I have answered the question but not sure how to check for extraneous solution.
x/(5x+4)=3
The answer is -6/7
X=-6/7
This was answer to the problem
x/(5x+4)=3
x=3(5x+4)
x=15x+12
x-15x=12
-14x=12
-7x=6
x=-6/7
there you go find algebra difficult?
I have that but it tells me to check for extraneous solutions and that is where I am lost for this problem
forget about that grammer just smile with your solution
and learn to use your dictionary although i am not a fan of english
But
In mathematics, an extraneous solution (or spurious solution) is a solution, such as that to an equation, that emerges from the process of solving the problem but is not a valid solution to the original problem.
Extraneous roots mostly occur after squaring an equation.
Since you verified your solution by subbing it back into the original equation, there are no extraneous roots
LS = x/(5x+4)
= (-6/7) / (5(-6/7) + 4)
= (-6/7) / (-2/7)
= (-6/7)(-7/2) = 42/14 = 3
= RS
To check for extraneous solutions in this case, you need to substitute the solution you found (-6/7) back into the original equation and see if it still holds true.
Let's plug in -6/7 for x in the original equation:
(-6/7) / (5(-6/7) + 4) = 3
Simplify the equation:
(-6/7) / (-30/7 + 4) = 3
(-6/7) / (-30/7 + 28/7) = 3
(-6/7) / (-2/7) = 3
-6/7 * (-7/2) = 3
(36/14) = 3
Finding the least common denominator (14) to add the fractions, we get:
18/7 = 3
As you can see, the left side of the equation does not equal the right side. Therefore, the solution -6/7 is not an extraneous solution in this case.
Hope this helps!