Solve the equations 2a3b=5, 4a+3b=17 and 5c2a=1?

2 Answers
Jan 19, 2017

a=2, b=3 and c=1

Explanation:

2a3b=5 ....................................(A)
4a+3b=17 ....................................(B)
5c2a=1 ....................................(C)

Adding (A) and (B), we get 2a3b+4a+3b=5+17

i.e. 6a=12 i.e. a=126=2

Putting this in (B), we get 4×2+3b=17

or 8+3b=17 i.e. 3b=17=8=9 and b=93=3

and putting value of a in (C), we get

5c2×2=1

or 5c4=1 i.e. 5c=1+4=5 i.e. c=1

and a=2, b=3 and c=1

Jan 19, 2017

a=2,b=3,c=1

Explanation:

If we add the first 2 equations, we will eliminate b and be able to solve for a

2a3b=5×x(1)
4a+3b=×17×(2)
---------------------
6a+0x=×12a=2

Substitute a = 2 into (2)

(4×2)+3b=178+3b=17

3b=178=9b=3

Substitute a = 2 into the third equation and solve for c

5c(2×2)=15c4=1

5c=1+4=5c=1

Thus a=2,b=3 and c=1