 2 \cdot y = 42 + \frac{4}{3}$ $10*y - 2*y = 42 + 4/3$

$8 \cdot y = \frac{126 + 4}{3}$ $8*y = (126 + 4)/3$

$8 \cdot y = \frac{130}{3}$ $8*y = 130/3$

$y = \frac{130}{8 \cdot 3}$ $y = 130 /(8*3)$

$y = \frac{65}{12}$ $y = 65/12$

After this step, you can solve x:

$x = \frac{65}{12} + \frac{2}{3}$ $x = 65/12 + 2/3$

or $x = \frac{65 + 8}{12}$ $x= (65+8)/12$

$x = \frac{73}{12}$ $x= 73/12$

How do you solve the following system: 10y=42+2x , 3x - 3y = 2 10y=42+2x,3x−3y=210y=42+2x , 3x - 3y = 2 ?