How can you tell if a solution is a viable?
If you are asking how to tell if an answer is viable; a viable answer, means that our result is a possible solution to the question being asked. After solving a question or an equation in Mathematics, we can use the result to prove that it is a possible, or viable, answer. For example, if we have the following equation:
if we solve for X we will have:
Then to show that it is a viable answer, we have to "plug in" the value that we calculated for X into the original equation:
Since `7=7`, x=2 is a viable answer to this question.
If we have these two equations:
`y=x+1` and `y+1=2x`
then by using substitution, we can solve the equations for x first, then y, and finally prove that the answers are correct by plugging them in:
`x+2=2x` combine like terms
`2=x` Subtract X from both sides
Then we can use the value that we calculated for X to find what Y is:
`y=(2)+1` OR `y+1=2(2)`
No matter which one we plug X into, we will get the following result for Y:
So in order to tell if the ordered pair `(2,3)` satisfies both equations, or is a viable answer for both, we must plug in the x and y values into one of the original equations to test it:
Since, 3=3 is a valid statement of equality, we know that the answers that we have calculated are both viable answers to this problem.
I will give two examples that determine whether if an answer is viable.
Example 1: `x^2-1 = 0`
Add one on both sides and take the square root.
`x =+- sqrt1`
If we plugged these two solutions back to the original equation, we can satisfy that:
`sqrt1^2-1 = 0`
`(-sqrt1)^2-1 = 0`
Therefore, are both viable solutions.
Example 2: `(x^2-1)/(x-1) =2`
The numerator can be factored to:
The (x-1) terms cancel, and we have:
If we substitute back to , we have:
`0/0 = 2`
The left side becomes indeterminate, which means that cannot be a solution as our domain cannot be one.