Intro. to College Algebra and Trigonometry questions!

[Littleone98]

So we are starting off the year by reviewing Algebra 2..which I haven't forgotten a lot about over the summer! Please help if you can with at least one and please explain it to me thanks

Solve the equation. Check for extraneous solutions.

6 over x-2 is equal to 5 over x-3


9 over x-3 is equal to 7 over x-5

K over (K+4) + 4 over (K-4) is equal to 6K + 8 over (K+4)(K-4)

 

[Loren]

6 over x-2 is equal to 5 over x-3
\(\displaystyle \frac{6}{x-2}=\frac{5}{x-3}\)

There is a very well known shortcut to solve this. But a very useful rule that applies to all fractional equations is...

1) Find the least common denominator of all the fractions in the equation.
2) Multiply both sides of the equation by this LCD.
3) Solve the resulting equation.
4) Check your answer to eliminate any extraneous roots.

Can you take it from there?

 

[Denis]

6 over x-2 is equal to 5 over x-3

If a/b = c/d then ad = bc ; so 6(x - 3) = 5(x - 2) ; wrap it up!