How do solve the following linear system?: # 4x-2y=2 , x-4y=4 #?

2 Answers
Jun 11, 2017

Answer:

See a solution process below:

Explanation:

Step 1) Solve the second equation for #x#:

#x - 4y = 4#

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

#x - 0 = 4 + 4y#

#x = 4 + 4y#

Step 2) Substitute #(4 + 4y)# for #x# in the first equation and solve for #y#:

#4x - 2y = 2# becomes:

#4(4 + 4y) - 2y = 2#

#(4 xx 4) + (4 xx 4y) - 2y = 2#

#16 + 16y - 2y = 2#

#16 + (16 - 2)y = 2#

#16 + 14y = 2#

#-color(red)(16) + 16 + 14y = -color(red)(16) + 2#

#0 + 14y = -14#

#14y = -14#

#(14y)/color(red)(14) = -14/color(red)(14)#

#(color(red)(cancel(color(black)(14)))y)/cancel(color(red)(14)) = -1#

#y = -1#

Step 3) Substitute #-1# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = 4 + 4y# becomes:

#x = 4 + (4 * -1)#

#x = 4 + (-4)#

#x = 4 - 4#

#x = 0#

The solution is: #x = 0# and #y = -1# or #(0, -1)#

Jun 11, 2017

Answer:

#x=0# and #y=-1#

Explanation:

#color(white)(-)4x-2y=2#
#color(white)(-)#
#color(white)(-)x-4y=4#

Let's try elimination! If we multiply the first equation by #2#, the #y#s will have equal coefficients, and we'll be able to eliminate them from the set:

#color(white)(-)2(4x-2y=2)#
#color(white)(-)#
#color(white)(-2..)x-4y=4#

#color(white(--)#

#color(white)(-)8x-4y=4#
#color(black)(-)#
#color(white)(-)x-4y=4#
...........................................
#7x=0#

divide by #7# on both sides

#x=0#

Now let's solve for #y#!

#x-4y=4#

substitute #0# for #x#

#(0)-4y=4#

#-4y=4#

divide by #-4# on both sides

#y=-1#

To check our work, let's substitute #x# and #y# for #0# and #-1# in the other equation.

#4x-2y=2#

#4(0)-2(-1)# should equal #2#, if we did our math right...

#0--2#

#2=2#
We were right!!!

#x=0, y=-1#

Just to triple check, let's graph the equations and see where they intercept:

enter image source here

They cross at #(0,-1)#, where #x=0# and #y=-1#