3.1 Straight lines, gradients and straight-line models — Practise with help

1. Subtract $3x$ from both sides: $2y = 12 - 3x$. Divide by $2$: $y = 6 - \dfrac{3}{2}x$.
2. $\dfrac{6}{5}$.
3. Add $1$ to both sides: $2x = 10$, so $x = 5$.

If question 1 was hard, revisit rearranging a formula before you continue.

The gradient is the change in $y$ divided by the change in $x$. Take the second point's coordinates minus the first point's, on the top and on the bottom.

The change in $y$ is $14 - 2 = 12$. The change in $x$ is $5 - 1 = 4$. Divide.

The line rises $3$ units for each unit across, so the gradient is $3$.

Substitute carefully: the change in $y$ is $-4 - 6$, and the change in $x$ is $3 - (-2)$.

Work out the top and the bottom separately before you divide: $-4 - 6 = -10$ and $3 - (-2) = 5$.

The gradient is negative, so the line falls as it moves right.

Use $y - y_1 = m(x - x_1)$ with $x_1 = 1$, $y_1 = 5$ and $m = 3$.

Write $y - 5 = 3(x - 1)$, then expand the bracket on the right.

Add $5$ to both sides:

Check with the point $(1, 5)$: $3 \times 1 + 2 = 5$. The point fits.

This is two jobs. Work out the gradient first, then use one of the points.

The gradient is $\dfrac{9 - 1}{6 - 2} = 2$. Now put $m = 2$ and the point $(2, 1)$ into $y - y_1 = m(x - x_1)$.

Add $1$:

Check the other point $(6, 9)$: $2 \times 6 - 3 = 9$. Both points fit.

Which number is paid once, whatever the time? That is the intercept. Which number is paid for each hour? That is the gradient.

Let $h$ be the number of hours and $C$ the cost in pounds. The model starts from $C = 40 + 25h$. Substitute $h = 3$.

The intercept $40$ is the fixed charge, paid before any work starts. The gradient $25$ is the charge for each extra hour, in pounds per hour.

For a visit of $3$ hours,

The charge is £115.

The fraction is the only obstacle. Multiply every term by $3$ first.

Multiplying by $3$ gives $3y = 2x - 12$. Now move every term to one side so that the equation reads $= 0$.

Multiply both sides by $3$:

Subtract $3y$ from both sides:

which is the same as

So $a = 2$, $b = -3$ and $c = -12$.

The $x$-intercept is the point with $y = 0$. The $y$-intercept is the point with $x = 0$.

Put $y = 0$ to get $4x = 24$. Then put $x = 0$ to get $-3y = 24$.

Put $y = 0$:

Put $x = 0$:

The line crosses the $x$-axis at $(6, 0)$ and the $y$-axis at $(0, -8)$.