Suppose you want to solve the equation 2a+b=2a, where a and b are nonzero real numbers. describe the solution.
1 answer:
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Answer:
Step-by-step explanation:
hello :
Part A : x+6y =6 means : 6y = - x+6
so : y = (-1/6)x+1 an equation for the line (D)
y = (1/3)x -2 is the line (D')
PartB : solution of the system : y = (-1/6)x+1 ....(1) color red
y = (1/3)x -2 ....(2) color bleu
is the intersection point : (6 ; 0)
PartC : Algebraically by (1) and (2) : (-1/6)x+1 = (1/3)x -2
(-1/6)x - (1/3)x = -2-1
(-x-2x)/6= -3
-3x = -18
so : x = 6 put this value in (1) or (2) : y = (-1/6)(6)+1 =0 the solution is : (6 ;0)
Answer:
The graph has a removable discontinuity at x=-2.5 and asymptoe at x=2, and passes through (6,-3)
Step-by-step explanation:
A rational equation is a equation where
where both are polynomials and q(x) can't equal zero.
1. Discovering asymptotes. We need a asymptote at x=2 so we need a binomial factor of
in our denomiator.
So right now we have
2. Removable discontinues. This occurs when we have have the same binomial factor in both the numerator and denomiator.
We can model -2.5 as
So we have as of right now.
Now let see if this passes throught point (6,-3).
So this doesn't pass through -3 so we need another term in the numerator that will make 6,-3 apart of this graph.
If we have a variable r, in the numerator that will make this applicable, we would get
Plug in 6 for the x values.
So our rational equation will be
or
We can prove this by graphing
