Linear and Quadratic Equations for CUET General Aptitude

Q: When does a pair of linear equations have one solution, no solution or infinitely many?

Compare the ratios of the coefficients. If the x and y ratios differ, the lines cross once. If they are equal but the constant ratio differs, the lines are parallel and there is no solution. If all three ratios are equal, the two equations describe the same line and there are infinitely many solutions.

Q: How do you solve a quadratic equation by factoring?

Write the equation as ax^2 + bx + c = 0. Split the middle term into two parts whose product is a × c and whose sum is b. Factorise in pairs, then set each factor equal to zero.

Q: How do you use the quadratic formula?

Put the coefficients a, b and c into the formula below. Use it when the equation does not factorise neatly, or when you cannot see the split quickly.

Q: How do you form a quadratic equation from its roots?

If the roots are p and q, the equation is x^2 - (p + q)x + pq = 0. For roots 4 and -7, the sum is -3 and the product is -28, so the equation is x^2 + 3x - 28 = 0.

Q: What does the discriminant tell you about the roots?

The discriminant D = b^2 - 4ac tells you how many real roots a quadratic has, before you solve it. If D is positive there are two different real roots, if it is zero there are two equal roots, and if it is negative there are no real roots.

Q: How do you find the sum and product of roots without solving?

For ax^2 + bx + c = 0 with roots α and β, the sum of the roots is -b/a and the product is c/a. Many other expressions in the roots can be built from these two numbers alone.

About 15 min to learn · 66 questions

CUET's equations questions are Class 9–10 algebra asked at speed, with four options built from the usual slips: a sign left behind, a bracket half expanded, the wrong unknown given as the answer. This lesson covers linear equations in one and two variables, when a pair has no solution, word problems, quadratics by factoring and by formula, and what the discriminant and the sum and product of roots tell you.

Syllabus area: Reasoning

On this page · 10 sections

01The core idea
02Linear equations in one variable
03Pairs of linear equations
04Word problems
05Quadratic equations
06Roots and their nature
07Exam shortcuts
08Common mistakes
09Quick revision
10Check your understanding

The core idea

An equation is a balance. Whatever you do to one side, you do to the whole of the other side, and the balance holds. Every method in this chapter is a way of using that rule to get the unknown on its own.

Most lost marks come from two places. One is a small slip in moving terms around. The other is answering a different question from the one asked, such as giving $x$ when the question wants $x + y$ . A five-second check at the end catches both: put your answer back into the original equation, and reread the question.

Linear equations in one variable

How do you solve a linear equation in one variable?

Move every term with the variable to one side and every plain number to the other, then divide by the coefficient of the variable. When a term crosses the equals sign, its sign changes.

A reliable order of work:

Clear fractions by multiplying every term by the LCM of the denominators.
Expand brackets, watching the sign in front of each one.
Collect the variable terms on one side and the numbers on the other.
Divide by the coefficient.
Substitute the answer back into the original equation.

Example 1 · Collecting terms

Solve $7x - 4 = 2x + 21$ .

Move $2x$ to the left and $-4$ to the right, changing both signs: $7x - 2x = 21 + 4$ .
Simplify: $5x = 25$ , so $x = 5$ .

Answer

x = 5

To check: $7 \times 5 - 4 = 31$ and $2 \times 5 + 21 = 31$ .

Watch out

Forgetting to change the sign when a term crosses the equals sign is the most common error in this chapter. The wrong answer it produces is almost always one of the options.

How do you clear fractions and brackets without sign errors?

Multiply every term on both sides by the LCM of all the denominators, and put each numerator in brackets before you expand. A minus sign in front of a fraction applies to the whole of its numerator.

Example 2 · Fractions

Solve $\dfrac{x + 2}{3} - \dfrac{x - 4}{5} = 2$ .

The LCM of 3 and 5 is 15. Multiply every term by 15, including the 2 on the right: $5(x + 2) - 3(x - 4) = 30$ .
Expand. The minus sign multiplies both terms of the second bracket: $5x + 10 - 3x + 12 = 30$ .
Simplify: $2x + 22 = 30$ , so $2x = 8$ and $x = 4$ .

Answer

x = 4

To check: $\tfrac{6}{3} - \tfrac{0}{5} = 2$ .

Watch out

Two slips give tempting wrong answers here. One is multiplying only the fractions by 15 and leaving the 2 as 2. The other is writing $-3(x - 4)$ as $-3x - 12$ instead of $-3x + 12$ .

Exam shortcut

For decimals, multiply every term by 10 or 100. For $0.3x + 0.5 = 0.1x + 1.7$ , multiplying by 10 gives $3x + 5 = x + 17$ , so $x = 6$ .

Pairs of linear equations

How do you solve a pair of linear equations in two variables?

Use elimination when one variable has the same coefficient in both equations, or can easily be made to. Use substitution when one equation already gives a variable on its own, such as $y = 3x - 1$ . Either way, you end with one equation in one variable.

Example 3 · Elimination

Solve $3x + 2y = 16$ and $5x - 2y = 16$ .

The $y$ terms are $+2y$ and $-2y$ , so adding the equations removes $y$ : $8x = 32$ , so $x = 4$ .
Put $x = 4$ into the first equation: $12 + 2y = 16$ , so $y = 2$ .

Answer

$x = 4$ and $y = 2$

To check in the second equation: $20 - 4 = 16$ .

Example 4 · Substitution

Solve $y = 3x - 1$ and $2x + 3y = 19$ .

Replace $y$ in the second equation with $3x - 1$ : $2x + 3(3x - 1) = 19$ .
Expand and solve: $2x + 9x - 3 = 19$ , so $11x = 22$ and $x = 2$ .
Then $y = 3 \times 2 - 1 = 5$ .

Answer

$x = 2$ and $y = 5$

To check: $4 + 15 = 19$ .

Exam shortcut

When the coefficients swap places, add and subtract instead. For $5x + 3y = 29$ and $3x + 5y = 27$ , adding gives $8(x + y) = 56$ , so $x + y = 7$ . Subtracting gives $2(x - y) = 2$ , so $x - y = 1$ . Then $x = 4$ and $y = 3$ .

Common variation: given the sum and the difference

If $x + y = S$ and $x - y = D$ , then $x = \tfrac{S + D}{2}$ and $y = \tfrac{S - D}{2}$ . You can also go straight to $x^2 - y^2 = (x + y)(x - y) = SD$ without finding either number.

The same idea handles reciprocals. If $\tfrac{1}{x} + \tfrac{1}{y} = \tfrac{5}{6}$ and $\tfrac{1}{x} - \tfrac{1}{y} = \tfrac{1}{6}$ , treat $\tfrac{1}{x}$ and $\tfrac{1}{y}$ as the unknowns: $\tfrac{1}{x} = \tfrac{1}{2}$ and $\tfrac{1}{y} = \tfrac{1}{3}$ , so $x = 2$ and $y = 3$ .

When does a pair of linear equations have one solution, no solution or infinitely many?

Compare the ratios of the coefficients. If the $x$ and $y$ ratios differ, the lines cross once. If they are equal but the constant ratio differs, the lines are parallel and there is no solution. If all three ratios are equal, the two equations describe the same line and there are infinitely many solutions.

For $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ :

Condition	Lines	Solutions
$\tfrac{a_1}{a_2} \ne \tfrac{b_1}{b_2}$	Cross at one point	Exactly one
$\tfrac{a_1}{a_2} = \tfrac{b_1}{b_2} \ne \tfrac{c_1}{c_2}$	Parallel, never meet	None
$\tfrac{a_1}{a_2} = \tfrac{b_1}{b_2} = \tfrac{c_1}{c_2}$	The same line	Infinitely many

Example 5 · No solution

For what value of $k$ do $kx + 6y = 5$ and $2x + 3y = 7$ have no solution?

The lines are parallel when $\tfrac{k}{2} = \tfrac{6}{3} = 2$ , so $k = 4$ .
Check the constants: $\tfrac{5}{7} \ne 2$ , so the lines are parallel and separate, not the same line.

Answer

k = 4

Watch out

After finding $k$ from the first two ratios, always test the third. If it also matches, the lines coincide and the answer is infinitely many solutions, not none. And "exactly one solution" means every value of $k$ except the one that makes the first two ratios equal.

Why this works

Each equation is a straight line, and a solution is a point on both lines. Equal $x$ and $y$ ratios mean the lines have the same slope. Two lines with the same slope either never meet or lie on top of each other, and the constant ratio tells you which.

Word problems

How do you turn a word problem into an equation?

Name the unknown clearly, write every other quantity in terms of it, and translate the sentence one phrase at a time. Then solve, and check the answer against the story, not just the equation.

Phrase	Algebra
5 more than a number	$x + 5$
5 less than four times a number	$4x - 5$
Twice as many ₹10 notes as ₹50 notes	₹50 notes $= n$ , ₹10 notes $= 2n$
A two-digit number, tens digit $t$ , units digit $u$	$10t + u$
The same number with its digits reversed	$10u + t$
If A gives B 5, they are equal	$A - 5 = B + 5$

Coins, notes and tickets always give two facts: how many items there are, and how much they are worth. If you use one letter, write the other count in terms of it.

Example 6 · Coins

A jar has 25 coins, all ₹1 and ₹5 coins, worth ₹69 altogether. How many ₹5 coins are there?

Let there be $f$ coins of ₹5. Then there are $25 - f$ coins of ₹1.
Write the value equation: $5f + (25 - f) = 69$ , so $4f = 44$ .
Solve: $f = 11$ .

Answer

11 coins of ₹5.

To check: $11 \times 5 + 14 \times 1 = 55 + 14 = 69$ .

Example 7 · Fixed charge plus a rate

A plumber charges a fixed call-out fee plus a fixed amount per hour. A 3-hour job costs ₹1,150 and a 7-hour job costs ₹2,150. What is the call-out fee?

Write fee $+$ rate $\times$ hours for each job: $F + 3r = 1150$ and $F + 7r = 2150$ .
Subtract the first from the second. The fee cancels: $4r = 1000$ , so $r = 250$ .
Put $r = 250$ back: $F = 1150 - 750 = 400$ .

Answer

₹400

To check: $400 + 7 \times 250 = 2150$ .

Exam shortcut

In a transfer problem, a transfer of 5 changes the gap between the two people by 10. "If A gives B 5, they become equal" tells you A has 10 more than B.

Watch out

In a fraction problem, the fraction after the change is usually printed among the options. Answer with the original fraction, which is what the question asks for.

Common variation: digits

The digits of a two-digit number add up to 9, and reversing the digits increases the number by 27. Find the number.

Write it as $10t + u$ . Then $t + u = 9$ , and $(10u + t) - (10t + u) = 27$ , which simplifies to $9(u - t) = 27$ , so $u - t = 3$ . Adding the two equations gives $2u = 12$ , so $u = 6$ and $t = 3$ . The number is 36, and $63 - 36 = 27$ .

For age puzzles, see ages and number problems.

Quadratic equations

How do you solve a quadratic equation by factoring?

Write the equation as $ax^2 + bx + c = 0$ . Split the middle term into two parts whose product is $a \times c$ and whose sum is $b$ . Factorise in pairs, then set each factor equal to zero.

Example 8 · Splitting the middle term

Solve $3x^2 + 10x - 8 = 0$ .

Find $a \times c = 3 \times (-8) = -24$ . Two numbers with product $-24$ and sum 10 are 12 and $-2$ .
Split and factorise in pairs: $3x^2 + 12x - 2x - 8 = 3x(x + 4) - 2(x + 4) = (3x - 2)(x + 4)$ .
Set each factor to zero: $x = \tfrac{2}{3}$ or $x = -4$ .

Answer

$x = \tfrac{2}{3}$ or $x = -4$

Three rules that keep both roots

Get zero on the right first. In $(x - 1)(x + 2) = 4$ , you cannot set each bracket equal to 4. Expand and bring the 4 across.
Never divide by an expression containing $x$ . Dividing $x^2 = 6x$ by $x$ gives only $x = 6$ and loses $x = 0$ . Factorise instead: $x(x - 6) = 0$ .
A square root has two signs. $x^2 = 25$ means $x = 5$ or $x = -5$ .

How do you use the quadratic formula?

Put the coefficients $a$ , $b$ and $c$ into the formula below. Use it when the equation does not factorise neatly, or when you cannot see the split quickly.

The quadratic formula, for ax² + bx + c = 0

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Example 9 · Using the formula

Solve $x^2 - 6x + 4 = 0$ .

Here $a = 1$ , $b = -6$ and $c = 4$ . So $b^2 - 4ac = 36 - 16 = 20$ .
Put the values in: $x = \dfrac{6 \pm \sqrt{20}}{2} = \dfrac{6 \pm 2\sqrt{5}}{2}$ .
Simplify: $x = 3 \pm \sqrt{5}$ .

Answer

$x = 3 + \sqrt{5}$ or $x = 3 - \sqrt{5}$

Watch out

The $2a$ divides the whole numerator, not just the square root. When $b$ is negative, $-b$ is positive. And when $a$ and $c$ have opposite signs, $-4ac$ becomes an addition.

How do you form a quadratic equation from its roots?

If the roots are $p$ and $q$ , the equation is $x^2 - (p + q)x + pq = 0$ . For roots 4 and $-7$ , the sum is $-3$ and the product is $-28$ , so the equation is $x^2 + 3x - 28 = 0$ .

Exam shortcut

To get the equation whose roots are the reciprocals of the roots of $ax^2 + bx + c = 0$ , reverse the coefficients: $cx^2 + bx + a = 0$ .

Example 10 · A quadratic word problem

A cyclist rides 60 km at a steady speed. If she rode 5 km/h faster, the trip would take 1 hour less. What is her speed?

Let the speed be $v$ km/h. Time is distance ÷ speed, so $\dfrac{60}{v} - \dfrac{60}{v + 5} = 1$ .
Multiply through by $v(v + 5)$ : $60(v + 5) - 60v = v(v + 5)$ , so $300 = v^2 + 5v$ .
Rearrange and factorise: $v^2 + 5v - 300 = 0$ , which is $(v + 20)(v - 15) = 0$ .
Speed cannot be negative, so reject $v = -20$ .

Answer

15 km/h

To check: 60 km takes 4 hours at 15 km/h and 3 hours at 20 km/h.

Watch out

Reject a root only when it is impossible in the story, such as a negative length or a fraction of a person. Check it first rather than dropping the negative root by habit.

Roots and their nature

What does the discriminant tell you about the roots?

The discriminant $D = b^2 - 4ac$ tells you how many real roots a quadratic has, before you solve it. If $D$ is positive there are two different real roots, if it is zero there are two equal roots, and if it is negative there are no real roots.

Discriminant	Roots
$D > 0$	Two distinct real roots (rational if $D$ is a perfect square)
$D = 0$	Two equal real roots, each $-\tfrac{b}{2a}$
$D < 0$	No real roots

Example 11 · Equal roots

For what values of $k$ does $4x^2 + kx + 25 = 0$ have equal roots?

Equal roots need $D = 0$ : $k^2 - 4 \times 4 \times 25 = 0$ .
So $k^2 = 400$ , giving $k = 20$ or $k = -20$ .

Answer

$k = 20$ or $k = -20$

Both count unless the question restricts the sign.

Watch out

"Real roots" means $D \ge 0$ . "Distinct real roots" means $D > 0$ , strictly. "No real roots" means $D < 0$ . Mixing up the strict and non-strict inequalities is exactly what separates two of the options.

How do you find the sum and product of roots without solving?

For $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$ , the sum of the roots is $-\tfrac{b}{a}$ and the product is $\tfrac{c}{a}$ . Many other expressions in the roots can be built from these two numbers alone.

Sum and product of roots

\alpha + \beta = -\frac{b}{a} \qquad \alpha\beta = \frac{c}{a}

Expression	In terms of the sum and product
$\alpha^2 + \beta^2$	$(\alpha + \beta)^2 - 2\alpha\beta$
$(\alpha - \beta)^2$	$(\alpha + \beta)^2 - 4\alpha\beta$
$\tfrac{1}{\alpha} + \tfrac{1}{\beta}$	$\tfrac{\alpha + \beta}{\alpha\beta}$
$\tfrac{\alpha}{\beta} + \tfrac{\beta}{\alpha}$	$\tfrac{\alpha^2 + \beta^2}{\alpha\beta}$

Example 12 · Sum of squares of the roots

The roots of $x^2 - 8x + 13 = 0$ are $\alpha$ and $\beta$ . Find $\alpha^2 + \beta^2$ .

Sum: $\alpha + \beta = 8$ . Product: $\alpha\beta = 13$ .
Use the identity: $\alpha^2 + \beta^2 = 8^2 - 2 \times 13 = 64 - 26 = 38$ .

Answer

Common variation: roots in a given relation

If one root is three times the other, call them $r$ and $3r$ . For $x^2 - 12x + k = 0$ , the sum gives $4r = 12$ , so $r = 3$ , and the product gives $k = 3 \times 9 = 27$ . To check: $x^2 - 12x + 27 = (x - 3)(x - 9)$ .

If the roots differ by $d$ , use $(\alpha - \beta)^2 = d^2$ together with the sum and product.

Exam shortcuts

Try the options. In a one-variable question, putting each option back into the equation is often faster than solving.
A pair must satisfy both equations. Many wrong options satisfy only one of them.
Use the sum and product to test roots. A pair of options is right only if it has the right sum and the right product.
Swapped coefficients: add, then subtract. You get $x + y$ and $x - y$ at once.
Fixed charge plus rate: subtract the two bills. The fixed charge cancels and the rate is left.

Common mistakes

Not changing the sign when a term crosses the equals sign.
Applying a minus sign to the first term of a bracket only. $-2(x - 3)$ is $-2x + 6$ .
Clearing fractions on one side only. Every term on both sides gets multiplied by the LCM.
Answering with the wrong unknown. Reread what the question asks for.
Not testing the third ratio. Equal $x$ and $y$ ratios mean no solution only when the constant ratio differs.
Dividing by $x$ or by a bracket. You lose a root.
Factorising before the right side is zero.
Dropping the minus signs in the root formulas. The sum of the roots is $-\tfrac{b}{a}$ , and the equation from roots is $x^2 - (\text{sum})x + \text{product} = 0$ .
Dividing by $a$ instead of $2a$ in the quadratic formula.
Losing the negative value. $k^2 = 144$ gives $k = \pm 12$ .
Using a strict inequality where equality is allowed. Real roots include $D = 0$ . Distinct real roots do not.

Quick revision

Linear equations

One variable: clear fractions, expand, collect, divide, check
A term changes sign when it crosses the equals sign
Pairs: eliminate when coefficients match, substitute when a variable stands alone

How many solutions?

$\tfrac{a_1}{a_2} \ne \tfrac{b_1}{b_2}$ : exactly one
$\tfrac{a_1}{a_2} = \tfrac{b_1}{b_2} \ne \tfrac{c_1}{c_2}$ : none
All three ratios equal: infinitely many

Word problems

Coins and tickets: one count equation and one value equation
Fixed charge plus rate: subtract the two totals
Digits: the number is $10t + u$ , reversed is $10u + t$

Quadratics

Factoring: split $b$ into two parts with product $ac$
Formula: $x = \tfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
From roots $p$ and $q$ : $x^2 - (p + q)x + pq = 0$
Reciprocal roots: reverse the coefficients

Roots

$D > 0$ : two distinct real roots. $D = 0$ : equal roots. $D < 0$ : no real roots
$\alpha + \beta = -\tfrac{b}{a}$ and $\alpha\beta = \tfrac{c}{a}$
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

Traps

Test the third ratio before saying no solution
Never divide by $x$ : factorise
Two signs from a square root, and reject a root only when the story rules it out

These methods come back in ages and number problems, time and work and time, speed and distance.

Next · 09Quick revision

Check your understanding

66 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Linear equations in one variableQ1 of 66Q1

Linear equations in one variable

12 questions

Check your understanding

Linear equations in one variable

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Pairs of linear equations

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Word problems into equations

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Quadratic equations

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Roots and their nature

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