SOLUTION: If a + b = 11, what are two possible values for c that satisfy the equation below: (ax - 4)(bx + 7) = 30x^2 + cx - 28

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Question 1111142: If a + b = 11, what are two possible values for c that satisfy the equation below:
(ax - 4)(bx + 7) = 30x^2 + cx - 28
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
%28ax+-+4%29%28bx+%2B+7%29+=+30x%5E2+%2B+cx+-+28
abx%5E2-4bx%2B7ax-28=30x%5E2%2Bcx-28
abx%5E2%2B%287a-4b%29x-28=30x%5E2%2Bcs-28

Compare the corresponding parts to show
system%28ab=30%2C7a-4b=c%29

Can you continue from that?

Answer by ikleyn(52809) About Me  (Show Source):
You can put this solution on YOUR website!
.
After opening the parentheses, you have  ab = 30.


So, actually, you have this system of two equations

a + b = 11,
ab    = 30.


There are two pairs of solutions:  (a,b) = (5,6)   and   (a,b) = (6,5).


I found them MENTALLY.


If you want ALGEBRAIC solution, express a = 11-b  from one of the equations and substitute it into the other.


You will get (11-b)*b = 11.


It is a quadratic equation.  Write it in the standard form and solve by any method you want / (you know) / (you like).



After finding "a" and "b",  

c = 7a - 4b = 7*5-4*6 = 35-24 = 11   or

c = 7a - 4b = 7*6-4*5 = 42-20 = 22.


So, the two possible values for "c" are  11  and  22.

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Ignore writing by  @josgarithmetic,  since it will lead you to  N O W H E R E.