Solve the equations #2a-3b=-5#, #4a+3b=17# and #5c-2a=1#?

2 Answers
Jan 19, 2017

Answer:

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

Explanation:

#2a-3b=-5# ....................................(A)
#4a+3b=17# ....................................(B)
#5c-2a=1# ....................................(C)

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

i.e. #6a=12# i.e. #a=12/6=2#

Putting this in (B), we get #4xx2+3b=17#

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

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

#5c-2xx2=1#

or #5c-4=1# i.e. #5c=1+4=5# i.e. #c=1#

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

Jan 19, 2017

Answer:

#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

#2a-3b=-5color(white)(xxx)to(1)#
#4a+3b=color(white)(xx)17color(white)(xx) to(2)#
#"---------------------"#
#6a+0color(white)(x)=color(white)(xx)12rArra=2#

Substitute a = 2 into (2)

#(4xx2)+3b=17rArr8+3b=17#

#rArr3b=17-8=9rArrb=3#

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

#rArr5c-(2xx2)=1rArr5c-4=1#

#rArr5c=1+4=5rArrc=1#

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