Why no solution?

[englsinger_J]

I have two problems where the answers are no solution.

(2x)/(x-3)=7+(6/(x-3)) I just solved an equation just like this but the 6 was replaced with a 4 and the answer was attainable. even in this problem I was able to "solve" it.

another problem is: the square root of (x+5)+7=0. I was also able to "solve" this one also but the answer is evidently no answer, why is that?

any help on either or both of these problems would be greatly appreciated!

 

[stapel]

Please show your steps and the answers you got, so that we may check your work.

Thank you.

Eliz.

 

[englsinger_J]

the main problem with these questions is just that the answer that I am supposed to get is no answer

for the first question I subtracted 6/(x-3) to the other side. next I times (x-3) to both sides and distributed it with the 7. at this point I have 2x-6=7x-21. then I added 21 to each side and subtract 2x from each side. divide each side by 5 and I got x=3.

for the second problem I subtracted the 7 from both sides and then squared both sides so I got x+5=49. subtract 5 from each side and I got x=44

 

[Mrspi]

the main problem with these questions is just that the answer that I am supposed to get is no answer

for the first question I subtracted 6/(x-3) to the other side. next I times (x-3) to both sides and distributed it with the 7. at this point I have 2x-6=7x-21. then I added 21 to each side and subtract 2x from each side. divide each side by 5 and I got x=3.

for the second problem I subtracted the 7 from both sides and then squared both sides so I got x+5=49. subtract 5 from each side and I got x=44

Did you try to check your answer to either of these problems? You'll find that the answers that you got DO NOT check. If you have used "legal procedures" in solving an equation, and come up with a solution that doesn't work, then the original equation has no solution.

In the first case, using 3 for x will make the denominator of a fraction 0, and you can't have 0 as the denominator of a fraction. When you multiplied both sides of the equation by x -3, you were in fact multiplying by 0.....

When you multiply both sides of an equation by an expression containing the variable, or when you raise both sides of an equation to an even power, you may produce a new equation whose solutions are not necessarily solutions to the original equation.

That's why is is extremely important to check all solutions you obtain this way.