How do solve the following linear system?: # -3x -8y = -9 , 4 x+ 16y = 2 #?

2 Answers
Sep 3, 2017

Answer:

#color(red)((x,y)=(8,-15/8))#

Explanation:

Given
[1]#color(white)("XXX")-3x-8y=-9#
[2]#color(white)("XXX")4x+16y=2#

Multiplying [1] by #2#
[3]#color(white)("XXX")-6x-16y=-18#

Adding [2] and [3]
#color(white)("XXX+(-")4x+16y=color(white)("xxxx")2#
#color(white)("XXX")+ul(-6x-16y=-18)#
[4]#color(white)("XXX")-2xcolor(white)("xxxxx")=-16#

Dividing both sides of [4] by #(-2)#
[5]#color(white)("XXX")x=8#

Substituting #8# for #x# in [1]
[6]#color(white)("XXX")(-3)xx8-8y=-9#
#color(white)([6])##color(white)("XXX")-24-8y=-9#
#color(white)([6])##color(white)("XXX")-8y=15#
[7]#color(white)("XXX")y=-15/8#

So we have #(x,y)=(8,-15/8)#

We can verify this result by substituting these values for #x# and #y# in equation [2]

Sep 3, 2017

Answer:

See a solution process below:

Explanation:

**Step 1) Solve both of the equations for #8y#:

Equation 1:

#-3x - 8y = -9#

#color(red)(3x) - 3x - 8y = color(red)(3x) - 9#

#0 - 8y = 3x - 9#

#-8y = 3x - 9#

#color(red)(-1) xx -8y = color(red)(-1)(3x - 9)#

#8y = (color(red)(-1) xx 3x) - (color(red)(-1) xx 9)#

#8y = -3x + 9#

Equation 2:

#4x + 16y = 2#

#(4x + 16y)/color(red)(2) = 2/color(red)(2)#

#(4x)/color(red)(2) + (16y)/color(red)(2) = 1#

#2x + 8y = 1#

#-color(red)(2x) + 2x + 8y = -color(red)(2x) + 1#

#0 + 8y = -2x + 1#

#8y = -2x + 1#

Step 2) Because the left side of each equation are equal we can now equate the right side of each equation and solve for #x#:

#-3x + 9 = -2x + 1#

#color(red)(3x) - 3x + 9 - color(red)(1) = color(red)(3x) - 2x + 1 - color(red)(1)#

#0 + 8 = (color(red)(3) - 2)x + 0#

#8 = 1x#

#8 = x#

#x = 8#

Step 3) Substitute #8# for #x# into the solution for either equation in Step 1 and solve for #y#:

#8y = -2x + 1# becomes:

#8y = (-2 xx 8) + 1#

#8y = -16 + 1#

#8y = -15#

#(8y)/color(red)(8) = -15/color(red)(8)#

#(color(red)(cancel(color(black)(8)))y)/cancel(color(red)(8)) = -15/8#

#y = -15/8#

The Solution Is: #x = 8# and #y = -15/8# or #(8, -15/8)#