How do you solve the system #x^2+y^2=17# and #y=x+3#?

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Answer 1 · 2016-08-30T19:45:58

The Soln. Set # ={(1,4),(-4,-1)}#.

Sub.ing #y=x+3# in the first eqn. to get.

#x^2+(x+3)^2=17, or, 2x^2+6x+9-17=0#, i.e.,

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

#rArr (x-1)(x+4)=0#

#rArr x=1, x=-4#. Then, from #y=x+3#,

#y=4, y=-1#.

These roots satisfy the eqns.

Thus, the Soln. Set # ={(1,4),(-4,-1)}#.

Answer 2 · 2016-08-30T20:26:42

#(x, y) = (-4, -1)# or #(1, 4)#

Here's an alternative method that tries to use symmetry to help...

Let #t = (x+y)/2#

Then #x = t-3/2# and #y = t+3/2#

So we have:

#17 = x^2+y^2#

#= (t-3/2)^2+(t+3/2)^2#

#= t^2-color(red)(cancel(color(black)(3t)))+9/4+t^2+color(red)(cancel(color(black)(3t)))+9/4#

#= 2t^2+9/2#

Subtract #9/2# from both ends and transpose to get:

#2t^2 = 17-9/2 = 25/2#

Divide both ends by #2# to get:

#t^2 = 25/4 = 5^2/2^2#

Hence:

#t = +-5/2#

If #t = -5/2# then #x = -5/2-3/2 = -4# and #y = -5/2+3/2 = -1#

If #t = 5/2# then #x = 5/2-3/2 = 1# and #y = 5/2+3/2 = 4#