From 1/5x = (y-3)^3 , which is the inverse and correct, you now have to solve for y
So, take the cube root of both sides
(x/5)^(1/3) = y-3
y = (x/5)^(1/3) + 3
in original let x = 4
y = 5(1)^3 = 5
now use x=5 in your inverse
y = (5/5)^(1/3) + 3 = 4 , which is what we started with.
now another, x = 8
y = 5(5^3) = 625
now use x=625 in inverse:
y = (625/5)^(1/3) + 3
= (125)^(1/3) + 3
= 5 +3 = 8
all looks good!!!
Where did you get 4 from? For the first one? Where you wrote in original let x = 4?
And I do not get this one
y = x-2/5x-3
x = y-2/5y-3
I could have picked any x I wanted, since it must be true for all values of x
I simply picked x = 4, because it gave me a "nice" answer.
Same reason I picked x=8
I could have picked x = 45.678
As to your second post
is that a new equation??
did you mean:
y = (x-2)/(5x-3)
those brackets are critically important.
Oh okay that makes sense, Im just sort of still confused with how the answer to the first one needs to be layed out
I got y = ³√(1/5x) +3
And yes that is a new equation on my second post, and yeah I meant that, sorry will remember to put brackets next time.
y = ³√(1/5x) +3 is good
Just make sure when you type the bracket part (1/5x) , it is only 5 that is divided and not the 5 as well
that is why I wrote it as (x/5)
Your new problem:
y = (x-2)/(5x-3) -----> x = (y-2)/(5y-3) as the inverse
5xy - 3x = y-2
5xy - y = 3x - 2
y(5x - 1) = 3x - 2
y = (3x-2)/(5x - 1)
test by taking any suitable value for x in the original
I suggest x = 2, (worked for me)
Could you explain why it wouldnt be 1/5x again? Because isnt there an invisible 1 beside the x, and so when we divide the 5 doesnt it become 1/5x?
For the second equation to get rid of 5y-3 on the bottom you multiply it on the other side? Could you explain step by step the whole cross multiplication?
I didn't say it shouldn't be 1/5x
I just wanted you to be aware of the difference between
1/5x and 1/(5x)
for your second part of your question ....
x = (y-2)/(5y-3) , so far ok ?
x/1 = (y-2)/(5y-3)
the common denomator is 5y-3 , so multiply each side by that
x(5y-3) = (y-2)/(5y-3) * (5y-3) , the * means multiply
so after you cancel the 5y-3 on the right, you get
x(5y-3) = 1(y-2)
now look at that pattern.
Did we not multiply the top of the left fraction by the bottom of the second fraction, and the bottom of the first fraction by the top of the 2nd fraction?
That is why we call it cross-multiplying
(use it only if you have a single fraction on the left, and a single fraction on the right of the equal sign)
How is 5y-3 is a common denominator?
Answer this Question
Precalculus - 2=-4+4 square root x^3 I added the four to both sides divided four...
Math - Find all solutions to the equation sqrt(2-3x)= 9. Express your answer as...
math,correction - Is this correct: Problem #13 Solve: 9x^2 = 14 I did the ...
Pre Calculus - square root of 3x-4 = (x-4)^2-3 Normally, I either take the ...
Pre-Calculus - (not sure how to type square root of) square root of 3x - 4 = (x-...
Math - I can not find the solutions to the following problems using 2 methods: ...
math-volume - I'm confused with how to solve this problem, so help would be ...
Pre Algebra - Please help me with some problems. Thank you! Solve the equation ...
Pre Calculus - Solve 2cos^2(2x)=1 I divided both sides by 2 and got this cos^2(...
math - I need help figuring out these two questions on euations with integers. -...