How do you solve #1 / 5x + 1/3y = - 1 / 15# and #1 / 5x + 2y = 8 / 5# using substitution?

We can solve each equation for #1/5x# and set them equal to each other:

#1/5x+1/3y=-1/15=>1/5x=-1/15-1/3y#

#1/5x+2y=8/5=>1/5x=8/5-2y#

#:.-1/15-1/3y=8/5-2y#

We can now solve for #y#. Let's first multiply through by 15:

#15(-1/15-1/3y)=15(8/5-2y)#

#-1-5y=24-30y#

#25y=25#

#y=1#

And now we can substitute back into one of the original equations to solve for #x#:

#1/5x+2(1)=8/5#

#1/5x=8/5-2#

#1/5x=8/5-10/5=-2/5#

#x=-2#

Which gives us the point #(-2,1)# which we can see on the graph:

graph{(1/5x+1/3y+1/15)(1/5x+2y-8/5)=0}

I would usually simplify the equations to get rid of the fractions first, but in this case we have identical terms in each equation, so I will solve for those first.

#1/5x = -1/15-1/3y" and "1/5x =8/5-2y#

#1/5x = 1/5x#

We can substitute the other sides of each equation for #1/5x#

#-1/15-1/3y = 8/5-2y" "xx15#

#-1-5y= 24-30y#

#30y-5y = 24+1#

#25y = 25#

#y =1#

Substitute into #1/5x =8/5-2y#

#1/5x = 8/5 -2(1)" "xx 5#

#x = 8-10#

#x =-2#

Check:
#1/5x +1/3y = -1/15" " xx5#

#x+5/3y = -1/3#

#-2 +1 2/3 = -1/3#

#-1/3 = -1/3#