Write as a single fraction in the simplest form $\frac{3x}{2x^2-50}-\frac{1}{x-5}$. $x = \sqrt[3]{\frac{25y}{7z+2}}$ Find x when y = 25 and z = -1. Answer x = ___ Rearrange the formula to make y the subject. Answer y = ___ Solve the equation $\frac{2x+3}{(3x+1)(x-1)} - 2 = 0$. Answer x = ___ or ___

1. Factorise the quadratic denominator.

$2x^2-50=2(x^2-25)=2(x+5)(x-5)$

So
$\frac{3x}{2x^2-50}-\frac{1}{x-5}=\frac{3x}{2(x+5)(x-5)}-\frac{1}{x-5}.$

2. Use a common denominator to write both terms over one denominator.

The common denominator is $2(x+5)(x-5)$.

\frac{3x}{2(x+5)(x-5)}-\frac{1}{x-5}=rac{3x}{2(x+5)(x-5)}-\frac{2(x+5)}{2(x+5)(x-5)}.

3. Combine the numerators and simplify.

$\frac{3x}{2(x+5)(x-5)}-\frac{2(x+5)}{2(x+5)(x-5)}$
$=\frac{3x-2(x+5)}{2(x+5)(x-5)}$
$=\frac{3x-2x-10}{2(x+5)(x-5)}$
$=\frac{x-10}{2(x+5)(x-5)}.$

1. Substitute $y=25$ and $z=-1$ into the formula and evaluate.

$x=\sqrt[3]{\frac{25(25)}{7(-1)+2}}=\sqrt[3]{\frac{625}{-5}}=\sqrt[3]{-125}=-5$

1. Cube both sides to remove the cube root.

$x=\sqrt[3]{\dfrac{25y}{7z+2}}$

$x^3=\dfrac{25y}{7z+2}$

2. Isolate $y$ as the subject.

$x^3(7z+2)=25y$

$y=\dfrac{x^3(7z+2)}{25}$

1. Clear the denominator to form an equation without fractions.

$\frac{2x+3}{(3x+1)(x-1)}-2=0$
$\frac{2x+3}{(3x+1)(x-1)}=2$
Multiply both sides by $(3x+1)(x-1)$:
$2x+3=2(3x+1)(x-1)$

2. Expand and rearrange into a quadratic equation.

$(3x+1)(x-1)=3x^2-2x-1$
So
$2x+3=2(3x^2-2x-1)$
$2x+3=6x^2-4x-2$
$6x^2-6x-5=0$

3. Solve the quadratic equation and state the valid solutions.

$6x^2-6x-5=0$
Using the quadratic formula,
$x=\frac{-(-6)\pm\sqrt{(-6)^2-4(6)(-5)}}{2(6)}$
$x=\frac{6\pm\sqrt{36+120}}{12}$
$x=\frac{6\pm\sqrt{156}}{12}$
$x=\frac{6\pm2\sqrt{39}}{12}$
$x=\frac{3\pm\sqrt{39}}{6}$
Since $x\neq -\dfrac{1}{3}$ and $x\neq 1$, both solutions are valid.