plug your solution back into the original equation and work it out again
how can the reflexive property be applied to check the accuracy of a solution to equation?
You can check an answer by substituting it into the equation. x - 17 = 11 25 - 17 = 11 8 = 11 Since this is not true, 25 is not the answer.
By substitution.
An "extraneous solution" is not a characteristic of an equation, but has to do with the methods used to solve it. Typically, if you square both sides of the equation, and solve the resulting equation, you might get additional solutions that are not part of the original equation. Just do this, and check each of the solutions, whether it satisfies the original equation. If one of them doesn't, it is an "extraneous" solution introduced by the squaring.
It really depends on the type of equation, but in the simpler cases - those that you are likely to encounter in high school algebra - you will usually need to replace the purported solution into the original equation, then simplify the equation as appropriate. If this results in a true statement (for example, "5 = 5"), then the solution is correct; if you get a false statement (for example, "1 = 0"), then the purported solution is not correct.
Solution. A solution of an equation is a number that satisfy the equation. This means that if you replace this number on the equation and check it, the equation will be true. When you solve an equation you can find some roots, but not all of them satisfy the equation. Thus always check your answers after resolving your equation, and eliminate as solution the answers that don't make the equation true or undefined.
That's an extraneous solution. You need to check for these when algebraically solving equations, especially when you take both sides of an equation to a power.
If you found the value of x that is a solution to an equation, you want to substitute that value back into the original equation, to check that it indeed satisfies the equation. If it does not satisfy the equation, then you made an error in your calculations, and you need to rework the problem.
Substitute that value in the equation, and then check to see if the resulting statement is TRUE.
No, it is not. Substitute and check. You have 128 + 16r. So now, if r=9, then the equation would be 128 = 16x9. But, 16x9 is 145 and is not equal to 128. Thus, r=9 is not the solution. The solution is r+ 128/16 which is equal to 8. Thus, the actual solution is r = 8.
insert the answer in the equation, replacing the variable, and see if it still makes sense.