+0

x-3)²+(y+3)²=5

0
778
3

x-3)²+(y+3)²=5

x-2y=9

thats what i did:

x²-6x+9+y²+6y+9=5

x=2y+9

(2y+9)²-6(2y+9)+9+y²+6y+9=5

4y²+36y+81-12y-54+9+y²+6y+9-5=0

5y²+30y+40=0

can i solve it in shorter way?

Guest Dec 3, 2014

#3
+27223
+8

Your second equation is x - 2y = 9

If we set a = x - 3 then by adding 3 to each side we have x = a + 3

If we set b = y + 3 then by subtracting 3 from each side we have y = b - 3

Now we replace x and y in x - 2y = 9,

so (a + 3) - 2(b - 3) = 9

or a + 3 -2b + 6 = 9

or a - 2b + 9 = 9

or a - 2b = 0

.

Alan  Dec 3, 2014
#1
+27223
+5

You could re-write the second equation as  x-3 -2(y+3) = 0

Let a = x-3,   b = y+3

a2 + b2 = 5

a - 2b = 0  so a = 2b

so

(2b)2 + b2 = 5  or b2 = 1  so  b = ±1  so  a = ±2

Therefore we must have x = a+3 = ±2 + 3  and  y = b-3 = ±1 - 3

or x = 5 and 1  with  y = -2 and -4

.

Alan  Dec 3, 2014
#2
0

i dont understand why a - 2b = 0

Guest Dec 3, 2014
#3
+27223
+8

Your second equation is x - 2y = 9

If we set a = x - 3 then by adding 3 to each side we have x = a + 3

If we set b = y + 3 then by subtracting 3 from each side we have y = b - 3

Now we replace x and y in x - 2y = 9,

so (a + 3) - 2(b - 3) = 9

or a + 3 -2b + 6 = 9

or a - 2b + 9 = 9

or a - 2b = 0

.

Alan  Dec 3, 2014