Question

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

Answer 1

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*

Answer 2

If 3c+4d=2 then c=(2-4d)/3
And if you put that into the other equation
${\frac{2 - 4d}{3}}^{2} + {d}^{2} = 5$
Then you multiply everything by 3, getting

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

$2 - 4d + \sqrt{3} d = \sqrt{15}$
Leaving
$d( \sqrt{3} - 4) = \sqrt{15} - 2$
I dont have a calc with me right now but you just pot this into the calculator:
$d = \frac{ \sqrt{15} - 2 }{ \sqrt{3} - 4}$
And when you get the number you insert it into the nest eqation to get c

$c = \frac{2 - 4d}{3}$
So basically

$c = \frac{2 - 4 \frac{ \sqrt{15} - 2}{ \sqrt{3} - 4} }{3}$