Need help with Algebra 1
okay
(a) Solve the system of equations using the method of your choice. SHOW ALL WORK. (B) Explain why you chose that method a+c=9 8a+4.5c=58
what did you do?
We can use the substitution method.
yeah you can do that
What is the first step?
our 1st equation is this-> a+c=9 so subtract c from both sides a+c-c=9-c a=9-c now you have this-> a=9-c just put this value of a in the 2nd equation :) then try to solve
so in the second equation will look like : 8(9-c)+4.5c=58
u are in 9 grade and in flvs
no hes not neither am i
8(9-c)+4.5c=58 8(9-c)+4.5/10c=58 \[8(-c+9)+\frac{ 3^{2} *5}{ 2*5 }\]
c=4?
yeah
now use this- a=9-c
to get a
im so confused
lets start over
no wait
do you understand how we get c?
kind of
when i sub it in for a then I got c?
9-c
okay we just wanted to get rid of "a" from the 2nd equation so we substituted the value of a as (9-c) which we got from the 1st equation now after we did the substitution in the 2nd equation we solved it to get c now we have c we just put the value of c in our 1st equation to get the value of a
Did i solve for c correctly?
well your answer is correct c is 4 but i didn't get your method it looks messed up
okay so how do I do it correctly?
\[8(-c+9)+\frac{ 3^{2}*5 }{ 2*5 }c=58\]
I just simplify that to get c=4
oh okay :) well you can see it all explained here- http://www.cymath.com/answer.php?q=8(9-c)%2B4.5c%3D58
Okay we have c=4 now we need a which im guessing is a=5
yeah thats correct
but how do we solve for a?
sub in c?
yeah
a+c=9 a+4=9 a=5
correct
Thank you pogo
:D npo prbl yuko
*np
was this substitution method?
yes it was substitution method :)
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