A. ASA B. CPCTC C. AAS D. SAS
2 answers:
You might be interested in
Answer:
Infinite many solutions. Any x-value can satisfy the equation.
Step-by-step explanation:
Let's work on simplifying the equation a little to investigate which x-values satisfy it. Start by combining like terms on the left side (6x +4x=10x),
then distribute the factor "10" into the binomial (x+10), obtaining 10x +30.
Now we have the same expression on the left and the right of the equal sign:
10x +30=10x+30. We may subtract 30 from both sides, and obtain 10x=10x, and at this point divide by 10 both sides, and we obtain: x=x
The process is shown below.
x=x is an equation that is verified by absolutely ANY x value on the number line, and there are infinite x-values in the number line.
Therefore there are infinite many solutions to this equation (any x-value will satisfy it).
Let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3
Then the equation is (x-x₁)(x-x₂) = 0
If you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.
You can see that a=1 (there is just an x²)
b = (-5 -2√7 + 2√7 - 5)/3 = -10/3
c = ((-5-2√7)/3 ) * (-5+2√7)/3) = -1/3
So the equation you're looking for is:
Then the equation is (x-x₁)(x-x₂) = 0
If you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.
You can see that a=1 (there is just an x²)
b = (-5 -2√7 + 2√7 - 5)/3 = -10/3
c = ((-5-2√7)/3 ) * (-5+2√7)/3) = -1/3
So the equation you're looking for is: