How do you solve the following linear system: # x + 3y= 8, 2x + y = 15 #?

1 Answer
Mar 9, 2016

Answer:

#color(blue)(x=37/5" ; "y=1/5)#

Explanation:

Given:
#x+3y=8#............................(1)
#2x+y=15#..........................(2)

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The most straight forward way of dealing with this one is to make the coefficients of y the same and then subtract so that you only have 1 variable.

Multiply equation (2) by 3 then subtract, giving:

#6x+3y=45......................(2_a)#
#underline(color(white)(.)x+3y=8).........................(1)#
#5x+0color(white)(.)=37#

Divide both sides by 5 giving:

#color(blue)(x=37/5)#................................(3)
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Substitute for #x# in equation (1)

#37/5+3y=8#

Subtract #37/5# from both sides

#3y=8-37/5 = 3/5#

Divide both sides by 3

#color(blue)(y=1/5)#
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Check:
#color(brown)(x+3y=8)color(green)(-> 37/5+3(1/5) = 8) color(red)( " True")#

#color(brown)(2x+y=15)color(green)(->2(37/5)+1/5=15)color(red)(" True")#