3x-1/5-2/9x=124/5how do you solve this?
1 answer:
So first you would combine the like-terms so you get:
and combine the second like-terms:
now you only have a two step equation and you continue by doing the reciprocal and you add the -1/5 to the 24.8 where you get:
7/9x=24.6
then because the 7/9 is being multiplied by the x and your trying to isolate it you divide it by the 7/9 on both sides and you'll end up getting
x=31.628571429....
and combine the second like-terms:
now you only have a two step equation and you continue by doing the reciprocal and you add the -1/5 to the 24.8 where you get:
7/9x=24.6
then because the 7/9 is being multiplied by the x and your trying to isolate it you divide it by the 7/9 on both sides and you'll end up getting
x=31.628571429....
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Answer:
2/5
Step-by-step explanation:
Hi! Whenever you find a limit, you first directly substitute x = 5 in.
Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.
For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.
From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.
From x²-25, you can factor as (x+5)(x-5) via differences of two squares.
After factoring the expressions, we get a new Limit.
We can cancel x-5.
Then directly substitute x = 5 in.
Therefore, the limit value is 2/5.
L’Hopital Method
I wouldn’t recommend using this method since it’s <em>too easy</em> but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.
The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.
So from the given function:
Differentiate numerator and denominator, apply power rules.
<u>Differential</u> (Power Rules)
<u>Differentiation</u> (Property of Addition/Subtraction)
Hence from the expressions,
<u>Differential</u> (Constant)
Therefore,
Now we can substitute x = 5 in.
Thus, the limit value is 2/5 same as the first method.
Notes:
- If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.