The problem consists of several sub-problems covering various topics in algebra. These include expressing numbers in standard and engineering notation, simplifying expressions, using the remainder theorem, solving simultaneous equations, solving logarithmic and linear equations, finding the gradient and y-intercept of a line, finding the equation of a line given two points, and determining the partial fraction decomposition of a rational expression.

1. Problem Description

The problem consists of several sub-problems covering various topics in algebra. These include expressing numbers in standard and engineering notation, simplifying expressions, using the remainder theorem, solving simultaneous equations, solving logarithmic and linear equations, finding the gradient and y-intercept of a line, finding the equation of a line given two points, and determining the partial fraction decomposition of a rational expression.

2. Solution Steps

a) (i) Express 5630 in standard form.

Standard form is expressing a number as $a \times 10^n$, where $1 \le |a| < 10$ and $n$ is an integer.

$5630 = 5.630 \times 10^3$

(ii) Express 16346000000 in engineering notation.

Engineering notation expresses a number as $a \times 10^n$, where $1 \le |a| < 1000$ and $n$ is a multiple of

3. $16346000000 = 16.346 \times 10^9$

b) Simplify the following expression: $(-5x^{-2}y) (-2x^{-3}y^2)$

$(-5x^{-2}y) (-2x^{-3}y^2) = (-5)(-2) (x^{-2}x^{-3})(y y^2)$

$= 10 x^{-2-3} y^{1+2} = 10 x^{-5} y^3 = \frac{10y^3}{x^5}$

c) Use the remainder theorem to find the remainder when $2x^3 + x^2 - 7x - 6$ is divided by $x-2$.

Let $f(x) = 2x^3 + x^2 - 7x - 6$.

The remainder theorem states that when a polynomial $f(x)$ is divided by $x-a$, the remainder is $f(a)$.

In this case, $a=2$, so we need to find $f(2)$.

$f(2) = 2(2)^3 + (2)^2 - 7(2) - 6 = 2(8) + 4 - 14 - 6 = 16 + 4 - 14 - 6 = 20 - 20 = 0$

d) Solve the following pair of simultaneous equations:

$7x - 3y = 23$

$2x - 4y = -8$

Multiply the first equation by 4 and the second by -3:

$4(7x - 3y) = 4(23) \Rightarrow 28x - 12y = 92$

$-3(2x - 4y) = -3(-8) \Rightarrow -6x + 12y = 24$

Add the two equations:

$28x - 12y + (-6x + 12y) = 92 + 24$

$22x = 116$

$x = \frac{116}{22} = \frac{58}{11}$

Substitute $x = \frac{58}{11}$ into the first equation:

$7(\frac{58}{11}) - 3y = 23$

$\frac{406}{11} - 3y = 23$

$3y = \frac{406}{11} - 23 = \frac{406 - 253}{11} = \frac{153}{11}$

$y = \frac{153}{33} = \frac{51}{11}$

e) (i) Solve the following equation: $log_3 x = -2$

$log_3 x = -2$

$x = 3^{-2} = \frac{1}{3^2} = \frac{1}{9}$

(ii) Solve the following equation: $4x - 7(2-x) = 3x + 2$

$4x - 7(2-x) = 3x + 2$

$4x - 14 + 7x = 3x + 2$

$11x - 14 = 3x + 2$

$8x = 16$

$x = 2$

Question Two

a) Determine the gradient and y-intercept of the line $x + 2y = 14$.

Rewrite the equation in the form $y = mx + c$, where $m$ is the gradient and $c$ is the y-intercept.

$2y = -x + 14$

$y = -\frac{1}{2}x + 7$

Gradient $m = -\frac{1}{2}$

y-intercept $c = 7$

b) Find the equation of the line that passes through $(-3,1)$ and $(2, -14)$.

The gradient $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-14 - 1}{2 - (-3)} = \frac{-15}{5} = -3$

Using the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$

$y - 1 = -3(x - (-3))$

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

$y - 1 = -3x - 9$

$y = -3x - 8$

c) Determine the partial fraction decomposition of the following expression: $\frac{17x - 53}{x^2 - 2x - 15}$

Factor the denominator: $x^2 - 2x - 15 = (x-5)(x+3)$

Now we express the fraction as a sum of partial fractions:

$\frac{17x - 53}{(x-5)(x+3)} = \frac{A}{x-5} + \frac{B}{x+3}$

$17x - 53 = A(x+3) + B(x-5)$

To find A, let $x = 5$:

$17(5) - 53 = A(5+3) + B(5-5)$

$85 - 53 = 8A$

$32 = 8A$

$A = 4$

To find B, let $x = -3$:

$17(-3) - 53 = A(-3+3) + B(-3-5)$

$-51 - 53 = -8B$

$-104 = -8B$

$B = 13$

Therefore, $\frac{17x - 53}{x^2 - 2x - 15} = \frac{4}{x-5} + \frac{13}{x+3}$

3. Final Answer

a) (i) $5.630 \times 10^3$

(ii) $16.346 \times 10^9$

b) $\frac{10y^3}{x^5}$

c) 0

d) $x = \frac{58}{11}$, $y = \frac{51}{11}$

e) (i) $x = \frac{1}{9}$

(ii) $x = 2$

Question Two

a) Gradient: $-\frac{1}{2}$, y-intercept: $7$

b) $y = -3x - 8$

c) $\frac{4}{x-5} + \frac{13}{x+3}$