Question

Y=A(0.025k-0.5N)N

A=2\3

K=2000

N=-18+(18/5)w

5-c=200+(2/3)(y-T)-300r

6-T=-75+(1/4)Y

I=100-100r

G=100

L=0.5Y-200i

M=6300

π=0.10

What are the values of C, I, L and N

Answer 1

$Y=\frac{2}{3}[(0.025×2000)-0.5N]N\\Y=\frac{2}{3}(50-0.5N)N\\Y=\frac{100N}{3}-\frac{1}{3}N^2$

$N=-18+\frac{18}{5}W\\Y=\frac{100}{3}(-18+\frac{18}{3}W)-\frac{1}{3}(-18+\frac{18}{3})^2$

$Y=-600+120W+6-\frac{6W}{5}\\Y=\frac{594}{5}W-594$

$6-T=-75+\frac{1}{4}Y\\T=6+75-\frac{1}{4}Y\\T=6+75-\frac{1}{4}(\frac{594}{5}W-594)\\T=81-\frac{594}{5}W+594\\T=229.5-29.7W$

$5-C=200+\frac{2}{3}(Y-T)-300r\\5-C=200+\frac{2}{3}Y-\frac{2}{3}T-300r\\5-C=200+\frac{2}{3}(\frac{594}{5}-594)-\frac{2}{3}T-300r\\C=-\frac{396W}{5}+\frac{2}{3}t+300r-201$

$C=-\frac{396W}{5}+\frac{2}{3}(2295-29.7W)+300r-201\\C=-99W+300r+354$

$N=-18+\frac{18}{5}W\\Y=\frac{594}{5}W-594\\Y-594=\frac{594}{5}W\\W=\frac{5(Y-594)}{594}\\N=-18+\frac{18}{5}(\frac{5(Y-594)}{594})\\N=\frac{Y-1188}{33}$

$I=100-100r\\L=0.5-200i\\L=0.5(\frac{594}{5}W-594)-200(100-100r)\\L=59.4W-297-20000+20000r\\L=20000r+59.4W-20297$

$I=100-100r\\L=20000r+59.4W-20297\\L-59.4W+20297=20000r\\r=\frac{L}{20000}-0.00297W+\frac{20297}{20000}\\I=100-100(\frac{L}{20000}-0.00297W+\frac{20297}{20000})$

$I=-\frac{L}{200}+0.297W-\frac{297}{200}$