Find the gradient, y-intercepts and x-intercepts of the following: a) y = -(2x-2) b) y - 1/4 − 6 = 0 c) −3x + y = 2 d) 2+y=x e) −(6x+y) = 2 f) 6-2y/3 = 2. Find the equations of lines parallel to x−2y=6 that pass through each of the following points. a) (-4,1) b) (1/2, 0)

The gradient $m$ is the coefficient of $x$. The y-intercept $c$ is the constant term. To find the x-intercept, set $y = 0$ and solve for $x$.

$$0 = -2x + 2$$ $$2x = 2$$ $$x = 1$$

• Gradient: $\boxed{-2}$
• Y-intercept: $\boxed{2}$
• X-intercept: $\boxed{1}$

b) $y - \frac{1}{4}x - 6 = 0$

To find the x-intercept, set $y = 0$ and solve for $x$.

$$0 = \frac{1}{4}x + 6$$ $$-\frac{1}{4}x = 6$$ $$x = 6 \times (-4)$$ $$x = -24$$

• Gradient: $\boxed{\frac{1}{4}}$
• Y-intercept: $\boxed{6}$
• X-intercept: $\boxed{-24}$

c) $-3x + y = 2$

To find the x-intercept, set $y = 0$ and solve for $x$.

$$0 = 3x + 2$$ $$-3x = 2$$ $$x = -\frac{2}{3}$$

• Gradient: $\boxed{3}$
• Y-intercept: $\boxed{2}$
• X-intercept: $\boxed{-\frac{2}{3}}$

d) $2 + y = x$

To find the x-intercept, set $y = 0$ and solve for $x$.

$$0 = x - 2$$ $$x = 2$$

• Gradient: $\boxed{1}$
• Y-intercept: $\boxed{-2}$
• X-intercept: $\boxed{2}$

e) $-6x + y = 2$

To find the x-intercept, set $y = 0$ and solve for $x$.

$$0 = 6x + 2$$ $$-6x = 2$$ $$x = -\frac{2}{6}$$ $$x = -\frac{1}{3}$$

• Gradient: $\boxed{6}$
• Y-intercept: $\boxed{2}$
• X-intercept: $\boxed{-\frac{1}{3}}$

f) $\frac{6 - 2y}{3} = 2$

Multiply both sides by 3:

$$6 - 2y = 6$$

Subtract 6 from both sides:

$$-2y = 0$$

Divide by -2:

$$y = 0$$

This is a horizontal line along the x-axis.

• Gradient: $\boxed{0}$
• Y-intercept: $\boxed{0}$
• X-intercept: $\boxed{\text{All real numbers}}$

g) $-(x - 2y) - (-x - 2y) = -2$

$$-x + 2y + x + 2y = -2$$ $$4y = -2$$ $$y = -\frac{2}{4}$$ $$y = -\frac{1}{2}$$

This is a horizontal line.

• Gradient: $\boxed{0}$
• Y-intercept: $\boxed{-\frac{1}{2}}$
• X-intercept: $\boxed{\text{None}}$

h) $\frac{x + 1}{4 + 1} = \frac{3}{2 + 2}$

$$\frac{x + 1}{5} = \frac{3}{4}$$

Multiply both sides by 5:

$$x + 1 = \frac{15}{4}$$

Subtract 1 from both sides:

$$x = \frac{15}{4} - 1$$ $$x = \frac{15}{4} - \frac{4}{4}$$ $$x = \frac{11}{4}$$

This is a vertical line.

• Gradient: $\boxed{\text{Undefined}}$
• Y-intercept: $\boxed{\text{None}}$
• X-intercept: $\boxed{\frac{11}{4}}$

Part 2: Find the equations of lines parallel to $x - 2y = 6$ that pass through each of the following points.

Rearrange the equation into slope-intercept form $y = mx + c$:

$$-2y = -x + 6$$ $$y = \frac{1}{2}x - 3$$

The gradient of this line is $m = \frac{1}{2}$. Since the new lines are parallel, their gradient will also be $m = \frac{1}{2}$.

a) Point: $(-4, 1)$

$$y - 1 = \frac{1}{2}(x - (-4))$$ $$y - 1 = \frac{1}{2}(x + 4)$$

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

The equation of the line is $\boxed{y = \frac{1}{2}x + 3}$.

b) Point: $(\frac{1}{2}, 0)$

$$y = \frac{1}{2}x - \frac{1}{4}$$

The equation of the line is $\boxed{y = \frac{1}{2}x - \frac{1}{4}}$.