Question

12=4k-2m and 27 = 6k - 2m solve for value k and m
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Answer 1

Answer:

it is very simple

4k-2m=12

6k-2m=27

on transposing constants we will get

4k-2m-12

6k-2m-27

now both these equations are equalent to 0 so we can substitute these

which would give

4k-2m-12=6k-2m-27

we will cancel -2m on both sides and we will get 2k=15

k=15/2

and by substituting value of k in equation 1 we will get 30-2m=12

-2m=-18

m=9

hence k=15/2 and m=9

hope it will help