Systems of Equations?!?!?!?

Question

Solve.

1.) 30R+3C+9M=66

2.) 35R+16C+13M=94.5

3.) 200R+10C+300M=910

PLEASE SHOW ALL WORK!!!

Answer 1

Multiply the first equation by (200/30)

(200/30)(30R + 3C + 9M = 66)

200R + 20C + 60M = 440

Subtract this equation from the third equation and solve for C.

200R + 10C + 300M = 910

-(200R + 20C + 60M = 440)

-10C + 240M = 470   ---   Call this equation *1*

Multiply the first equation by (35/30)

(35/30)(30R + 3C + 9M = 66)

35R + (7/2)C + (21/2)M = 77

Subtract this equation from the second equation and solve for C.

35R + 16C + 13M = 94.5

-(35R + (7/2)C + (21/2)M = 77)

(25/2)C + (5/2)M = 35/2   ---   Call this equation *2*

Multiply *2* by 96

96((25/2)C + (5/2)M = 35/2)

1200C + 240M = 1680

Subtract this from *1*

-10C + 240M = 470

-(1200C + 240M = 1680)

-1210C = -1210

C = 1

Substitute this value into *1* and solve for M.

-10 + 240M = 470

240M = 480

M = 2

Substitute these values into the first equation and solve for R.

30R + 3C + 9M = 66

30R + 3 + 9*2 = 66

30R + 18 = 63

30R = 45

R = 3/2

Answer:  R = 3/2, C = 1 and M = 2

Hope this helps you!