Question 219999
subt (3c+1,d+5) first to make a new equation
2(3c+1)+5(d+5) = 10
6c+2+5d+5-10=0
6c+5d=3 --(Eqn1)


Subt (c+4,2d+6) to make another equation
2(c+4)+5(2d+6)=10
2c+8+10d+30=10
2c+10d+38-10=0
2c+10d=-28 --(Eqn2)


Solve it simultaneously


6c+5d=3  --(Eqn1)
2c+10d=-28 --(Eqn2) 


(eqn1)x2


12c+10d=6 --(eqn 3)


(eqn3) - (eqn2)


(12c+10d)-(2c+10d)=6-(-28)
10c=34
c=3.4


subt c=3.4 to eqn 2 
2(3.4)+10d= -28
10d=-28-6.8
10d= -34.8
d= -3.48


c=3.4 , d= -3.48 (ans)