Solve the simultaneous equations #2x+y=8.......................................(1)# #4x^2+3y^2=52.......................(2)# ?

2 Answers
Pallu
Dec 20, 2017

# x= 3.5 and y =1 #
OR
#x = 2.5 and y = 3 #

Explanation:

#2x+y=8.......................................(1)#

#4x^2+3y^2=52.......................(2)#

(1) #=> y=8-2x#

(2) #=> 4x^2+3(8-2x)^2=52#

#=> 4x^2 +3(64 - 32x +4x^2) =52#

#=> 4x^2 + 192 - 96x + 12x^2 = 52#

#=> 16x^2 -96x + 140 = 0 #

#=> 4(4x^2 - 24x +35) = 0 #

#=> 4x^2 -24x +35 = 0#
Solving this quadratic equation, we get:

#=> (x-3.5)(x-2.5) = 0 #

#=> x= 3.5 or x = 2.5#

Substitute this value of #x# in equation (1):

Case 1: Taking #x= 3.5#

#=> 2x+y=8#

#=> 2(3.5) + y = 8#

# => y = 8-7 =1 #

OR

Case 2 : Taking #x= 2.5#

# 2(2.5) + y = 8#

# => y = 8- 5 = 3#

#therefore x= 3.5 and y =1 #

OR

#x = 2.5 and y = 3 #

José F.
Dec 20, 2017

#y=3 and x=5/2 or y=1 and x=7/2#

Explanation:

#2x+y=8#
#4x^2+3y^2=52#

#2x=8-y=>4x^2=(8-y)^2=color(blue)(64-16y+y^2)#

#color(blue)(64-16y+y^2)+3y^2=52#
#4y^2-16y+12=0#

dividing everything by 4(for symplifiying calculations):

#y^2-4y+3=0#

#y=(4+-sqrt(4^2-4*1*3))/2#

#y=(4+-sqrt(4))/2#

#y=(4+-2)/2#

#y=3 and x=5/2 or x=1 and x=7/2#

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