$b = \dfrac{a}{a-1} + \dfrac{2}{c}$ Find the value of $b$ when $a = 3$ and $c = 4$. Answer b = ... Rearrange the formula to make $a$ the subject. Answer ... Solve these simultaneous equations. $4x - y = -11$ $5x + 3y = -1$ Answer x = ... y = ... Solve the equation $\dfrac{x}{3x-1} - \dfrac{5}{2x+3} = 1$. Give your solutions correct to three decimal places. Answer x = ... or ...

1. Substitute $a=3$ and $c=4$ into the formula for $b$ and simplify.

$b=\dfrac{a}{a-1}+\dfrac{2}{c}$

$=\dfrac{3}{3-1}+\dfrac{2}{4}$

$=\dfrac{3}{2}+\dfrac{1}{2}=2$

1. Clear the denominator $c$ from the term $\dfrac{2}{c}$.

$b=\dfrac{a}{a-1}+\dfrac{2}{c}$

$bc=\dfrac{ac}{a-1}+2$

2. Clear the denominator $a-1$ and expand.

$bc-2=\dfrac{ac}{a-1}$

$(bc-2)(a-1)=ac$

$abc-2a-bc+2=ac$

3. Collect the terms containing $a$, factorise, and divide to isolate $a$.

$abc-2a-ac=bc-2$

$a(bc-2-c)=bc-2$

$a(bc-c-2)=bc-2$

$a=\dfrac{bc-2}{bc-c-2}$

1. Make one variable the subject using the first equation.

From $4x-y=-11$,
$-y=-11-4x$
$y=11+4x$

2. Substitute the expression for $y$ into the second equation and solve for $x$.

Substitute $y=11+4x$ into $5x+3y=-1$:
$5x+3(11+4x)=-1$
$5x+33+12x=-1$
$17x=-34$
$x=-2$

3. Substitute $x=-2$ back to find $y$.

Using $y=11+4x$,
$y=11+4(-2)$
$y=3$

1. Multiply throughout by the common denominator to remove the algebraic fractions.

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

2. Expand and simplify to form a quadratic equation.

$2x^2+3x-15x+5=6x^2+7x-3$
$2x^2-12x+5=6x^2+7x-3$
$4x^2+19x-8=0$

3. Solve the quadratic equation using the quadratic formula.

For $4x^2+19x-8=0$,
$x=\frac{-19\pm\sqrt{19^2-4(4)(-8)}}{2(4)}$
$x=\frac{-19\pm\sqrt{489}}{8}.$

4. Evaluate the two solutions and round to three decimal places.

$x=\frac{-19+\sqrt{489}}{8}=0.8159\ldots$
$x=\frac{-19-\sqrt{489}}{8}=-5.5659\ldots$
The value $x=-5.5659\ldots$ is valid since it does not make either denominator zero.