A) Given two lines y = 2x + 1 and 3x − y = 5, find the intersection point of the two lines.

$$84 = 2 \times 42 = 2 \times 2 \times 21 = 2 \times 2 \times 3 \times 7 = 2^2 \times 3 \times 7$$

The prime factors are:

• $126 = 2 \times 3^2 \times 7$
• $84 = 2^2 \times 3 \times 7$

10.b) Hence, find the Highest Common Factor (HCF) of 126 and 84.

Common prime factors are 2, 3, and 7.

• For 2: The lowest power is $2^1$ (from $2 \times 3^2 \times 7$).
• For 3: The lowest power is $3^1$ (from $2^2 \times 3 \times 7$).
• For 7: The lowest power is $7^1$ (from both).

$$\text{HCF} = 2^1 \times 3^1 \times 7^1 = 2 \times 3 \times 7 = 42$$

The HCF of 126 and 84 is $\boxed{42}$.

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1. (i) a) Given two lines $y = 2x + 1$ and $3x - y = 5$, find the intersection point of the two lines.

Given:

1. $y = 2x + 1$
2. $3x - y = 5$

Substitute (1) into (2):

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

$$3x - 2x - 1 = 5$$ $$x - 1 = 5$$ $$x = 5 + 1$$ $$x = 6$$

$$y = 2(6) + 1$$ $$y = 12 + 1$$ $$y = 13$$

The intersection point is $\boxed{(6, 13)}$.

1. (i) b) Find the distance of the point A $(-3, 4)$ from B $(1, 7)$.

Let $(x_1, y_1) = (-3, 4)$ and $(x_2, y_2) = (1, 7)$.

$$d = \sqrt{(1 - (-3))^2 + (7 - 4)^2}$$

$$d = \sqrt{(1 + 3)^2 + (3)^2}$$ $$d = \sqrt{(4)^2 + (3)^2}$$ $$d = \sqrt{16 + 9}$$ $$d = \sqrt{25}$$

$$d = 5$$

The distance between A and B is $\boxed{5 \text{ units}}$.

1. (i) c) Find the midpoint between L $(5, 6)$ and N $(1, 2)$.

Let $(x_1, y_1) = (5, 6)$ and $(x_2, y_2) = (1, 2)$.

$$M = \left(\frac{5 + 1}{2}, \frac{6 + 2}{2}\right)$$

$$M = \left(\frac{6}{2}, \frac{8}{2}\right)$$ $$M = (3, 4)$$

The midpoint between L and N is $\boxed{(3, 4)}$.

1. (ii) a) Complete the table for $y = x^2 - 4x + 3$.

• For $x = -1$: $y = (-1)^2 - 4(-1) + 3 = 1 + 4 + 3 = 8$
• For $x = 0$: $y = (0)^2 - 4(0) + 3 = 0 - 0 + 3 = 3$
• For $x = 1$: $y = (1)^2 - 4(1) + 3 = 1 - 4 + 3 = 0$
• For $x = 2$: $y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$
• For $x = 3$: $y = (3)^2 - 4(3) + 3 = 9 - 12 + 3 = 0$
• For $x = 4$: $y = (4)^2 - 4(4) + 3 = 16 - 16 + 3 = 3$

The completed table is:

$$\begin{array}{|c|c|c|c|c|c|c|} \hline x & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & 8 & 3 & 0 & -1 & 0 & 3 \\ \hline \end{array}$$

1. (ii) b) Draw the graph of the quadratic and the linear function on the same axes (1cm on both axes).

To draw the graph:

• For the quadratic function $y = x^2 - 4x + 3$: Plot the points from the completed table: $(-1, 8)$, $(0, 3)$, $(1, 0)$, $(2, -1)$, $(3, 0)$, $(4, 3)$. Draw a smooth curve through these points.
• For the linear function $y = x - 1$: Choose a few $x$ values and calculate $y$. For example:
  • If $x = 0$, $y = 0 - 1 = -1$. Plot $(0, -1)$.
  • If $x = 1$, $y = 1 - 1 = 0$. Plot $(1, 0)$.
  • If $x = 4$, $y = 4 - 1 = 3$. Plot $(4, 3)$. Draw a straight line through these points.

1. (ii) c) Mark roots and the y-intercepts of the curve and the line.

• Roots of the curve ($y = x^2 - 4x + 3$): These are the $x$-values where $y=0$. From the table, the roots are $x=1$ and $x=3$. Mark the points $(1, 0)$ and $(3, 0)$ on your graph.
• Y-intercept of the curve ($y = x^2 - 4x + 3$): This is the $y$-value when $x=0$. From the table, the y-intercept is $y=3$. Mark the point $(0, 3)$ on your graph.
• Y-intercept of the line ($y = x - 1$): This is the $y$-value when $x=0$. For $y=x-1$, when $x=0$, $y=-1$. Mark the point $(0, -1)$ on your graph.

1. (ii) d) Using the graph, state the values of x for which $x^2 - 4x + 3 = x - 1$.

From the graph, observe where the curve and the line intersect.
Alternatively, solve the equation algebraically:

$$x^2 - 4x + 3 = x - 1$$

$$x^2 - 4x - x + 3 + 1 = 0$$ $$x^2 - 5x + 4 = 0$$

$$x - 1 = 0 \implies x = 1$$ $$x - 4 = 0 \implies x = 4$$

The values of $x$ for which $x^2 - 4x + 3 = x - 1$ are $\boxed{x=1 \text{ and } x=4}$.