Simplify $\frac{1}{6x-4} - \frac{x+1}{3x^2-5x+2}$. $\sqrt{2x^2 + x + z} = y$ Evaluate $y$ when $x = -\frac{1}{2}$ and $z = \frac{5}{2}$. Express $x$ in terms of $y$ and $z$. Express $x^2 - 5x - 3$ in the form $(x + h)^2 + k$. Hence, state the coordinates of the minimum point of the graph of $x^2 - 5x - 3$. Solve $\displaystyle \frac{1}{5x+1} + \frac{3}{3x-1} = 2$.

1. Factorise the denominators to identify the common denominator.

$6x-4=2(3x-2)$
$3x^2-5x+2=(3x-2)(x-1)$
So
$\frac{1}{6x-4}-\frac{x+1}{3x^2-5x+2}=\frac{1}{2(3x-2)}-\frac{x+1}{(3x-2)(x-1)}.$

2. Write both fractions over a common denominator and subtract the numerators.

The lowest common denominator is $2(3x-2)(x-1)$.

$\frac{1}{2(3x-2)}-\frac{x+1}{(3x-2)(x-1)}$
$=\frac{x-1}{2(3x-2)(x-1)}-\frac{2(x+1)}{2(3x-2)(x-1)}$
$=\frac{(x-1)-2(x+1)}{2(3x-2)(x-1)}$
$=\frac{x-1-2x-2}{2(3x-2)(x-1)}$
$=\frac{-x-3}{2(3x-2)(x-1)}.$

1. Substitute the given values of $x$ and $z$ into the expression for $y$.

Since $\sqrt{2x^2+x+z}=y$,
$y=\sqrt{2\left(-\frac12\right)^2+\left(-\frac12\right)+\frac52}$

2. Simplify the value inside the square root.

$y=\sqrt{2\left(\frac14\right)-\frac12+\frac52}$
$y=\sqrt{\frac12-\frac12+\frac52}$
$y=\sqrt{\frac52}$

1. Square both sides to remove the square root.

$\sqrt{2x^2+x+z}=y$
$2x^2+x+z=y^2$

2. Rearrange into standard quadratic form in $x$.

$2x^2+x+z-y^2=0$

3. Apply the quadratic formula to express $x$ in terms of $y$ and $z$.

For $2x^2+x+(z-y^2)=0,$
$a=2,\quad b=1,\quad c=z-y^2.$
Using $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},$
$x=\frac{-1\pm\sqrt{1-4(2)(z-y^2)}}{2(2)}$
$x=\frac{-1\pm\sqrt{1-8z+8y^2}}{4}$
$x=\frac{-1\pm\sqrt{8y^2-8z+1}}{4}$

1. Use half the coefficient of $x$ to form the square term.

For $x^2-5x-3$, half of $-5$ is $-\frac{5}{2}$.

$x^2-5x-3=\left(x-\frac{5}{2}\right)^2-\left(\frac{5}{2}\right)^2-3$

2. Simplify the constant term to obtain the required form.

$\left(x-\frac{5}{2}\right)^2-\frac{25}{4}-3$

$=\left(x-\frac{5}{2}\right)^2-\frac{25}{4}-\frac{12}{4}$

$=\left(x-\frac{5}{2}\right)^2-\frac{37}{4}$

1. Read off the minimum point from the completed-square form.

From part (c)(i),
$x^2-5x-3=(x-\frac{5}{2})^2-\frac{37}{4}.$
Since $(x-\frac{5}{2})^2\geq 0$, the minimum occurs when
$x-\frac{5}{2}=0 \Rightarrow x=\frac{5}{2}.$
The minimum value is
$-\frac{37}{4}.$
Hence, the minimum point is
$(\frac{5}{2},-\frac{37}{4}).$

1. Clear the fractions by multiplying by the common denominator.

$\frac{1}{5x+1}+\frac{3}{3x-1}=2$
Multiply throughout by $(5x+1)(3x-1)$:
$(3x-1)+3(5x+1)=2(5x+1)(3x-1)$

2. Expand and rearrange into a quadratic equation.

$3x-1+15x+3=2(15x^2-2x-1)$
$18x+2=30x^2-4x-2$
$30x^2-22x-4=0$
$15x^2-11x-2=0$

3. Factorise and solve the quadratic equation.

$15x^2-11x-2=0$
$(5x-1)(3x+2)=0$
Hence,
$5x-1=0 \quad \text{or} \quad 3x+2=0$
$x=\frac{1}{5} \quad \text{or} \quad x=-\frac{2}{3}$