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proyaop

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 #1
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y is a function of x, so y=f(x):

f(3)=9a3b+c=8                               Equation 1

f(1)=ab+c=5                                   Equation 2

f(2)=4a+2b+c=11                                Equation 3

 

This is a system of 3 equations and 3 variables, so it is solvable (and there is no redundancy).

Let's get solving!

 

First we want to change it into a system of 2 variables and 2 equations by manipulating to get rid of the variable c via elimination.

Equation 1 - Equation 2: 9a3b+c(ab+c)=85=8a2b=3                 Equation 1.2

Equation 3 - Equation 2: 4a+2b+c(ab+c)=115=3a+3b=6a+b=2       Equation 3.2

Then, we can eliminate b to find the value of a: Equation 1.2 + 2 * (Equation 3.2):

8a2b+2(a+b)=3+22=10a=7a=710                                 Value of a

Substitute this back into our equation a+b=2:

710+b=2=2010b=2010710=1310                                    Value of b

Substitute both variables' values into the equation ab+c=5 (it doesnt matter which equation you choose; take the easier one).

7101310+c=5=5010c=5010+1310710=5610                       Value of c

 

a+b+c=710+1310+5610=7610=7.6

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Feb 13, 2024