How do you solve this system of equations: #3x + 6y = 105; 14x + 12y = 314#?

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Answer 1 · 2018-02-09T12:22:35

#x=13; y =11#

#3x+6y=105;14x+12y=314#

dividing the entire #2^(nd)# equation by 2, the two equations now are,
#3x+6y=105# and #7x+6y=157#

subtracting the #1^(st)# equation from the #2^(nd)#,
#4x=52#
#x=13#

now, substitue the #x# in (any equation, here im going for) the #1^(st)# one.

#3(13)+6y=105#
#6y=105-39#
#6y=66#
#y=11#.

-Sahar

Answer 2 · 2018-02-09T12:30:13

#color(green)(x = 13, color(blue)(y=11)#

#3x + 6y = 105# Eqn(A)

#14x + 12y = 314# Eqn (B)

(B) - 2(A) gives

#14x - 6x + 12y - 12y = 314 - 210#

#8x = 104, or x = 13#

Substituting value of x in Eqn (A)

#(3*13) + 6y = 105#

#6y = 105-39 = 66, y = 66/6 = 11#