Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.
(i) y = –3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x – 10
(iv) 3y = 6x – 11
Are any of the lines parallel?

In a linear equation of the form y = ax + b, 'a' is the slope and 'b' is the y-intercept. The line cuts the y-axis at the point (0, b).

(i) y = -3x + 4

This is already in the form y = ax + b with a = -3 and b = 4.

To draw the graph, we need two points:

When x = 0, y = -3(0) + 4 = 4. Point: (0, 4).

When x = 1, y = -3(1) + 4 = 1. Point: (1, 1).

Slope = -3, y-intercept = 4. The line cuts the y-axis at (0, 4).

(ii) 2y = 4x + 7

Dividing both sides by 2:

y = $\dfrac{4x + 7}{2} = 2x + \dfrac{7}{2}$

So, a = 2 and b = $\dfrac{7}{2}$ = 3.5

To draw the graph:

When x = 0, y = 3.5. Point: (0, 3.5).

When x = 1, y = 2(1) + 3.5 = 5.5. Point: (1, 5.5).

Slope = 2, y-intercept = $\dfrac{7}{2}$ (or 3.5). The line cuts the y-axis at $\left(0, \dfrac{7}{2}\right)$.

(iii) 5y = 6x - 10

Dividing both sides by 5:

y = $\dfrac{6x - 10}{5} = \dfrac{6}{5}x - 2$

So, a = $\dfrac{6}{5}$ and b = -2.

To draw the graph:

When x = 0, y = -2. Point: (0, -2).

When x = 5, y = $\dfrac{6}{5}(5)$ - 2 = 6 - 2 = 4. Point: (5, 4).

Slope = $\dfrac{6}{5}$, y-intercept = -2. The line cuts the y-axis at (0, -2).

(iv) 3y = 6x - 11

Dividing both sides by 3:

y = $\dfrac{6x - 11}{3} = 2x - \dfrac{11}{3}$

So, a = 2 and b = $-\dfrac{11}{3}$.

To draw the graph:

When x = 0, y = $-\dfrac{11}{3} = -3.6$. Point: (0, -3.6)

When x = 3, y = 2(3) - $\dfrac{11}{3} = 6 - \dfrac{11}{3} = \dfrac{18 - 11}{3} = \dfrac{7}{3} = 2.3$. Point: (3, 2.3)

Slope = 2, y-intercept = $-\dfrac{11}{3}$ or (-3.6). The line cuts the y-axis at $\left(0, -\dfrac{11}{3}\right)$.

Are any of the lines parallel?

Comparing the slopes:

Slope of (i) = -3

Slope of (ii) = 2

Slope of (iii) = $\dfrac{6}{5}$

Slope of (iv) = 2

Lines (ii) and (iv) have the same slope (2), but different y-intercepts ($\dfrac{7}{2}$ and $-\dfrac{11}{3}$).

Hence, the lines 2y = 4x + 7 and 3y = 6x - 11 are parallel to each other.