http://www.mathsgenie.co.uk/papers/EDEXCELS22H.pdf
question 21 explain how please
Given that
2x−1 : x−4 = 16x+1 : 2x−1
find the possible values of x.
(We can see that x ≠ 4 and x ≠ 1/2 , because those values cause a zero in the denominator.)
\(\frac{2x-1}{x-4}=\frac{16x+1}{2x-1}\) Cross multilpy.
\((2x-1)(2x-1)=(16x+1)(x-4) \)
\(4x^2-4x+1=16x^2-63x-4\) Subtract 4x2 from both sides of the equation.
\(-4x+1=12x^2-63x-4\) Add 4x to both sides of the equation.
\(1=12x^2-59x-4\) Subtract 1 from both sides of the equation.
\(0=12x^2-59x-5\) Use the quadratic formula to solve for x .
\(x = {59 \pm \sqrt{(-59)^2-4(12)(-5)} \over 2(12)} \\~\\ x = {59 \pm 61 \over 24} \\~\\ x=5\qquad\text{and}\qquad x=-\frac1{12}\)
Also..I did work on the question 20 at first if you need help on that one...I had a whole answer ready for it then I saw you changed it! Haha
2x -1 : x - 4 = 16x + 1 : 2x -1
So....we can write this as
[ 2x - 1 ] / [x - 4 ] = [ 16x + 1 ] / [2x -1] cross-multiply
[2x -1 ] [2x -1 ] = [x - 4] [ 16x + 1 ] simplify
4x^2 - 4x + 1 = 16x^2 - 64x + x - 4
4x^2 - 4x + 1 = 16x^2 -63x - 4 subtract the left side from the right
12x^2 - 59x - 5 = 0 factor
(12x + 1) ( x - 5 ) = 0
Set both factors to 0 and solve.....and x = -1/12 or x = 5