Asked by Olayemi
Solve these simultaneous linear equations using elimination method :
3a = 2b + 1 and 3b = 5a - 3
3a = 2b + 1 and 3b = 5a - 3
Answers
Answered by
Reiny
first arrange them in "standard" form:
3a - 2b = 1
5a - 3b = 3
1st times 3, ----> 9a - 6b = 3
2nd times 2, ---> 10a - 6b = 6
subtract them,
a = 3
sub back into 2nd
3b= 5(3) - 3
b = 4
a = 3 , b = 4
3a - 2b = 1
5a - 3b = 3
1st times 3, ----> 9a - 6b = 3
2nd times 2, ---> 10a - 6b = 6
subtract them,
a = 3
sub back into 2nd
3b= 5(3) - 3
b = 4
a = 3 , b = 4
Answered by
Mary
Thanks for the solution . It's now clear to me.
Though in the solution "first arrange in standard form ", it's to be
3a-2b=1 and 3b-5a=-3 not 3b-5a=3.
Though in the solution "first arrange in standard form ", it's to be
3a-2b=1 and 3b-5a=-3 not 3b-5a=3.
Answered by
Muhammad n Abubakar
gud job my friend
Answered by
Anonymous
I don't understand
Answered by
KEMI
Yes
Answered by
Timothy
thanks
Thanks you! I now no that i can make it
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