Linear Equations — SSC CGL Quantitative Aptitude Formulas and Traps

Solving Linear Systems

Solving two equations in two unknowns

Sum and difference of two numbers

x + y = S,\; x - y = D \;\Rightarrow\; x = \frac{S + D}{2},\; y = \frac{S - D}{2}

$S$ the sum of the two numbers
$D$ the difference, larger minus smaller

Solving three equations in three unknowns

Symmetric set of three equations

x + y = p,\; y + z = q,\; z + x = r \;\Rightarrow\; x + y + z = \frac{p + q + r}{2}

$p, q, r$ the three pair sums

Common traps

Stopping at the first unknown

Finding

x

is half the job. An option may print only the other number, or a pair with the right

x

and a wrong

y

. Find both values and check the pair in the equation you did not use last.

Subtracting only part of an equation

When you subtract one equation from another, subtract every term, the constant included:

(2x + 3y) - (2x - y) = 4y

, not

2y

, and the right sides are subtracted too.

Removing a different unknown from each pair

If one pair of equations loses

x

and the other pair loses

y

, you still have all three unknowns spread over two equations. Remove the same unknown from both pairs.

Using the same pair twice

Two combinations of the same two equations give only one new fact. Each of the two pairs must bring in the third equation, or the system cannot be finished.

Accepting an option after one equation

An option that satisfies one equation is not yet a solution. Options are often built to pass the first equation, so check the survivor in every equation of the system.

Testing the pair before fixing the constant

When an option says "if a = 6, then x = 4, y = 2", the equations change with

a

. Put the value of

a

in first, then test the pair in the new equations.

Number of Solutions

Learn this subtopic in the notes

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})

$a_1, b_1, c_1$ 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

Common traps

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.

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.

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.

Word Problems and Applications

Learn this subtopic in the notes

Turning a word problem into one equation in one unknown

Steady change

\text{value after } t \text{ steps} = a + dt \qquad d = \frac{\text{later value} - \text{earlier value}}{\text{steps between them}}

$a$ the starting value
$d$ the change in each step (negative for a fall)
$t$ the number of steps

Two unknowns from two statements

Equal after a transfer of k

A - k = B + k \;\Rightarrow\; A - B = 2k

$A, B$ the amounts the giver and the receiver hold before the transfer
$k$ the number of items handed over

Swapped quantities and the change in the bill

Swapped quantities

rq_1 + q_2 \;\to\; rq_2 + q_1 \qquad (a + b)(x + y) = P + Q \qquad (a - b)(x - y) = P - Q

$r$ how many times the cheaper price the dearer price is
$q_1, q_2$ the ordered quantities of the dearer and the cheaper item
$P, Q$ the amounts before and after two kinds of notes are interchanged

Whole-number answers: counting coins and notes

Coins of ₹5, ₹2 and ₹1

x + y + z = N,\quad 5x + 2y + z = V \;\Rightarrow\; 4x + y = V - N

$x, y, z$ the numbers of ₹5, ₹2 and ₹1 coins
$N$ the total number of coins
$V$ the total value in rupees

Common traps

Using two letters where one will do

When the total count is given, the second kind is the total minus the first. Two letters with one equation cannot be solved; one letter with one equation can.

Miscounting the steps between two values

From the value after year 6 to the value after year 15 there are 9 yearly steps, not 15 and not 10. Divide the change by the difference of the two step numbers.

Answering x instead of the question

Questions often ask for something built from x: half the number, the other kind of item, the smallest side. Read the last line again before picking an option, since x itself is usually one of the wrong options.

Taking the transfer as the difference

If giving

k

items makes two people equal, they differed by

2k

, not

k

: the giver loses

k

and the receiver gains

k

.

Comparing totals over different numbers of days

One group earning ₹4,200 in 6 days and another earning ₹6,300 in 7 days cannot go into the same pair of equations as they stand. Divide each total by its own number of days to get earnings per day first.

Swapping the prices instead of the quantities

In a delivery mix-up each item keeps its own price; only the counts change places. The swapped bill is (dearer price × the cheaper item's count) plus (cheaper price × the dearer item's count).

Taking the percentage of the new bill

"The bill rose by 40%" compares with the bill that was ordered: new bill

= 1.4 \times

ordered bill, not ordered bill

= 0.6 \times

new bill.

Checking only one count for a negative value

Every count must be 0 or more, and each gives its own limit. In a three-coin question one count sets the upper limit and another the lower limit; an option outside either one is impossible.

Age Problems

Learn this subtopic in the notes

The gap between two ages never changes

Age-gap shortcut

\text{younger then} = \frac{\text{gap}}{k - 1} \qquad \text{one ratio part} = \frac{\text{gap}}{m - n}

$k$ how many times as old the elder is at that moment
$m : n$ the ratio of the elder's age to the younger's at that moment

Common traps

Moving only one person in time

"In 5 years, A will be twice B" means

A + 5 = 2(B + 5)

. Adding the 5 years to A alone,

A + 5 = 2B

, is the most common slip in age questions.

Expecting the ratio of ages to stay fixed

If a mother is twice her son's age today, she will be less than twice his age in 5 years. The difference of the ages stays fixed; the ratio moves towards 1 as both grow older.

Using today's ages inside a past condition

"Eight years ago he was 3 times as old as his son" is

F - 8 = 3(S - 8)

. Writing

F - 8 = 3S

moves the father back in time and leaves the son where he is.

Reading "times more than" one way only

"Five times more than" strictly means six times as much, but question setters also use it for five times. Try both readings; usually only one of them lands on a printed option.

Stopping at one age when the question asks for a combination

Age questions often ask for the sum of the present ages, their ratio or their difference. Solving for one person and choosing that number is a common way to pick a printed wrong option.

Deciding who is older when the question does not say

"Their ages differ by 7 years" allows either person to be the elder. Settle the order from the other facts, or try both and keep the one that gives whole, sensible ages.

Linear Equations formulas

Solving Linear Systems

Solving two equations in two unknowns

Solving three equations in three unknowns

Common traps

Number of Solutions

One, none or infinitely many solutions from the coefficient ratios

Common traps

Word Problems and Applications

Turning a word problem into one equation in one unknown

Two unknowns from two statements

Swapped quantities and the change in the bill

Whole-number answers: counting coins and notes

Common traps

Age Problems

The gap between two ages never changes

Common traps

More SSC CGL Quantitative Aptitude formula sheets