How do you solve the system #5x= -25+5y# and #10y=42+2x#?

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Answer 1 · 2015-06-13T22:30:15

Divide the first equation by #5# to get an expression for #x# to substitute in the second equation. Solve to find #y=4# and hence #x=-1#

Divide both sides of the first equation by #5# to get:

#x = -5+y = y-5#

Substitute #y-5# for #x# in the second equation:

#10y = 42 + 2x = 42 + 2(y - 5)#

#=42+2y - 10 = 32+2y#

Subtract #2y# from both ends to get:

#8y = 32#

Divide both sides by #8# to get:

#y = 4#

Then

#x = y-5 = 4-5 = -1#