Algebra

high importance~3 Q in Tier 146 formulas⚡ 18 shortcuts6 subtopics

Overview Learn (6)Formulas (46)Examples (4)Practice (78)

Linear equations, graphs and polynomials

⏱ 4 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

Two linear equations are two lines; the ratios of their coefficients say whether the lines cross once, never, or lie on each other. A single line with the axes cuts a triangle whose area is half the product of intercepts. Polynomials reduce to substitutions through the remainder theorem.

Overview

For the pair $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ , compare three ratios:

Ratios	Outcome
$a_1/a_2 \ne b_1/b_2$	one solution, lines cross
$a_1/a_2 = b_1/b_2 \ne c_1/c_2$	no solution, parallel
all three ratios equal	same line, infinite solutions

Find- $k$ questions fix the known ratios first, then force the third. For $3x+4y=10$ and $6x+ky=20$ : infinite solutions needs $6/3 = 2$ , so $k/4 = 2$ gives $k = 8$ , and $20/10 = 2$ agrees.

Rule: The constant ratio decides between parallel and coincident. Never answer before checking $c_1/c_2$ .

Area with the axes

The line $ax + by = c$ cuts the $x$ -axis at $\dfrac{c}{a}$ (put $y = 0$ ) and the $y$ -axis at $\dfrac{c}{b}$ (put $x = 0$ ). The axes are perpendicular, so the triangle area is half the product:

\text{area} = \frac{1}{2} \times \frac{c}{a} \times \frac{c}{b}

For $4x + 5y = 40$ : intercepts $10$ and $8$ , area $\dfrac{1}{2} \times 10 \times 8 = 40$ . For $5x - 4y = 20$ : intercepts $4$ and $-5$ , area $\dfrac{1}{2} \times 4 \times 5 = 10$ .

Watch: The $x$ -intercept divides by $a$ and the $y$ -intercept by $b$ . Swapping them is the classic slip. Take absolute values for the area.

Remainder and factor theorems

Dividing $p(x)$ by $(x - \alpha)$ leaves remainder $p(\alpha)$ : just substitute. For divisor $x + 1$ , substitute $x = -1$ . Two remainders give two linear equations in unknown coefficients. Factor questions are the same with the remainder forced to zero: $(x - \alpha)$ is a factor exactly when $p(\alpha) = 0$ .

Tip: For a divisor like $2x - 1$ , substitute $x = \frac{1}{2}$ , the zero of the divisor, not $1$ .

Quadratic roots

For $ax^2 + bx + c = 0$ with roots $\alpha, \beta$ : sum $= -\frac{b}{a}$ , product $= \frac{c}{a}$ . Build targets from these two: $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$ , and $\alpha^4+\beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha\beta)^2$ . For $x^2 - 3x + 1$ : sum $3$ , product $1$ , so $\alpha^2+\beta^2 = 7$ and $\alpha^4+\beta^4 = 47$ . The rebuild also works backwards: roots $4$ and $8$ give the equation $x^2 - 12x + 32 = 0$ , the sum as the flipped middle coefficient and the product as the constant.

Discriminant for root nature

Real distinct roots need $b^2 > 4ac$ ; equal roots need $b^2 = 4ac$ ; no real roots when $b^2 < 4ac$ . A question saying 'always positive' translates to $a > 0$ together with $b^2 < 4ac$ .

A quick habit

For 'where do the lines meet' questions with options, substitute each option into both equations. Testing options is usually faster than solving the pair.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1common2 practice Q

Consistency of a linear pair, find k

How to spot it:

Two linear equations with an unknown coefficient $k$ ; the question fixes the number of solutions.

\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \text{ for infinite solutions}

Method

Compute the ratio of the coefficients that are fully known.
Set the $k$ -ratio equal to it and solve for $k$ .
Check the constant ratio: equal for infinite, different for none.

Why it works:

Coincident lines must agree in all three ratios, which pins $k$ .

Try this

For what value of $k$ do $3x + 4y = 12$ and $9x + ky = 36$ have infinitely many solutions?

Show solution

$9/3 = 3$ , so $k/4 = 3$ gives $k = 12$ .
Check constants: $36/12 = 3$ . All three ratios equal.
So $k = 12$ .

Answer

Type 2common2 practice Q

Area of the triangle cut from the axes

How to spot it:

A line with the coordinate axes bounds a triangle; its area is asked.

\text{area} = \dfrac{c^2}{2ab}

Method

Put $y = 0$ and read the $x$ -intercept $c/a$ .
Put $x = 0$ and read the $y$ -intercept $c/b$ .
Halve the product of the absolute lengths.

Why it works:

The axes are perpendicular, so the intercepts are exactly base and height.

Try this

Find the area of the triangle formed by $4x + 5y = 40$ and the coordinate axes.

Show solution

$x$ -intercept: $40/4 = 10$ .
$y$ -intercept: $40/5 = 8$ .
Area $= \dfrac{1}{2} \times 10 \times 8 = 40$ .

Answer

40 square units

Type 3very common3 practice Q

Remainder theorem for unknown coefficients

How to spot it:

A polynomial with unknown coefficients is divided by two linear factors; the two remainders are given.

p(x) \div (x-a) \Rightarrow R = p(a)

Method

Substitute the zero of each divisor into the polynomial.
Set each result equal to the given remainder.
Solve the two linear equations for the unknowns.
Compute the asked quantity.

Why it works:

Each substitution turns a whole division into one evaluation, giving clean linear equations.

Try this

When $x^3 - 2x^2 + ax + b$ is divided by $(x-1)$ and $(x+1)$ the remainders are $3$ and $5$ . Find $ab$ .

Show solution

$p(1) = 1 - 2 + a + b = 3$ , so $a + b = 4$ .
$p(-1) = -1 - 2 - a + b = 5$ , so $b - a = 8$ .
Solving: $b = 6$ , $a = -2$ .
$ab = -12$ .

Answer

-12

Type 4very common3 practice Q

Quadratic roots and symmetric functions

How to spot it:

A quadratic is given, or its root sum and product; an expression in the roots is asked.

\alpha+\beta = -\frac ba,\qquad \alpha\beta = \frac ca

Method

Read the sum and product from the coefficients.
Write the target using the sum and product.
Substitute and finish.
For fourth powers, square the square-sum and subtract twice the product squared.

Why it works:

Every symmetric function of the roots is built from their sum and product.

Try this

If $\alpha, \beta$ are the roots of $x^2 - 12x + 32 = 0$ , find $\alpha^2 + \beta^2$ .

Show solution

$\alpha + \beta = 12$ and $\alpha\beta = 32$ .
$\alpha^2+\beta^2 = 144 - 64$ .
$= 80$ .

Answer

Type 5very common

Factor theorem for an unknown constant

How to spot it:

A stated linear factor like $(x-2)$ appears with a polynomial holding one unknown coefficient.

(x-a) \text{ is a factor} \iff p(a) = 0

Method

Write the zero of the stated factor.
Substitute it into the polynomial.
Set the result to zero and solve for the unknown.
Verify by substituting a nearby point if time allows.

Why it works:

A factor means zero remainder, and the remainder is just the substitution value.

Try this

If $(x - 2)$ is a factor of $x^3 - 3x^2 + kx + 10$ , find $k$ .

Show solution

$p(2) = 8 - 12 + 2k + 10 = 6 + 2k$ .
Factor means $p(2) = 0$ , so $2k = -6$ .
$k = -3$ .

Answer

-3

Formula sheet

Unique solution

\frac{a_1}{a_2} \ne \frac{b_1}{b_2}

lines cross once

No solution

\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}

parallel lines

Infinite solutions

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

same line

Area with the axes

\text{Area} = \frac{1}{2}\cdot\left|\frac{c}{a}\right|\cdot\left|\frac{c}{b}\right|

Remainder theorem

p(x) \div (x-a) \Rightarrow R = p(a)

for px - q, substitute q/p

Roots of a quadratic

\alpha+\beta = -\frac ba,\quad \alpha\beta = \frac ca

Discriminant

b^2 - 4ac \gtrless 0

decides root nature

Shortcuts that save time

⚡ Area from intercepts

Put y = 0 for the x-intercept and x = 0 for the y-intercept, then halve the product of the absolute values.

Example

Find the area of the triangle formed by 5x - 4y = 20 and the axes.

Show solution

$y = 0$ : $x = 4$ .
$x = 0$ : $y = -5$ , length $5$ .
Area $= \dfrac{1}{2} \times 4 \times 5 = 10$ .

Answer

10 square units

⚡ Remainder equals substitution at the zero

No long division: substitute the zero of the divisor into the polynomial.

Example

Find the remainder when x^3 - 2x^2 - 2x + 6 is divided by x - 2.

Show solution

Zero of the divisor: $x = 2$ .
$p(2) = 8 - 8 - 4 + 6$ .
$= 2$ .

Answer

⚡ Test the options for the intersection point

For a meeting-point question, plug each option into both equations instead of solving the pair.

Example

Where do 2x + 3y = 13 and 3x - y = 3 meet?

Show solution

Try $(2, 3)$ : $4 + 9 = 13$ holds.
Second check: $6 - 3 = 3$ holds.
Both equations pass, so the point is $(2, 3)$ .

Answer

(2, 3)

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Answering parallel without checking the constant ratio.

The c-ratio separates parallel from coincident; check it last, always.

Mistake 02

Taking c/b as the x-intercept.

The x-intercept is c/a; put y = 0 to see it.

Mistake 03

Substituting the wrong sign for divisor x + 3.

Its zero is -3; substitute minus 3.

Mistake 04

Writing the root sum as b/a.

It is minus b over a; the sign flips.

Mistake 05

Using a negative intercept as a negative area.

Lengths are absolute; the area is always positive.

Quick revision

Read this the night before the exam.

Ratios: unequal gives one solution; all equal gives infinite.
Parallel needs the first two ratios equal and the third different.
Intercepts $c/a$ and $c/b$ ; area is half their product.
Remainder $= p(\alpha)$ for divisor $x - \alpha$ .
$\alpha+\beta = -b/a$ , $\alpha\beta = c/a$ .
$b^2 - 4ac$ compared with 0 decides root nature.

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 7 min · wrong answers go to your mistake notebook automatically.

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