How do you solve the following linear system: # y=x-4, 4x+y=26 #?

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Answer 1 · 2018-05-05T11:19:48

Solution: # x=6 , y=2#

#y= x-4 or x - y = 4 ; (1); 4 x + y =26 ; (2) # Adding equation

(1) and equation (2) we get, #5 x =30 :. x =30/5=6# Putting

#x=6# in equation (1) we get , # 6- y = 4 or y = 6- 4=2#

Solution: # x=6 , y=2# [Ans]

graph{(y-x+4)(4x+y-26)=0 [-10, 10, -5, 5]}

Answer 2 · 2018-05-05T12:36:13

#x =6 and y=2#

We have two equations, each with two variables.

#color(red)(y = x-4) and color(blue)(4x+y =26)" "rarr# rewrite as #color(blue)(y = 26-4x)#

Now we have two equations for #y#

As #color(red)(y)=color(blue)(y)#, we can also write an equation in #x#.

#color(red)(x-4) = color(blue)(26-4x)#

#x+4x = 26+4#

#5x = 30#

#x=6#

Now find #y#

#y = 6-4=2#