Question

3c+4u=33 6c+3u=36 by elimination method

Answer 1

Answer:

6c+3u=36

6c-6c+3u=36-6c

3u=36-6c

3u/3=36-6c/3

u=12-2c

3c+4u=33

3c+4(12-2c)=33

3c+4(12) (4)(-2c)=33

3c+48-8c=33

48-5c=33

48-5c-48=33-48

-5c=-15

-5c/-5=-15/-5

c=3

u=12-2c

u=12-2(3)

u=12-6=6

Hiopw i did it ryt=)

Answer 2

3c+4u=33.....(1)

6c+3u=36.....(2)

multiply 1 equation with 2

6c+8u=66......(3)

now subtract equation 2 and 3

5u=30

therefore u=6

now put the value of u in any of the equation and find the value of c let us put it in 1 equation

3c+4(6)=33

3c=9

therefore c = 3

thank you