1. The problem is to find the gradients (slopes) of the lines given by the equations j: 4x - y = 8, k: x + 2y - 2 = 0, and l: 4x = 2 + 3y by rearranging them into the form $y = mx + c$ where $m$ is the gradient. 2. Recall the slope-intercept form of a line is: $$y = mx + c$$ where $m$ is the gradient and $c$ is the y-intercept. 3. For line j: $4x - y = 8$ Rearrange to solve for $y$: $$-y = -4x + 8$$ Multiply both sides by $-1$: $$y = \cancel{-1} \times (-4x + 8) = 4x - 8$$ So the gradient $m_j = 4$. 4. For line k: $x + 2y - 2 = 0$ Rearrange to solve for $y$: $$2y = -x + 2$$ Divide both sides by 2: $$y = \frac{-x + 2}{2} = -\frac{1}{2}x + 1$$ So the gradient $m_k = -\frac{1}{2}$. 5. For line l: $4x = 2 + 3y$ Rearrange to solve for $y$: $$3y = 4x - 2$$ Divide both sides by 3: $$y = \frac{4x - 2}{3} = \frac{4}{3}x - \frac{2}{3}$$ So the gradient $m_l = \frac{4}{3}$. Final answers: - Gradient of j is $4$ - Gradient of k is $-\frac{1}{2}$ - Gradient of l is $\frac{4}{3}$