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^2=1`

`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:

`((x-1)(x+1))/(x-1) =2`

The (x-1) terms cancel, and we have:

`x+1=2`

`x=1`

If we substitute back to , we have:

`(1^2-1)/(1-1) =2`

`0/0 = 2`

The left side becomes indeterminate, which means that cannot be a solution as our domain cannot be one.

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:

`5+x=7`

if we solve for X we will have:

`x=2`

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:

`5+(2)=7`

Since `7=7`, x=2 is a viable answer to this question.

**Another example**:

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+1)+1=2x`

`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:

`y=3`

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:

`(3)=(2)+1`

`3=3`

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.

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