SSC CGL Maths · Linear Equations

Number of Solutions

Two linear equations have one solution, no solution or infinitely many, and comparing the ratios of their coefficients tells you which without solving anything.

All Linear Equations notes

Why this matters

Fifteen questions in the bank, all from the 2024 shifts and eleven of them EASY. None needs the system solved: five ask which case a system is in, seven find one constant that forces a case, and three find two constants. Each is three fractions and one comparison, a few seconds once the rule is sure.

Concept 1 of 3: One, none or infinitely many solutions from the coefficient ratios

Each linear equation in x and y is a straight line. Two lines either cross at one point, run parallel and never meet, or lie on top of each other and share every point. The ratios of the coefficients tell you which, with no drawing and no solving.

Definition

Write both equations in the same form, $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ , or both with the constant on the right.
One solution (the lines cross): $\frac{a_1}{a_2} \ne \frac{b_1}{b_2}$ .
No solution (parallel lines): $\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$ .
Infinitely many solutions (the same line, called coincident lines): all three ratios are equal, so one equation is a multiple of the other.
A system with at least one solution is consistent; a system with none is inconsistent.
Fast route: if the left side of one equation is a multiple of the other's, multiply the other's constant by the same number. The same constant means infinitely many solutions; a different one means none.
Two straight lines can never meet at exactly two points, so "more than one but finitely many" is never the answer.

Number of solutions of two linear equations

\frac{a_1}{a_2} \ne \frac{b_1}{b_2} \;(\text{one}) \qquad \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \;(\text{none}) \qquad \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \;(\text{infinitely many})

$a_1, b_1, c_1$ the coefficients of x and y and the constant in the first equation
$a_2, b_2, c_2$ the same three numbers in the second equation

Worked example

How many solutions does the system

6x - 4y = 10

9x - 6y = 12

have?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 24 Sep 2024, Shift 1 · Q55Easy

Example 1 · Linear Equations · Number of Solutions

What conclusion can be drawn about the solution of the following system of linear equations in two variables:

3x + 2y = 7

2x + 3y = 7

Mixing forms flips the sign of the constant ratio

Take

4x + 6y - 10 = 0

and

2x + 3y = 5

. Comparing

-10

with 5 gives a constant ratio of

-2

against coefficient ratios of 2, which looks like no solution, yet the first equation is just 2 times the second: the same line. Move both constants to the same side before comparing

\frac{c_1}{c_2}

Calling a system with all three ratios different inconsistent

No solution needs the two coefficient ratios to be equal. If

\frac{a_1}{a_2}

and

\frac{b_1}{b_2}

differ, the lines cross and there is exactly one solution, whatever the constants are.

Choosing "finitely many, but more than one"

Two straight lines share no point, one point or every point. A pair of linear equations never has exactly two or three solutions, so that option is always wrong.

Drill 4 more on one, none or infinitely many solutions from the coefficient ratios

Concept 2 of 3: Finding a constant that gives a required number of solutions

The case is given and one coefficient is unknown. Write the ratio that holds the unknown, set it against a ratio made only of numbers, and the constant falls out of a one-step equation.

Definition

Infinitely many solutions: set the unknown's ratio equal to a ratio of known numbers and solve. Then check that the third ratio matches as well; if it does not, no value works.
No solution: set the unknown's ratio equal to the other coefficient ratio, then confirm that the ratio of the constants is different.
One solution: the answer is a "not equal" statement. Find the value at which the two coefficient ratios become equal, and exclude it: $k \ne$ that value.
Cross-multiply to clear fractions: $\frac{4}{6} = \frac{6}{k + 4}$ becomes $4(k + 4) = 36$ .

Worked example

For what value of

m

does the system

3x + my = 7

9x + 12y = 20

have no solution?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 17 Sep 2024, Shift 2 · Q53Easy

Example 2 · Linear Equations · Number of Solutions

For what value of k, the system of equations 4x + 12y + 36 = 0 and 5x + ky + 45 = 0 has an infinite number of solutions?

Skipping the constant check for no solution

The value that makes the coefficient ratios equal can also make the constant ratio equal. Then the lines are the same, not parallel, and the system has infinitely many solutions. For

kx + 2y = 3

and

6x + 4y = 6

k = 3

gives the same line, so no value of

k

gives no solution.

Giving a single value for a unique solution

A unique solution holds for every value except one. The answer reads

k \ne 5

, and an option saying

k = 5

is the one value that fails.

Drill 6 more on finding a constant that gives a required number of solutions

Concept 3 of 3: Finding two unknown constants from the ratio conditions

Infinitely many solutions means three ratios are equal, and that is two equations at once: enough to fix two unknown constants.

Definition

Infinitely many solutions: write first ratio $=$ second ratio and second ratio $=$ third ratio. Two equations fix two constants.
When one ratio holds only numbers, set each of the other two equal to it: that is the quickest pair.
No solution: the coefficient ratios give one equation, which may be a quadratic with two roots. The constant ratio then gives a "not equal" condition, which can rule out one of the roots or limit the second constant.
Check the answer by writing out both equations: one should be an exact multiple of the other, left side and right side.

Worked example

Find

a

and

b

so that

2x + 3y = 7

and

(a + b)x + (2a - b)y = 21

have infinitely many solutions.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 25 Sep 2024, Shift 1 · Q71Moderate

Example 3 · Linear Equations · Number of Solutions

Find the values of 'a' and 'b' for which the system of equations 3x+y=3 and (a-b)x+ (a+b)y=3a+b-3 has infinite solutions.

Keeping both roots of the coefficient condition

Setting two coefficient ratios equal can give a quadratic such as

a^2 - 9 = 0

with two roots. One root may also make the constants proportional, which gives the same line instead of parallel lines. Test the constant ratio at each root before choosing.

Drill 2 more on finding two unknown constants from the ratio conditions

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

One, none or infinitely many solutions from the coefficient ratios
Number of solutions of two linear equations
$\frac{a_1}{a_2} \ne \frac{b_1}{b_2} \;(\text{one}) \qquad \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \;(\text{none}) \qquad \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \;(\text{infinitely many})$

Watch out for (6)

Mixing forms flips the sign of the constant ratio
→ One, none or infinitely many solutions from the coefficient ratios
Calling a system with all three ratios different inconsistent
→ One, none or infinitely many solutions from the coefficient ratios
Choosing "finitely many, but more than one"
→ One, none or infinitely many solutions from the coefficient ratios
Skipping the constant check for no solution
→ Finding a constant that gives a required number of solutions
Giving a single value for a unique solution
→ Finding a constant that gives a required number of solutions
Keeping both roots of the coefficient condition
→ Finding two unknown constants from the ratio conditions

Test yourself on Linear Equations

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Linear Equations chapter test Drill all 15 questions