How do you solve the following system: #4x+y=-7, 10y=42+2x #?

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Answer 1 · 2017-05-23T12:03:20

Follow explanation. #x=-8/3# and #y=11/3#

From the second equation you can get
#y=4.2+(x/5)# when you divide the equation by 10.

Now put this value on the first equation:

#4x + 4.2 + x/5 = -7#

You have one unknown (x) and one equation.

#21x/5 = -7-4.2#

#21x = 5*(-11.2)#
#21x=-56#
#x=-56/21#
#x=-8/3#

Now you can put this value on the second equation:

#10y=42+2*(-8/3)#
#10y=42 - 16/3#
#10y=110/3#
#y=110/30#
or
#y=11/3#