Alan

x + 5 + 4/x = 0

Multiply all terms by x

x² + 5x + 4 = 0

This factorises as:

(x + 1)(x + 4) = 0

So the two solutions are:  x = -1  and  x = -4

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$$\sqrt{x^3y^4z^6}=\sqrt{x\times x^2\times (y^2)^2\times (z^3)^2}=xy^2z^3\sqrt{x}$$

Note:  There aren't any denominators!

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Yes, using: a*logb = logb^a  and loga + logb = log(a*b)

2log₃x + log₃5 = log₃x² + log₃5 = log₃(5x²)

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b = number of bikes,  t = number of trikes

2*b + 3*t = 57       wheels

b = 3*t - 3          

Substitute the 2nd equation into the first

2*(3*t - 3) + 3*t = 57 or

9t - 6 = 57 or

9t = 63  or

t = 7

Put this back into the 2nd equation;

b = 3*7 - 3 or

b = 18

So, in total: b + t = 25

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Here's my algebraic approach, though it's even longer than heureka's!

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If you mean 

$$x^2-10=x^2-98$$

There is no solution, it's a false statement.

If you mean

$$x^{2-10}=x^{2-98}$$

this can be written as:

$$x^{88}=1$$

There are two real solutions here, namely x = ±1

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Oops!  Yes, the book is correct.  I answered without thinking it through carefully enough.  I should have drawn a picture!

I've taken a point off myself!

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To address Melody's question:

2²ⁿ = 4ⁿ = (3 + 1)ⁿ

Expand

(3 + 1)ⁿ = 3ⁿ + n×3^n-1 + n(n-1)×3^n-2/2 + ... + n×3 + 1

So 2²ⁿ - 1 = (3 + 1)ⁿ - 1 = 3n + n×3^n-1 + n(n-1)×3^n-2/2 + ... + n×3

Every term on the far right is exactly divisible by 3, hence 2²ⁿ - 1 is exactly divisible by 3.  The only prime number that can be exactly divisible by 3 is 3 itself. Hence there are no further examples of the type asked for in the original question.

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Here's a Geogebra visual solution (sorry, I used D for the orthocenter, not H):

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Perhaps this graph will help:

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