Could I solve (800+x)tan33=xtan35 like xtan33=(x/800)tan35?
x tan 33 = x tan 35 - 800 tan 33
I do not know how you got what you got, but I just did this problem a few minutes ago and will go find it I hope.
NOTE - Steve left a tan 33 out of 800 tan 33
http://www.jiskha.com/display.cgi?id=1419896043
No
you have to expand (800+x)tan33 to get at the x inside the bracket.
so ...
800tan33 + xtan33 = xtan35
now get all the x terms to one side
800tan33 = xtan35 - xtan33
x is a common factor:
800tan33 = x(tan35-tan33)
divide both sides by tan35-tan33
800tan33/(tan35-tan33) = x
with a good scientific calculator you can now this in one sequence of keystrokes:
800
x
tan33
รท
(
tan35
-
tan33
)
=
to get 10226.9023
This answer satisfies the original equation, it does not satisfy your equation.
How did you possibly come up with that ?
agree with Reiny. Note the problem here is really not the algebra but the problem of keeping significant figures when subtracting numbers that are close to each other. However, you must start by doing the algebra correctly :)
I divided by 800 on both sides to get xtan33=(x/800)tan 35.
To solve the equation (800+x)tan33=xtan35, you cannot directly simplify it as xtan33 = (x/800)tan35.
However, you can solve the equation step by step. Here's how:
1. Distribute the tangent function to both terms on the left side of the equation:
800tan33 + xtan33 = xtan35
2. Group the terms involving "x" on one side of the equation:
xtan33 - xtan35 = -800tan33
3. Factor out "x" from both terms:
x(tan33 - tan35) = -800tan33
4. Divide both sides of the equation by (tan33 - tan35) to isolate "x":
x = -800tan33 / (tan33 - tan35)
So, the solution to the equation is x = -800tan33 / (tan33 - tan35).