How do I solve and check this system of equations?

solve and check algebraically

#2x^2-4x=y+1#
#x+y=1#

1 Answer
Shiva Prakash M V
Mar 7, 2018

#(x,y)=(-1/2,3/2)#
and
#(x,y)=(2,-1) form solutions#

Explanation:

Given:
#2x^2-4x=y+1#
#x+y=1#
#y=1-x#
#2x^2-4x=1-x+1#
#2x^2-4x=2-x#
#2x^2-4x+x-2=0#
#2x^2-3x-2=0#
#2x^2-4x+x-2=0#
#2x(x-2)+1(x-2)=0#
#(2x+1)(x-2)=0#

#2x+1=0#
#2x=-1#
#x=-1/2#
#y=1-x#
#y=1-(-1/2)#
#1+1/2#
#y=3/2#
Check:
#2x^2-4x=y+1#----(1)
#2(-1/2)^2-4xx(-1/2)=3/2+1#
#2xx1/4+4xx1/2=3/2+2/2#
#1/2+4/2=3/2+2/2#
#(1+4)/2=(3+2)/2#
#5/2=5/2#
lhs=rhs
#x+y=1#-----(2)
#-1/2+3/2=1#
#(-1+3)/2=1#
#2/2=1#
#1=1#
lhs=rhs

#x-2=0#
#x=2#
#y=1-x#
#y=1-2#
#y=-1#
Check:
#2(2)^2-4(2)=-1+1#-----(1)
#2(4)-8=0#
#8-8=0#
#0=0#
lhs=rhs
#x+y=1#-----(2)
#2+(-1)=1#
#2-1=1#
#1=1#
lhs=rhs

Thus,
#(x,y)=(-1/2,3/2)#
and
#(x,y)=(2,-1) form solutions#

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