How do you solve the following system?: #23x +31y =1, -7x +5y = -2#

1 Answer
Aug 15, 2017

Answer:

#x= 67/332 and y = -39/332#

Explanation:

The given equations are really uncomfortable and do not lend themselves to either substitution or elimination without using big numbers.

Let's equate one of the variables: solve for #y# in each.

#23x+31y =1color(white)(xxxxx)and -7x+5y =-2#

#31y = 1-23xcolor(white)(xxxxxxxxxxxxxx)5y = 7x-2#

#y = (1-23x)/31color(white)(xxxxxxxxxxxxxxxx)y = (7x-2)/5#

Now as #y=y#, we can equate the right sides of each equation:

#(1-23x)/31 = (7x-2)/5" "larr# cross multiply

#31(7x-2) =5(1-23x)#

#217x-62 = 5-115x#

#217x+115x = 5+62#

#332x = 67#

#x = 67/332#

Now substitute this value for #x# into either equation above.

#y= (7(67/332)-2)/5#

#y = (469/332 -2)/5#

#y = -195/332 div 5#

#y = -cancel195^39/332 xx1/cancel5#

#y = -39/332#

Check by substituting the values for #x and y# into the the other equation.

#23x +31y#

#23(67/332) +31(-39/332)#

#=1#

The equation checks out.