How do you solve the following system?: #4x + 5y = 2 , 17 x + 2y = 6 #

2 Answers
Apr 5, 2017

Answer:

#x=26/77#
#y=10/77#

Explanation:

There are a couple of ways we can solve this. I'm going to use the elimination method
#4x+5y=2#
#17x+2y=6#

I need to make two of the variables equal, so I'm going to multiply the second equation by #2.5#. That will change #2y# into #5y#.

#4x+5y=2#
#2.5(17x+2y=6)# or #42.5x+5y=15#

Now we subtract the two equations:
#4x+5y=2#
#-#
#42.5+5y=15#
#color(black)(------)#
#-38.5x+0y=-13#

If we simplify our equation, we find that #x=(-13)/-38.5# or #x=26/77#

Now we just solve for #y#. We can use either of the two equations. I like the first one (it has nicer numbers than #17#).

#4(color(purple)(x))+5y=2#
#4(color(purple)(26/77))+5y=2#
#cancel(104/77)+5y=2#
#cancel(-104/77)color(white)(+5y)-104/77#

Now we have
#5y=50/77#
or
#y=10/77#.

To double check our work, we need to plug our values into one (or both) of the equations.
#4(26/77)+5(10/77)# should equal #2#
#104/77+50/77#
#2=2#, so we were right!
Just to be safe, let's look at the other equation.

#17(26/77)+2(10/77)# should equal #6#
#442/77+20/77#
#6=6#
Good job, we got it right! Nice work

Apr 5, 2017

Answer:

Using the process of elimination, we find that #x=0.338# and #y=0.130#. (See explanation)

Explanation:

We can solve this system of equations using the elimination method. First, we can set up the system in such a way that the #y# terms cancel out. To do this, we need both the #y# terms in each equation to have the same coefficient (one negative and one positive):

#2(4x+5y=2)#
#-5(17x+2y=6)#

So we get:
#8x+10y=4#
#-85x-10y=-30#

Adding these two equations together gives us:
#-77x=-26#

#x=26/77=0.338#

Plug this value of #x# back into one of the given equations:
#4(0.338)+5y=2#

#y=(2-4(0.338))/5=0.130#

So, #x=0.338# and #y=0.130#