How do solve the following linear system?: # 8x+2y=3 , 2x+7=-5y #?

2 Answers
Jun 27, 2018

#8x+2y = 3#

#2x+7=-5y#

#* * * * * * * * * * * * * * * * * * * #

Let's isolate for #y# in the second equation:

#(-2x-7)/5 = y#

Now plug that into the first equation for #y#:

#8x+(2(-2x-7))/5 = 3#

#color(gray)(5/5) xx 8x - (4x)/5 - 14/5 = 3#

common denominator

#( 40x - 4x -14 )/5 = 3 #

multiply both sides by #5#

#36x =29#

#color(green)(x = 29/36 = 0.806#

#color(white)(..)#

Now let's solve for #y#:

#8(29/36) + 2y = 3 xx color(gray)(36/36)#

#232/36 + 2y = 108/36#

#2y = 108/36 - 232/36#

#2y = -124/36#

#y = -124/72 = -1.722#

#* * * * * * * * * * * * * * * * * * * * * * * *#

Let's check our work by graphing the two equations and seeing where they intersect

Looks right! Good job

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Jun 27, 2018

Answer:

#x=29/36, color(white)("xx")y=-31/18#

Explanation:

Given
[1]#color(white)("XXX")8x+2y=3#
[2]#color(white)("XXX")2x+7=-5y#

Converting [2] into standard form:
[3]#color(white)("XXX")2x+5y=-7#

We note that if we multiply [3] by #4# the coefficient of #x# becomes the same as that of [1]
[4]#color(white)("XXX")8x+20y=-28#

Subtracting [1] from [4] (to get rid of the #x# variable
[5]#color(white)("XXX")18y=-31#

Dividing both sides of [5] by #18#
[6]#color(white)("XXX")y=-31/18#

Substituting #(-31/8)# for #y# in [1]
[7]#color(white)("XXX")8x+2 *(-31/18)=3#

[8]#color(white)("XXX")8xcolor(white)("xxxxxxxxxxxx")=3+31/9=58/9#

[9]#color(white)("XXX")x=29/36#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Verifying by substituting #(-31/18)# for #y# and #29/36# for #x# in [2]
[10]#color(white)("XXX")2 * (29/36)+7 color(white)("xxx")?=?color(white)("xxx") -5 * (-31/18)#

[11]#color(white)("XXX")29/18+126/18color(white)("xxxxx")?=?color(white)("xx")155/18#

Yes: results verified!