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?
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
.
You could re-write the second equation as x-3 -2(y+3) = 0
Let a = x-3, b = y+3
Now your two equations become
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
.
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
.