How do you solve the linear equation system #(4/x)-(3/y)=1# & #(6/x)-(3/y)=-4#?

1 Answer
May 4, 2016

Answer:

#(x,y)=(-2/5,-3/11)#

Explanation:

#4x-3y=1#
#color(white)("XXX")rarr 4y-3x=xy# (after multiplying everything by #xy#)

#color(white)("XXX")rarr 4y-xy=3x#

#color(white)("XXX")rarr y(4-x) = 3x#

[1]#color(white)("XXX")rarr y=(3x)/(4-x)#

#6/x-3/y=-4#
#color(white)("XXX")rarr 6y-3x=-4xy#

#color(white)("XXX")rarr 6y+4xy=3x#

#color(white)("XXX")rarr y(6+4x)=3x#

[2]#color(white)("XXX")rarr y=(3x)/(6+4x)#

Combining [1] and [2]
#color(white)("XXX")(3x)/(4-x)=y=(3x)/(6+4x)#

#color(white)("XXX")rarr 4-x=6+4x#

#color(white)("XXX")rarr -5x = 2#

#color(white)("XXX")rarr x= -2/5#

Substituting #(-2/5)# for #x# in [1]
#color(white)("XXX")y = (3*(-2/5))/(4-(-2/5))#

#color(white)("XXXX")=(-6/5)/((20+2)/5)#

#color(white)("XXXX")=-6/22 = -3/11#

(Substitute these values back into the original equations to verify).