How do you solve the system of equations #2x+8y=6# and #-5x-20y=-15#?

2 Answers
Nov 19, 2016

Answer:

This system of equation doesn't have any solutions.

Explanation:

#2x+8y=6#
#-5x-20y=-15#

Let's simplify these equations: divide the first one by #2# andd the second by #-5#:

#x+4y=3#
#x+4y=15#

Now, using any of the equations, we must express one variable by the other, let's say #x# by #y# from the second equation:

#x=15-4y#

Using this we solve the other equation:

#x+4y=3#
#15-4y+4y=3#
#15=3#

We got #15=3# which obviously isn't true so this equation has no solution. It follows that this system of equation doesn't have any solutions as well.

Nov 19, 2016

Answer:

#x# can have any value and the equations will work.
The y- values will depend on the x chosen.

Explanation:

#2x+8y =6" and "-5x -20y = -15#

There are different options available to solve the equations.
#rarr" "# elimination
#rarr" "# substitution #rarr# single variable
#rarr" "# equating
#rarr" "#matrices
#rarr" "# graphically

To be able to use substitution, a single variable in one of the equations is a useful indicator

#2x+8y =6 " "div2 rarr " "x+4y=3#

Make x the subject of the equation:

#color(red)(x = 3-4y)#

Replace #color(red)x# in the second equation by #color(red)((3-4y))#

#" "-5color(red)(x) -20y = -15#
#color(white)(xxxx)darr#
#-5color(red)((3-4y))-20y =-15#

#-15+20y -20y = -15#

#-15 = -15#

#x# can have any value