Answer:
D = -4/7 = - 0.57
C = 17/7 = 2.43
Explanation:
We have the following two equations:
[tex]3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)[/tex]
First, we isolate C from equation (2):
[tex]2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)[/tex]
using this value of C from equation (3) in equation (1):
[tex]3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}[/tex]
D = - 0.57
Put this value in equation (3), we get:
[tex]C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C = \frac{17}{7}\\[/tex]
C = 2.43