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grigory [225]
3 years ago
13

Determine whether the following equation defines y as a function of x.

Mathematics
1 answer:
Aleks04 [339]3 years ago
3 0

Answer:

The equation can define y as a function of x and it also can define x as a function of y.

Step-by-step explanation:

A relation is a function if and only if each value in the domain is mapped into only one value in the range.

So, if we have:

f(x₀) = A

and, for the same input x₀:

f(x₀) = B

Then this is not a function, because it is mapping the element x₀ into two different outputs.

Now we want to see it:

x + y = 27

defines y as a function of x.

if we isolate y, we get:

y = f(x) = 27 - x

Now, this is a linear equation, so for each value of x we will find an unique correspondent value of y, so yes, this is a function.

Now we also want to check if:

x +  y = 27

defines x as a function of y.

So now we need to isolate x to get:

x = f(y) = 27 - y

Again, this is a linear equation, there are no values of y such that f(y) has two different values. Then this is a function.

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1 step (B): raise both sides of the equation to the power of 2.

(\sqrt{x+3}-\sqrt{2x-1})^2=(-2)^2,\\   (x+3)-2\sqrt{x+3}\cdot \sqrt{2x-1}+(2x-1)=4,\\ 3x+2-2\sqrt{x+3}\cdot \sqrt{2x-1}=4.

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4 step (E): simplify to get a quadratic equation.

4(2x^2-x+6x-3)=(3x)^2-2\cdot 3x\cdot 2+2^2,\\ 8x^2+20x-12=9x^2-12x+4,\\ x^2-32x+16=0.

5 step (D): use the quadratic formula to find the values of x.

D=(-32)^2-4\cdot 16=1024-64=960, \\ \sqrt{D} =8\sqrt{5} ,\\ x_{1,2}=\dfrac{32\pm 8\sqrt{5}}{2} =16\pm 4\sqrt{5}.

6 step (C): apply the zero product rule.

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Additional 7 step: check these solutions, substituting into the initial equation.

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