How do you solve the system #x= 175+15y#, #.196x= 10.4y#, #z=10*y#?

1 Answer
Oct 18, 2016

Answer:

#x=53.05#
#y=-8.13#
#z=-81.3#

Explanation:

To begin, plug in a equation for a variable that is already isolated. There are two variables that are already isolated, and those are #x and z#. You could choose either to work with, doesn't matter, but I'm going to use #x#

So, plug the equation for #x# into another equation that has that same variable in it. Which would be:

#.196x=10.4y#

To begin, plug #x# equation into #.196x=10.4y#

#.196(175+15y)=10.4y#

Distribute #.196# throughout the set of parenthesis

#34.3+2.94y=10.4y#

Begin to isolate #y# by subtracting #34.3# on both sides of the equation

#2.94y=-23.9#

Isolate y by dividing #2.94# on both sides of the equation

#y=-8.13#

Now, we have solved for one of the variables. The next thing to do is plug #y# into another equation that contains #y# in it. An easy one to use would be #x=175+15y# because #x# is already isolated

So, plug #y=-8.13# into #x=175+15y#

#x=175+15(-8.13)#

Distribute #15# throughout the set of parenthesis

#x=175-121.95#

Subtract

#x=53.05#

And now we have solved for #y# and #x#. The only variable left is #z#

To solve for #z#, you have to plug #y=-8.13# into the equation

#z=10(-8.13)#

Multiply

#z=-81.3#

#x=53.05#
#y=-8.13#
#z=-81.3#