Linear Law Worked Examples (Medium)

Q: How do I decide between plotting \lg x and plotting x?

Look at where the variable sits. In a power law y=axny = ax^{n}y=axn the xxx is the base, so you plot lg⁡y\lg ylgy against lg⁡x\lg xlgx. In an exponential law y=abxy = ab^{x}y=abx the xxx is the exponent, so you plot lg⁡y\lg ylgy against plain xxx.

Q: Why divide by x for y = ax + bx^{2}?

Dividing every term by xxx gives yx=bx+a\frac{y}{x} = bx + axy=bx+a, which is a straight line in yx\frac{y}{x}xy and xxx. The gradient becomes bbb and the intercept becomes aaa, so both constants can be read from one line.

Q: The intercept came out as \lg a = 3. How do I get a?

Undo the logarithm by raising ten to that power: a=103=1000a = 10^{3} = 1000a=103=1000. Leaving your answer as lg⁡a\lg alga is a common way to lose the final mark, so always convert back to aaa or bbb.

What these examples cover

These medium Linear Law examples move past reading a ready-made straight line and into the reductions themselves. You will divide a relation through by $x$ to linearise $y = ax + bx^{2}$ , then use logarithms to straighten two of the most common exam relations: the power law $y = ax^{n}$ and the exponential law $y = ab^{x}$ .

Each one uses small, clean numbers so the algebra stays visible. The best way to use the set is to cover the solution, attempt the reduction and the arithmetic in full on paper, and only then check line by line against our working.

If your answer differs, find the exact step where the two solutions part company, deciding what to plot and where the logarithm lands is where most marks are won or lost.

Worked examples

Work through all three. Attempt each fully before you read the matching solution, and watch how the same discipline, rearrange into $Y = mX + c$ , name $Y$ and $X$ , then match gradient and intercept to the constants, carries every one.

Q1[4 marks]

The variables $x$ and $y$ are related by $y = ax + bx^{2}$ , where $a$ and $b$ are constants. A straight-line graph of $\frac{y}{x}$ against $x$ passes through the points $(2, 7)$ and $(5, 16)$ .

Find the values of $a$ and $b$ .

Show worked solution

First linearise the relation. Divide every term by $x$ so the equation matches $Y = mX + c$ :

\frac{y}{x} = a + bx = bx + a

So the vertical axis is $Y = \frac{y}{x}$ and the horizontal axis is $X = x$ . The gradient is $b$ and the intercept is $a$ .

Find the gradient from the two points:

b = \frac{16 - 7}{5 - 2} = \frac{9}{3} = 3

Substitute the point $(2, 7)$ to find the intercept $a$ :

7 = 3(2) + a \;\Rightarrow\; 7 = 6 + a \;\Rightarrow\; a = 1

Answer

$a = 1$ and $b = 3$ , so $y = x + 3x^{2}$ . Check with the second point: at $x = 5$ , $y = 5 + 3(25) = 80$ , and $\frac{y}{x} = \frac{80}{5} = 16$ , which matches $(5, 16)$ .

Q2[4 marks]

The variables $x$ and $y$ are related by $y = ax^{n}$ , where $a$ and $n$ are constants. When $\lg y$ is plotted against $\lg x$ , a straight line passes through $(\lg x, \lg y) = (1, 5)$ and $(3, 9)$ .

Find the values of $a$ and $n$ .

Show worked solution

Take $\lg$ of both sides and use the power and product laws of logarithms to straighten the relation:

\lg y = \lg\left(a x^{n}\right) = \lg a + n\lg x = (n)\lg x + \lg a

Comparing with $Y = mX + c$ , the vertical axis is $Y = \lg y$ and the horizontal axis is $X = \lg x$ . The gradient is $n$ and the intercept is $\lg a$ .

Find the gradient from the two points:

n = \frac{9 - 5}{3 - 1} = \frac{4}{2} = 2

Substitute the point $(1, 5)$ to find the intercept $\lg a$ , then undo the logarithm:

5 = 2(1) + \lg a \;\Rightarrow\; \lg a = 3 \;\Rightarrow\; a = 10^{3} = 1000

Answer

$n = 2$ and $a = 1000$ , so $y = 1000\,x^{2}$ . Check the second point: at $\lg x = 3$ , $\lg y = 2(3) + 3 = 9$ , which matches $(3, 9)$ .

Q3[4 marks]

The variables $x$ and $y$ are related by $y = ab^{x}$ , where $a$ and $b$ are constants. When $\lg y$ is plotted against $x$ , a straight line passes through $(0, 2)$ and $(3, 5)$ .

Find the values of $a$ and $b$ .

Show worked solution

Take $\lg$ of both sides. Here the variable $x$ sits in the exponent, so the power law brings it down as a coefficient:

\lg y = \lg\left(a b^{x}\right) = \lg a + x\lg b = (\lg b)x + \lg a

Comparing with $Y = mX + c$ , the vertical axis is $Y = \lg y$ and the horizontal axis is $X = x$ . The gradient is $\lg b$ and the intercept is $\lg a$ .

The point $(0, 2)$ sits on the vertical axis, so it gives the intercept directly:

\lg a = 2 \;\Rightarrow\; a = 10^{2} = 100

Now find the gradient from the two points and undo the logarithm to get $b$ :

\lg b = \frac{5 - 2}{3 - 0} = \frac{3}{3} = 1 \;\Rightarrow\; b = 10^{1} = 10

Answer

$a = 100$ and $b = 10$ , so $y = 100(10)^{x}$ . Check the second point: at $x = 3$ , $\lg y = 1(3) + 2 = 5$ , which matches $(3, 5)$ .

Q4[3 marks]

The variables $x$ and $y$ are related by $y = px^{2} + q$ , where $p$ and $q$ are constants. A straight-line graph of $y$ against $x^{2}$ passes through the points $(1, 7)$ and $(3, 17)$ .

Find the values of $p$ and $q$ .

Show worked solution

This relation is already linear once the horizontal axis is $x^{2}$ : the vertical axis is $Y = y$ and the horizontal axis is $X = x^{2}$ , matching $Y = mX + c$ . The gradient is $p$ and the intercept is $q$ .

Find the gradient from the two points:

p = \frac{17 - 7}{3 - 1} = \frac{10}{2} = 5

Substitute the point $(1, 7)$ to find $q$ :

7 = 5(1) + q \;\Rightarrow\; q = 2

Answer

$p = 5$ and $q = 2$ , so $y = 5x^{2} + 2$ . Check the second point: at $x^{2} = 3$ , $y = 5(3) + 2 = 17$ , which matches $(3, 17)$ .

Q5[4 marks]

The variables $x$ and $y$ are related by $xy = ax + b$ , where $a$ and $b$ are constants. A straight-line graph of $y$ against $\frac{1}{x}$ passes through the points $(2, 9)$ and $(5, 21)$ .

Find the values of $a$ and $b$ .

Show worked solution

Divide every term by $x$ so the equation matches $Y = mX + c$ :

\frac{xy}{x} = \frac{ax}{x} + \frac{b}{x} \;\Rightarrow\; y = a + b\cdot\frac{1}{x}

So the vertical axis is $Y = y$ and the horizontal axis is $X = \frac{1}{x}$ . The gradient is $b$ and the intercept is $a$ .

Find the gradient from the two points:

b = \frac{21 - 9}{5 - 2} = \frac{12}{3} = 4

Substitute the point $(2, 9)$ to find the intercept $a$ :

9 = 4(2) + a \;\Rightarrow\; 9 = 8 + a \;\Rightarrow\; a = 1

Answer

$a = 1$ and $b = 4$ , so $xy = x + 4$ , i.e. $y = 1 + \frac{4}{x}$ . Check the second point: at $\frac{1}{x} = 5$ , $y = 1 + 4(5) = 21$ , which matches $(5, 21)$ .

Q6[3 marks]

The variables $x$ and $y$ are related by $\frac{1}{y} = ax + b$ , where $a$ and $b$ are constants. A straight-line graph of $\frac{1}{y}$ against $x$ passes through the points $(1, 7)$ and $(4, 19)$ .

Find the values of $a$ and $b$ .

Show worked solution

This relation is already linear once the vertical axis is $\frac{1}{y}$ : $Y = \frac{1}{y}$ and $X = x$ , matching $Y = mX + c$ . The gradient is $a$ and the intercept is $b$ .

Find the gradient from the two points:

a = \frac{19 - 7}{4 - 1} = \frac{12}{3} = 4

Substitute the point $(1, 7)$ to find $b$ :

7 = 4(1) + b \;\Rightarrow\; b = 3

Answer

$a = 4$ and $b = 3$ , so $\frac{1}{y} = 4x + 3$ , i.e. $y = \frac{1}{4x + 3}$ . Check the second point: at $x = 4$ , $\frac{1}{y} = 4(4) + 3 = 19$ , which matches $(4, 19)$ .

Q7[4 marks]

The variables $x$ and $y$ are related by $y = ab^{x^{2}}$ , where $a$ and $b$ are constants. When $\lg y$ is plotted against $x^{2}$ , a straight line passes through $(0, 3)$ and $(2, 7)$ .

Find the values of $a$ and $b$ .

Show worked solution

Take $\lg$ of both sides. Because $x$ is squared inside the power, $x^{2}$ itself becomes the horizontal variable:

\lg y = \lg\left(ab^{x^{2}}\right) = \lg a + x^{2}\lg b = (\lg b)x^{2} + \lg a

Comparing with $Y = mX + c$ , the vertical axis is $Y = \lg y$ and the horizontal axis is $X = x^{2}$ . The gradient is $\lg b$ and the intercept is $\lg a$ .

The point $(0, 3)$ sits on the vertical axis, so it gives the intercept directly:

\lg a = 3 \;\Rightarrow\; a = 10^{3} = 1000

Now find the gradient from the two points and undo the logarithm to get $b$ :

\lg b = \frac{7 - 3}{2 - 0} = \frac{4}{2} = 2 \;\Rightarrow\; b = 10^{2} = 100

Answer

$a = 1000$ and $b = 100$ , so $y = 1000(100)^{x^{2}}$ . Check the second point: at $x^{2} = 2$ , $\lg y = 2(2) + 3 = 7$ , which matches $(2, 7)$ .

Q8[3 marks]

The variables $x$ and $y$ are related by $y^{2} = ax + b$ , where $a$ and $b$ are constants. A straight-line graph of $y^{2}$ against $x$ passes through the points $(2, 13)$ and $(5, 25)$ .

Find the values of $a$ and $b$ .

Show worked solution

This relation is already linear once the vertical axis is $y^{2}$ : $Y = y^{2}$ and $X = x$ , matching $Y = mX + c$ . The gradient is $a$ and the intercept is $b$ .

Find the gradient from the two points:

a = \frac{25 - 13}{5 - 2} = \frac{12}{3} = 4

Substitute the point $(2, 13)$ to find $b$ :

13 = 4(2) + b \;\Rightarrow\; 13 = 8 + b \;\Rightarrow\; b = 5

Answer

$a = 4$ and $b = 5$ , so $y^{2} = 4x + 5$ . Check the second point: at $x = 5$ , $y^{2} = 4(5) + 5 = 25$ , which matches $(5, 25)$ .

The three questions look different, one divides through, two take logarithms, yet the plan is identical: rearrange until the equation reads as $Y = mX + c$ , state exactly what $Y$ and $X$ are, and only then match the gradient and intercept to the constants. Naming the axes before you touch the numbers is what keeps a power law and an exponential law from being confused.

Key method points

These three examples rehearse the reductions that carry the harder half of Linear Law. Keep the following points in mind as you practise more.

To linearise $y = ax + bx^{2}$ , divide every term by $x$ : $\frac{y}{x} = bx + a$ , so plot $\frac{y}{x}$ against $x$ .
For a power law $y = ax^{n}$ , take $\lg$ of both sides: $\lg y = n\lg x + \lg a$ , so plot $\lg y$ against $\lg x$ .
For an exponential law $y = ab^{x}$ , take $\lg$ of both sides: $\lg y = (\lg b)x + \lg a$ , so plot $\lg y$ against $x$ .
The difference is where $x$ sits: $x^{n}$ makes the gradient $n$ and needs $\lg x$ on the horizontal axis; $b^{x}$ makes the gradient $\lg b$ and keeps $x$ itself.
When the intercept equals $\lg a$ or $\lg b$ , undo it with a power of ten: $\lg a = k \Rightarrow a = 10^{k}$ .
Because marking is analytic, a correct reduction earns method marks even before you reach the final values of the constants.

How a teacher helps

The slip we see most often here is a rushed logarithm, leaving the intercept as $\lg a$ instead of $a$ , or confusing a power law with an exponential one. In a one-to-one lesson our teacher slows the reduction down with you, so you can see exactly why $\lg x$ belongs on one graph and plain $x$ on the other.

Because our teachers are experienced, you work with someone who explains the why behind each step, not just the what. Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation reads the same to you either way.

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Frequently asked questions

How do I decide between plotting

\lg x

and plotting

x

Look at where the variable sits. In a power law $y = ax^{n}$ the $x$ is the base, so you plot $\lg y$ against $\lg x$ .

In an exponential law $y = ab^{x}$ the $x$ is the exponent, so you plot $\lg y$ against plain $x$ .

Why divide by

x

for

y = ax + bx^{2}

Dividing every term by $x$ gives $\frac{y}{x} = bx + a$ , which is a straight line in $\frac{y}{x}$ and $x$ . The gradient becomes $b$ and the intercept becomes $a$ , so both constants can be read from one line.

The intercept came out as

\lg a = 3

. How do I get

a

Undo the logarithm by raising ten to that power: $a = 10^{3} = 1000$ . Leaving your answer as $\lg a$ is a common way to lose the final mark, so always convert back to $a$ or $b$ .

How can I check my constants at the end?

Substitute the second given point into your straight-line equation. If it produces the correct value on the graph, and the original relation reproduces the data, both constants are secure.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Linear Law, Worked Examples (medium)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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