Agile Algebra

We thank Patrick from Woodbridge School for these solutions to parts (1) and (3). How about sending in your solution to part (2) or part (4)?    
 
1) Let us take $y = 2x - 5 + \frac{3}{x}$, then we have $\frac{2x}{xy} + \frac {13x}{xy + 6x} = 6$, so $\frac {2}{y} + \frac {13}{y+6} = 6.$

Expanding the fractions, we have $2(y+6) + 13y = 6y(y+6) = 15y + 12.$  Dividing both sides by 3, we have $2y(y+6) = 5y + 4.$  Expanding the bracket, we have $2y^2 + 12y - 5y - 4 = 0 = 2y^2 + 7y - 4$. We can factorise this to give $(2y - 1)(y + 4) = 0$, so $y = \frac {1}{2}$ or $-4$.

We know  $y = 2x - 5 + \frac{3}{x}$, so taking $y = \frac{1}{2}$ we have $\frac{1}{2} = 2x - 5 + \frac{3}{x}$ so we have $1 = 4x - 10 + \frac {6}{x}$ giving  $4x^2 - 10x + 6 = x$ and then  $4x^2 - 11x + 6 = 0.$  Factoring:  $(4x - 3)(x - 2) = 0$, so $x = 2$ or $\frac{3}{4}$.

Now, take $y = -4$ then $-4 = 2x - 5 + \frac{3}{x}$  which gives  $2x - 1 + \frac{3}{x} = 0 $ and $2x^2 - x + 3 = 0$ which has no real solutions.

Therefore, the solutions are $x = 2$ or $x = \frac{3}{4}$.
 
Alternatively, you could take advantage of the symmetry by substituting $y = 2x +\frac{3}{x}$ which simplifies the algebra a little.  Patrick also sent a solution to part (3).  Again substitutions turn the quartic equation into a quadratic which can be solved.      
 
(3) $(x-4)(x-5)(x-6)(x-7) = 1680.$  

Let us rewrite the equation with $a = x-7$, then we have $(a+3)(a+2)(a+1)a = 1680.$

Note that $a(a+3) = a^2 + 3a$, and $(a+2)(a+1) = a^2 + 3a + 2$, so $(a+2)(a+1)(a+3)a = (a^2 + 3a)(a^2 + 3a + 2).$  Let us then take $y = a^2 + 3a$, then $y(y+2) = 1680$ and $y^2 + 2y - 1680 = 0.$
Factorising, we find $(y-40)(y+42) = 0$, so $y = -42$ or $+40$. Since $y = a^2 + 3a$, then we have either: $40 = a^2 + 3a$, so $(a+8)(a-5) = 0$ and $a = -8$ or $5$.

Alternatively $-42 = a^2 + 3a$, which has no real roots. Therefore, we have $a = -8$ or $5$. Then, since $a = x - 7$, $x = a + 7$, so $x = -1$ or $12$.

We will test this by substitution: $(-1-4)(-1-5)(-1-6)(-1-7)$, so for each bracket (-a-b) we can take out the negative to give -(a+b). There are four such brackets, so there are four -1 factors removed, which multiply to give 1. Therefore, $(1+4)(1+5)(1+6)(1+7) = 5\times 6\times 7\times 8 = 1680.$ Similarly, $x=12$ gives us $8\times 7\times 6\times 5 = 1680.$
 
Well done Patrick. Again alternative substitutions include $t= x - \frac{11}{2}$ and a little more work will give the two complex solutions to this quartic equation, namely $\frac{11\pm i\sqrt159}{2}$.