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

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Answer 1 · 2016-08-31T11:23:08

#y=65/12 and x=73/12#

#x-y =2/3#

or #x=y + 2/3#

#10*y = 42 + 2*(y+(2/3))#

#10*y - 2*y = 42 + 4/3#

#8*y = (126 + 4)/3#

# 8*y = 130/3#

# y = 130 /(8*3)#

# y = 65/12#

After this step, you can solve x:

# x = 65/12 + 2/3#

or #x= (65+8)/12#

#x= 73/12#