Question

Solve the simultaneous equations C^2+d^2=5 and 3c+4d=2

Answer 1

If 3c+4d=2 then c=(2-4d)/3
And if you put that into the other equation

`{(2 - 4d)/(3)}^(2) + {d}^(2) = 5`
Then you multiply everything by 3, getting


`{(2 - 4d)}^(2) + 3 {d}^(2) = 15`
And then


`2 - 4d + √(3) d = √(15)`
Leaving

`d( √(3) - 4) = √(15) - 2`
I dont have a calc with me right now but you just pot this into the calculator:

`d = ( √(15) - 2 )/( √(3) - 4)`
And when you get the number you insert it into the nest eqation to get c


`c = (2 - 4d)/(3)`
So basically


`c = (2 - 4 ( √(15) - 2)/( √(3) - 4) )/(3)`

Answer 2

c²+d²=5----(1)
3c+4d=2----(2)

from (2):
3c=2-4d
c=(2-4d)/3

sub c into (1):
[(2-4d)/3]² + d² =5
(16d²-16d+4)/9 + d²=5
16d²-16d +4 +9d²=45
25d²-16d+4=45
25d²-16d+4-45=0
25d²-16d-41=0
(25d-41)(d+1)=0
25d-41=0 or d+1=0
d=41/25 or d=-1

If d=41/25;
Then c=[2-4(41/25)]/3
c= 38/25

If d=-1
Then c=[2-4(-1)]/3
c=2

Answer: *c=38/25, d=41/25 or c=2, d=-1*