How do you solve the following system: #6x + y = 2, 5x + 8y = -2#?

1 Answer
Apr 10, 2016

Answer:

#(x,y)=(-22/43,18/43)#

Explanation:

Solve by elimination substitution

#color(blue)(6x+y=2#

#color(blue)(5x+8y=-2#

We can eliminate #8y# in the second equation by #y# in the first equation if we multiply (whole equation) #y# with #-8# to get #-8y#

#rarr-8(6x+y=2)#

Use distributive property

#color(brown)(a(b+c=x)=ab+ac=ax#

#rarr-48x-8y=-16#

Now,add the above equation to the second equation to eliminate #8y#

#rarr(-48x-8y=-16)+(5x+8y=-2)#

#rarr-43x=-18#

Divide both sides by #-43#

#rarr(cancel(-43)x)/cancel(-43)=(-18)/-43#

#color(green)(rArrx=18/43#

Because,

#color(brown)((-a)/-b=a/b#

Now,substitute the value of #x# to the first equation

#rarr6(18/43)+y=2#

#rarr108/43+y=2#

#rarr(108+43y)/43=2#

Multiply both sides by #43#

#rarr(108+43y)/cancel43*cancel43=2*43#

#rarr108+43y=86#

Subtract #108# both sides

#rarr108+43y-108=86-108#

#rarr43y=-22#

Divide both sides by #43#

#rarr(cancel43y)/cancel43=-22/43#

#color(green)(rArry=-22/43#

#color(blue)(ul bar |(x,y)=(-22/43,18/43)|#