How do you solve the system of equations algebraically #2x+5y=4, 3x+6y=5#?

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Answer 1 · 2018-07-11T12:37:46

Solution: #x=1/3 , y=2/3#

#2 x+ 5 y = 4 ; (1) , 3 x+ 6 y = 5 ;(2)# . Multiplying equation (1) by

#3# and equation (2) by #2# we get ;

#6 x+ 15 y = 12 ; (3) , 6 x+ 12 y = 10 ;(4)# . Subtracting equation

(4) from equation (3) we get,

#(6 x+ 15 y)-(6 x+12 y)= 12-10; # or

#cancel(6 x)+ 15 y- cancel(6 x)-12 y=2 or 3 y = 2 :. y=2/3 # Putting

#y=2/3# in equation (2) we get, # 3 x +6 *2/3=5 or 3 x +4=5# or

#3 x= 1 or x =1/3 :. x=1/3 , y=2/3#

Solution: #x=1/3 , y=2/3# [Ans]