It is given that $y=\dfrac{5x-2}{3x+1}$. Find $y$ when $x=7$. Answer $y$ = ... Express $x$ in terms of $y$. Answer $x$ = ... Solve $\dfrac{x-5}{3} - \dfrac{x+3}{2} = -4$. Answer $x$ = ... Solve these simultaneous equations. $3x - 2y = 10$ $2x + y = 23$ Answer $x$ = ... $y$ = ... Simplify $\dfrac{4x^2 - 9}{6x^2 - x - 12}$. Answer ...

1. Substitute $x=7$ into the given expression for $y$ and evaluate.

$y=\frac{5(7)-2}{3(7)+1}=\frac{35-2}{21+1}=\frac{33}{22}=\frac{3}{2}$

1. Clear the denominator by multiplying both sides by $3x+1$.

$y=\dfrac{5x-2}{3x+1}$

$y(3x+1)=5x-2$

2. Expand, collect the $x$-terms, and factorise to express $x$ in terms of $y$.

$3xy+y=5x-2$

$3xy-5x=-2-y$

$x(3y-5)=-(y+2)$

$x=\dfrac{-(y+2)}{3y-5}=\dfrac{y+2}{5-3y}$

1. Clear the denominators by multiplying every term by $6$.

$\frac{x-5}{3}-\frac{x+3}{2}=-4$
Multiply throughout by $6$:
$2(x-5)-3(x+3)=-24$

2. Expand, simplify, and solve for $x$.

$2x-10-3x-9=-24$
$-x-19=-24$
$-x=-5$
$x=5$

1. Make the coefficients of $y$ equal in magnitude so that $y$ can be eliminated.

Given
$3x-2y=10 \quad \text{(1)}$
$2x+y=23 \quad \text{(2)}$
Multiply (2) by $2$:
$4x+2y=46 \quad \text{(3)}$

2. Add the equations to eliminate $y$ and solve for $x$.

Add (1) and (3):
$(3x-2y)+(4x+2y)=10+46$
$7x=56$
$x=8$

3. Substitute $x=8$ into one original equation to find $y$.

Substitute $x=8$ into (2):
$2(8)+y=23$
$16+y=23$
$y=7$

1. Factorise the numerator using the difference of two squares.

$4x^2-9=(2x)^2-3^2=(2x-3)(2x+3)$

2. Factorise the quadratic denominator.

$6x^2-x-12=(2x-3)(3x+4)$

since
$(2x-3)(3x+4)=6x^2+8x-9x-12=6x^2-x-12.$

3. Cancel the common algebraic factor and write the simplified fraction.

$\frac{4x^2-9}{6x^2-x-12}=\frac{(2x-3)(2x+3)}{(2x-3)(3x+4)}=\frac{2x+3}{3x+4}$