Linear Law Worked Examples (KBAT)

Q: How do I linearise a relation like y = \frac{x}{ax+b}?

Take the reciprocal of both sides to get 1y=ax+bx\frac{1}{y} = \frac{ax+b}{x}y1=xax+b, then split the right-hand side into a+b(1x)a + b\left(\frac{1}{x}\right)a+b(x1). Now it is a straight line in 1y\frac{1}{y}y1 and 1x\frac{1}{x}x1, with gradient bbb and intercept aaa.

Q: What does a fractional gradient mean in a power law?

It means the index nnn is a fraction, which corresponds to a root. A gradient of 12\frac{1}{2}21 gives y=axy = a\sqrt{x}y=ax. To find xxx from a value of yyy, isolate the root and square both sides.

What these examples cover

These hard Linear Law examples ask for two things at once: a reduction that is not obvious, and a second step that uses the straight line to answer a follow-up. You will take reciprocals of both sides to linearise $y = \frac{x}{ax+b}$ , handle a power law whose gradient is a fraction and then reverse it to find an $x$ -value, and convert an exponential law into an index equation you can solve exactly.

The numbers stay clean so every line is checkable. Cover the solution, attempt the full question on paper, reduction and follow-up, and only then compare line by line.

If your answer differs, the split is almost always in the reduction, so look there first.

Worked examples

Work through all three. Each one starts with a reduction into $Y = mX + c$ , finds the constants, and then does something with them.

Attempt each fully before reading the solution.

Q1[5 marks]

The variables $x$ and $y$ are related by $y = \frac{x}{ax + b}$ , where $a$ and $b$ are constants. A straight-line graph of $\frac{1}{y}$ against $\frac{1}{x}$ passes through $\left(\frac{1}{x}, \frac{1}{y}\right) = (1, 5)$ and $(3, 11)$ .

Find (a) the values of $a$ and $b$ , and (b) the value of $y$ when $x = 6$ .

Show worked solution

(a) The variable is trapped in a fraction, so take the reciprocal of both sides. Then split the single fraction into two terms:

\frac{1}{y} = \frac{ax + b}{x} = a + b\left(\frac{1}{x}\right)

Compared with $Y = mX + c$ , the vertical axis is $Y = \frac{1}{y}$ 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{11 - 5}{3 - 1} = \frac{6}{2} = 3

Substitute the point $(1, 5)$ to find the intercept $a$ :

5 = 3(1) + a \;\Rightarrow\; a = 2

(b) So $y = \frac{x}{2x + 3}$ . Substitute $x = 6$ :

y = \frac{6}{2(6) + 3} = \frac{6}{15} = \frac{2}{5}

Answer

$a = 2$ , $b = 3$ , and $y = \frac{2}{5}$ when $x = 6$ . Check via the line: at $X = \frac{1}{6}$ , $Y = a + bX = 2 + 3\left(\frac{1}{6}\right) = 2.5$ , so $y = \frac{1}{2.5} = \frac{2}{5}$ , which agrees.

Q2[5 marks]

The variables $x$ and $y$ are related by $y = ax^{n}$ , where $a$ and $n$ are constants. A straight-line graph of $\lg y$ against $\lg x$ has gradient $\frac{1}{2}$ and passes through the point $(\lg x, \lg y) = (2, 3)$ .

Find (a) the values of $a$ and $n$ , and (b) the value of $x$ when $y = 3000$ .

Show worked solution

(a) Take $\lg$ of both sides so the relation is straight:

\lg y = n\lg x + \lg a

The gradient is $n$ , so $n = \frac{1}{2}$ directly. Use the given point $(2, 3)$ to find the intercept $\lg a$ , then undo the logarithm:

3 = \tfrac{1}{2}(2) + \lg a \;\Rightarrow\; 3 = 1 + \lg a \;\Rightarrow\; \lg a = 2 \;\Rightarrow\; a = 100

(b) So $y = 100x^{1/2} = 100\sqrt{x}$ . Set $y = 3000$ and solve for $x$ , isolate the root, then square both sides:

100\sqrt{x} = 3000 \;\Rightarrow\; \sqrt{x} = 30 \;\Rightarrow\; x = 30^{2} = 900

Answer

$n = \frac{1}{2}$ , $a = 100$ , and $x = 900$ when $y = 3000$ . Check: $100\sqrt{900} = 100(30) = 3000$ , as required.

A fractional gradient simply means a root, here the square root, because $n = \frac{1}{2}$ .

Q3[5 marks]

The variables $x$ and $y$ are related by $y = ab^{x}$ , where $a$ and $b$ are constants. A straight-line graph of $\lg y$ against $x$ passes through $(0, 1)$ and $(2, 7)$ .

Find (a) the values of $a$ and $b$ , and (b) the value of $x$ for which $y = 10\,000\,000$ .

Show worked solution

(a) Take $\lg$ of both sides; the exponent $x$ comes down as a coefficient:

\lg y = (\lg b)x + \lg a

The point $(0, 1)$ lies on the vertical axis, so it gives the intercept: $\lg a = 1 \Rightarrow a = 10$ . Find the gradient $\lg b$ from the two points, then undo the logarithm:

\lg b = \frac{7 - 1}{2 - 0} = \frac{6}{2} = 3 \;\Rightarrow\; b = 10^{3} = 1000

(b) So $y = 10(1000)^{x} = 10 \times 10^{3x} = 10^{\,3x + 1}$ . Write the target as a power of ten and equate the indices:

10^{\,3x + 1} = 10\,000\,000 = 10^{7} \;\Rightarrow\; 3x + 1 = 7 \;\Rightarrow\; x = 2

Answer

$a = 10$ , $b = 1000$ , and $x = 2$ when $y = 10^{7}$ . Check on the line: at $x = 2$ , $\lg y = 3(2) + 1 = 7$ , so $y = 10^{7}$ , which agrees with the point $(2, 7)$ .

Q4[5 marks]

The variables $x$ and $y$ are related by $y = px^{2} + qx$ , where $p$ and $q$ are constants. A straight-line graph of $\frac{y}{x}$ against $x$ passes through the points $\left(x, \frac{y}{x}\right) = (1, -4)$ and $(4, 2)$ .

Find (a) the values of $p$ and $q$ , and (b) the non-zero value of $x$ for which $y = 0$ .

Show worked solution

(a) The right-hand side has two terms in $x$ , so divide both sides by $x$ ( $x \neq 0$ ) rather than taking logarithms:

\frac{y}{x} = px + q

Compared with $Y = mX + c$ , the vertical axis is $Y = \frac{y}{x}$ and the horizontal axis is $X = x$ , so the gradient is $p$ and the intercept is $q$ . Find the gradient from the two points:

p = \frac{2 - (-4)}{4 - 1} = \frac{6}{3} = 2

Substitute the point $(1, -4)$ to find the intercept $q$ :

-4 = 2(1) + q \;\Rightarrow\; q = -6

(b) So $y = 2x^{2} - 6x$ . Set $y = 0$ and factorise, do not simply divide by $x$ this time, since $x = 0$ is itself a genuine root:

2x^{2} - 6x = 0 \;\Rightarrow\; 2x(x - 3) = 0

This gives $x = 0$ or $x = 3$ ; the question asks for the non-zero value.

Answer

$p = 2$ , $q = -6$ , and the non-zero root is $x = 3$ . Check: at $x = 3$ , $y = 2(3)^{2} - 6(3) = 18 - 18 = 0$ , as required.

The $x = 0$ root is lost the moment you divide by $x$ to linearise, so it must be recovered by solving the original equation directly.

Q5[5 marks]

The variables $x$ and $y$ are related by $y = ax + \frac{b}{x}$ , where $a$ and $b$ are constants. A straight-line graph of $xy$ against $x^{2}$ passes through the points $(x^{2}, xy) = (1, -1)$ and $(4, 8)$ .

Find (a) the values of $a$ and $b$ , and (b) the positive value of $x$ for which $xy = 23$ .

Show worked solution

(a) The two terms are $ax$ and $\frac{b}{x}$ , so multiply both sides by $x$ to clear the fraction:

xy = ax^{2} + b

Compared with $Y = mX + c$ , the vertical axis is $Y = xy$ and the horizontal axis is $X = x^{2}$ , so the gradient is $a$ and the intercept is $b$ . Find the gradient from the two points:

a = \frac{8 - (-1)}{4 - 1} = \frac{9}{3} = 3

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

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

(b) So $xy = 3x^{2} - 4$ . Set $xy = 23$ and solve for $x^{2}$ , then take the positive square root:

3x^{2} - 4 = 23 \;\Rightarrow\; 3x^{2} = 27 \;\Rightarrow\; x^{2} = 9 \;\Rightarrow\; x = 3

Answer

$a = 3$ , $b = -4$ , and the positive value is $x = 3$ . Check: at $x = 3$ , $xy = 3(3)^{2} - 4 = 27 - 4 = 23$ , as required.

Because the reduced equation is quadratic in $x$ , remember $x = -3$ solves it too, only the positive root is asked for here.

Q6[5 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 $(1, 4)$ and $(5, 16)$ .

Find (a) the values of $a$ and $b$ , and (b) the two possible values of $y$ when $x = 8$ .

Show worked solution

(a) The variable $y$ appears squared, so no reciprocal or logarithm is needed, plot $y^{2}$ directly against $x$ :

y^{2} = ax + b

Compared with $Y = mX + c$ , the vertical axis is $Y = y^{2}$ and the horizontal axis is $X = x$ , so the gradient is $a$ and the intercept is $b$ . Find the gradient from the two points:

a = \frac{16 - 4}{5 - 1} = \frac{12}{4} = 3

Substitute the point $(1, 4)$ to find the intercept $b$ :

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

(b) So $y^{2} = 3x + 1$ . Substitute $x = 8$ , then take the square root of both sides, a squared variable always leaves two roots:

y^{2} = 3(8) + 1 = 25 \;\Rightarrow\; y = \pm 5

Answer

$a = 3$ , $b = 1$ , and $y = 5$ or $y = -5$ when $x = 8$ . Check: $3(8) + 1 = 25 = 5^{2} = (-5)^{2}$ , so both values satisfy the equation.

Squaring $y$ to linearise the relation hides its sign, so a follow-up asking for $y$ itself, not $y^{2}$ , always has two valid answers unless the question restricts $y$ to one sign.

Q7[6 marks]

The variables $x$ and $y$ are related by $y = ab^{x^{2}}$ , where $a$ and $b$ are constants. A straight-line graph of $\lg y$ against $x^{2}$ passes through the points $(1, 3)$ and $(4, 9)$ .

Find (a) the values of $a$ and $b$ , and (b) the positive value of $x$ for which $y = 10^{19}$ .

Show worked solution

(a) The variable $x$ sits inside the exponent, and squared besides, so take $\lg$ of both sides and let the new horizontal variable be $x^{2}$ itself:

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

Compared with $Y = mX + c$ , the vertical axis is $Y = \lg y$ and the horizontal axis is $X = x^{2}$ , so the gradient is $\lg b$ and the intercept is $\lg a$ . Find the gradient from the two points, then undo the logarithm:

\lg b = \frac{9 - 3}{4 - 1} = \frac{6}{3} = 2 \;\Rightarrow\; b = 10^{2} = 100

Substitute the point $(1, 3)$ to find the intercept, then undo the logarithm:

3 = 2(1) + \lg a \;\Rightarrow\; \lg a = 1 \;\Rightarrow\; a = 10

(b) So $y = 10(100)^{x^{2}} = 10^{1 + 2x^{2}}$ . Write the target as a power of ten and equate the indices:

10^{1 + 2x^{2}} = 10^{19} \;\Rightarrow\; 1 + 2x^{2} = 19 \;\Rightarrow\; x^{2} = 9 \;\Rightarrow\; x = 3

Answer

$a = 10$ , $b = 100$ , and the positive value is $x = 3$ . Check: $1 + 2(3)^{2} = 1 + 18 = 19$ , so $y = 10^{19}$ , as required.

The exponent here is $x^{2}$ , not $x$ , so recovering $x$ at the end needs a square root as well as an index equation, one extra layer on top of the usual exponential law.

Q8[5 marks]

The variables $x$ and $y$ are related by $y = p\sqrt{x} + \frac{q}{\sqrt{x}}$ , where $p$ and $q$ are constants. A straight-line graph of $y\sqrt{x}$ against $x$ passes through the points $(1, 8)$ and $(4, 14)$ .

Find (a) the values of $p$ and $q$ , and (b) the value of $y$ when $x = 36$ .

Show worked solution

(a) The two terms have $\sqrt{x}$ and $\frac{1}{\sqrt{x}}$ , so multiply both sides by $\sqrt{x}$ to clear the fraction:

y\sqrt{x} = px + q

Compared with $Y = mX + c$ , the vertical axis is $Y = y\sqrt{x}$ and the horizontal axis is $X = x$ , so the gradient is $p$ and the intercept is $q$ . Find the gradient from the two points:

p = \frac{14 - 8}{4 - 1} = \frac{6}{3} = 2

Substitute the point $(1, 8)$ to find the intercept $q$ :

8 = 2(1) + q \;\Rightarrow\; q = 6

(b) So $y\sqrt{x} = 2x + 6$ . Substitute $x = 36$ , then divide by $\sqrt{36}$ to isolate $y$ :

y\sqrt{36} = 2(36) + 6 = 78 \;\Rightarrow\; 6y = 78 \;\Rightarrow\; y = 13

Answer

$p = 2$ , $q = 6$ , and $y = 13$ when $x = 36$ . Check: $y\sqrt{x} = 2(36) + 6 = 78$ , and $13 \times \sqrt{36} = 13 \times 6 = 78$ , which agrees.

Multiplying by $\sqrt{x}$ rather than squaring keeps the relation linear in $x$ itself, so the follow-up is a direct substitution once $p$ and $q$ are known.

Each solution rests on the same two-part discipline: reduce cleanly into $Y = mX + c$ and name the axes, then read the constants and carry them into the follow-up. The follow-up is where hard questions test whether you truly rebuilt the original relation, a reciprocal, a root, or an index equation, rather than stopping at the straight line.

Key method points

These three examples show how a hard Linear Law question layers a follow-up on top of a reduction. Keep the following points in mind as you practise more.

When the variable is trapped in a fraction, take the reciprocal of both sides, then split the fraction into separate terms.
For $y = \frac{x}{ax+b}$ , the reciprocal gives $\frac{1}{y} = b\left(\frac{1}{x}\right) + a$ , so plot $\frac{1}{y}$ against $\frac{1}{x}$ .
A fractional gradient in a power law is just a root: $n = \frac{1}{2}$ means $y = a\sqrt{x}$ ; reverse it by isolating the root and squaring.
For an exponential law, rewrite the target as a power of ten and equate indices to solve for $x$ exactly.
Always rebuild the original relation with your constants, then substitute the given point back to confirm it.
Because marking is analytic, a clean reduction and a correct comparison earn method marks even before the follow-up is finished.

How a teacher helps

Hard questions rarely fail on the reduction alone, they fail on the join between the reduction and the follow-up, where a reciprocal or an index step gets rushed. In a one-to-one lesson our teacher works that join with you, so you can see how the constants you read from the line feed straight into the equation you still have to solve.

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 linearise a relation like

y = \frac{x}{ax+b}

Take the reciprocal of both sides to get $\frac{1}{y} = \frac{ax+b}{x}$ , then split the right-hand side into $a + b\left(\frac{1}{x}\right)$ . Now it is a straight line in $\frac{1}{y}$ and $\frac{1}{x}$ , with gradient $b$ and intercept $a$ .

What does a fractional gradient mean in a power law?

It means the index $n$ is a fraction, which corresponds to a root. A gradient of $\frac{1}{2}$ gives $y = a\sqrt{x}$ .

To find $x$ from a value of $y$ , isolate the root and square both sides.

How do I solve

y = ab^{x}

for

x

once I know

a

and

b

Substitute the constants and write both sides as powers of the same base. When $y = 10^{3x+1}$ and the target is $10^{7}$ , equate the indices: $3x+1 = 7$ , so $x = 2$ .

Why does the follow-up matter so much in hard questions?

The follow-up checks that you rebuilt the original relation, not just the straight line. Reading the gradient and intercept is only half the task; carrying them into a reciprocal, a root, or an index equation is what earns the later marks.

Source:SRC-DSKP-EN

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

Linear Law, Worked Examples (KBAT)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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