How do you solve #y=x^2# and #y=1/2x+5# using substitution?

Question

845 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-30T11:03:51

Solution: #( x= 5/2 , y= 25/4) or (x= -2 , y=4)#

#y= x^2 (1) and y =1/2 x +5 (2)#. Substituting # y =1/2 x +5# in equation (1) we get,

# x^2 =1/2 x +5 or 2x^2 =x+10 or 2x^2 -x -10 = 0 # or

#2x^2 -5x +4x -10 = 0 or x(2x-5) + 2(2x-5) =0 # or

# (2x-5)(x+2) = 0#. Either # 2x-5 = 0 or 2x= 5 :. x=5/2# or

#x+2=0 :. x=-2 :. x =5/2 or x= -2#

When #x=5/2 ; y= 1/2*5/2 +5 = 5/4+5 =25/4# ;

When #x=-2 ; y= 1/2*(-2) +5 = -1+5 =4# ;

Solution: #( x= 5/2 , y= 25/4) or (x= -2 , y=4)# [Ans]