How do you solve the following system of equations?: #13x+13y=16, 3x + 19y=19#?

1 Answer
Oct 22, 2017

Answer:

#x = 57/208, y = 199/208#

Explanation:

#13x + 13y = 16#.....#(I)#

#3x + 19y = 19#.....#(II)#

Rearranging #(II)# for x for simplicity,

#3x = 19 - 19y#

#implies##x = (19-19y)/3#

Substituting for #x# in #(I)#

#13( (19-19y)/3) + 13y = 16#

#implies##(247 - 247y)/3 + 13y = 16#

#implies##(247-247y+39y)/3 = 16#

#implies##247-208y = 16*3#

#implies##208y = 247 - 48#

#implies##y = 199/208#

Substituting for #y# in either equation (we'll choose #(II)#),

#3x + 19(199/208) = 19#

#implies##3x = 19 - 3781/208#

#implies##x = 1/3*((3952 - 3781)/208) = 1/3 * (171/208) = 171/624 = 57/208#