How do you solve the system of equations #3x + 4y = - 14# and #12x - 3y = 20#?

2 Answers
Aug 14, 2017

Solution : # x =2/3 , y = -4#

Explanation:

# 3x +4y = -14 (1)#

# 12x -3y = 20 (2)#

Multiplying equation(1) by #4# We get,

# 12x +16y = -56 (3) #

Subtracting equation (3) from equation (2) we get ,

#-19y= =76 or y = -4 #

Putting #y=-4# in equation (1) we get,

# 3x - 16 =-14 or 3x =2 or x = 2/3#

Solution : # x =2/3 , y = -4# [Ans]

Aug 14, 2017

#x = 2/3 and y =-4#

Explanation:

There are at least #5# methods to solve a system of equations:

  • eliminating one variable - create additive inverses
  • substituting for one variable
  • equating two variables
  • matrices
  • graphically

#color(white)(xxxxx)3xcolor(blue)(+4y) = -14color(white)(xxxxxxxxx)A#
#color(white)(xxxx)12xcolor(blue)(-3y) =+ 20color(white)(xxxxxxxxx)B#

I would choose to eliminate the #y# terms because they have opposite signs and are therefore easy to make into additive inverses. The sum of additive inverses is #0#

The LCM of #3 and 4 " is " 12#

#Axx3: color(white)(xxx)9xcolor(blue)(+12y) = -42color(white)(xxxxxxxxx)C#
#Bxx4:color(white)(xx)48xcolor(blue)(-12y) =+ 80color(white)(xxxxxxxxx)D#

#C+D:color(white)(xx)57x color(white)(xxx)= 38" "larr# the #y#-term is #0#

#color(white)(xxxxxxxxx)x color(white)(xxx)= 38/57#

#color(white)(xxxxxxxxxx)x color(white)(xxx)= 2/3#

Substitute #x = 2/3# into A

#color(white)(xxxxxx)3(2/3)+4y = -14color(white)(xxxxxxxxx)A#

#color(white)(xxxxxxxxxxxxx)4y = -14 -2#
#color(white)(xxxxxxxxxxxxx)4y = -16#
#color(white)(xxxxxxxxxxx.xx)y = -4#