How do you solve the following system: # 1/2x - 2y = 4 , 5x+3y=-1 #?

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Answer 1 · 2017-05-07T15:25:46

#x = 20/23 and y = -41/23#

The approach is either to use substitution or elimination.

For substitution, you need to have a single variable such as #x#

#1/2x-2y =4" and "5x+3y =-1#

Multiply the first equation by #2#

#2xx1/2x -2xx2y =2xx 4#

#x-4y =8#

#color(blue)(x = (4y+8))" "#Substitute for #x# in the other equation:

#color(white)(ml.m)5color(blue)(x)+3y=-1#
#color(white)(mmm)darr#
#color(white)(m)5color(blue)((4y+8)) +3y=-1#

#20y+40 +3y =-1#

#23y = -1-40#

#23y = -41#

#y = (-41)/23#

Now find #x" by using " x = 4y+8#

#x = 4((-41)/23)+8#

#x = 20/23#

Substitute to check:

#5x +3y#

#5(20/23)+3(-41 /23)#

#=100/23-123/23#

#=-1#