SOLUTION: hi, i am having trouble solving these , would you be so kind to explain how to determine whether the equation defines Y as a function of X: 1. 6x+3y^2=7 2. square

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Question 1050162: hi, i am having trouble solving these , would you be so kind to explain how to determine whether the equation defines Y as a function of X:
1. 6x+3y^2=7
2. square root (6x) +y=5
3. 5x^2+y^3=27
4. 4x+ IyI=3

Thanks in advance it would help tremendously to understand on how to solve these

Found 2 solutions by stanbon, advanced_Learner:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
explain how to determine whether the equation defines Y as a function of X:
1. 6x+3y^2=7
6x = -3y^2+7
x = -(1/2)y^2 + (7/6)
Parabola opening to the left.
Not a function.
---------------------------------------
2. square root (6x) +y=5
y = -sqrt(6x) + 5
Defined for x >= 0
As x gets larger y just gets more and more negative.
graph%28400%2C400%2C-10%2C100%2C-100%2C10%2C-sqrt%286x%29%2B5%29
Is a function
---------------------------------------
3. 5x^2+y^3=27
y = (-5x^2+27)^(1/3)
graph%28400%2C400%2C-10%2C10%2C-10%2C10%2C%28-5x%5E2%2B27%29%5E%281%2F3%29%29
Cubics tend to be functions.
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4. 4x+ IyI=3
Is that the "greatest integer" symbol?
If so, it looks like a step-function.
Is a function
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Cheers,
Stan H.
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Answer by advanced_Learner(501) About Me  (Show Source):
You can put this solution on YOUR website!
hi, i am having trouble solving these , would you be so kind to explain how to determine whether the equation defines Y as a function of X:
1. 6x+3y^2=7
2. square root (6x) +y=5
3. 5x^2+y^3=27
4. 4x+ IyI=3
solve for y
1. 6x%2B3y%5E2=7
1. 3y%5E2=7-6x
1. y%5E2=%287-6x%29%2F%283%29
1. y=+%2B-sqrt%28%287-6x%29%2F%283%29%29
2.sqrt%286x%29+%2By=5
2.y=5-sqrt%286x%29
3. 5x%5E2%2By%5E3=27
3. y%5E3=27-5x%5E2
3. y=cbrt%2827-5x%5E2%29
4. 4x%2B+IyI=3
4. IyI=%283-4x%29
meaning
y=3-4x+
or
-y=3-4x
y=4x-3