CBSE Class 10 Mathematics Linear Equations Assignment Set 02

Question. Find the value of k for which the given system of equations has infinitely many solutions:
i. (𝑘 − 3) 𝑥 + 3 𝑦 = 𝑘
ii. 𝑘 𝑥 + 𝑘 𝑦 = 12
Answer : K−3/𝑘 = 3/𝑘
k = 6

Question. Solve for x and y:
i. 𝑚 𝑥 – 𝑛 𝑦 = 𝑚²
+ 𝑛²;
ii. 𝑥 – 𝑦 = 2𝑛
Answer : mx – ny = m² + n²
mx – ny = 2 nm
(2) – (1)
(m – n) y = m² + n² - 2 nm = (m – n)²
Y = m – n
x – (m -n) = 2n
x = m + n

Question. Is the system of linear equations 2𝑥 + 3𝑦 – 9 = 0 𝑎𝑛𝑑 4𝑥 + 6𝑦 – 18 = 0 consistent? Justify your answer.
Answer : 2/4 = 1/2 , −9/−18 = 1/2 , 3/6 = 1/2
It has infinite number of solutions. It is consistent.

Question. Are the following pair of linear equations consistent? Justify your answer.
2𝑎𝑥 + 𝑏𝑦 = 𝑎; 4𝑎𝑥 + 2𝑏𝑦 – 2𝑎 = 0 ; 𝑎 ,𝑏 ≠0
Answer : 𝑎1/𝑎2 = 2𝑎/4𝑎 = 1/2
𝑏1/𝑏2 = 𝑏/2𝑏 = 1/2
𝑐1/𝑐2 = 𝑎/2𝑎 = 1/2
It has infinitely many solution, it is consistent

Question. For which value of a and b does the following pair of linear equations has infinite number of solutions?
i. 2𝑥 + 3𝑦 = 7
ii. 𝑎(𝑥 + 𝑦) – 𝑏 (𝑥 − 𝑦) = 3𝑎 + 𝑏 − 2
Answer : 2x + 3y = 7 𝑥(𝑎 −𝑏) + 𝑦 (𝑎 + 𝑏) = 3𝑎 + 𝑏 – 2
Since it has infinitely many solutions,
2/𝑎−𝑏 = 3/𝑎+𝑏 = 7/3𝑎+𝑏−2
After equating
a = 5b (1)
2a – 5b = 6 (2)
Solve (1) and (2)
a =5 and b= 1

Question. Find the solutions of the pair of linear equations 5𝑥 + 10𝑦 – 50 = 0 𝑎𝑛𝑑 𝑥 + 8𝑦 = 10. Hence find the value of m if 𝑦 = 𝑚𝑥 + 5.
Answer : 𝑆𝑜𝑙𝑣𝑒 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
𝑥= 10, 𝑦 = 0 𝑆
𝑢𝑏 𝑥 =10 𝑎𝑛𝑑 𝑦=0 𝑖𝑛 𝑦 = 𝑚𝑥 + 5
0 = 𝑚 ×10 + 5
𝑚 = −1/2

Question. Write a pair of linear equations which has a unique solution 𝑥 = 2 𝑎𝑛𝑑 𝑦 = − 1. How many such pairs are possible?
Answer : 3x + y = 5
Infinite number of solution

Question. If 𝑥 − 4 is a factor of 𝑥³ + 𝑎𝑥² + 2𝑏𝑥 – 24 and 𝑎 – 𝑏 = 8 , find the value of 𝑎 𝑎𝑛𝑑 𝑏.
Answer : Since x - 4 is a factor of x³ + ax² + 2bx – 24
4³ + a × 4² + 2b × 4 – 24 = 0
a + 2b + 10 =0 (i)
a -b = 8 (ii)
Solve (I and (ii)
a = 2 , b = -6

Question. If 2𝑥 + 𝑦 = 23 and 4𝑥 – 𝑦 = 19, find the values of 5𝑥 – 3𝑦 and 𝑦 – 2𝑥.
Answer : 𝑆𝑜𝑙𝑣𝑒 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
𝑥 = 7, 𝑦 = 9
𝑆𝑜,5𝑥 – 3𝑦 = 8
𝑦 – 2𝑥 = −5

Question. There are 20 vehicles – cars and motorcycles in a parking area. If there are 56 wheel together, how many cars and motorcycles are there.
Answer : Let no of cars = x and no of motor cycles = y
According to our condition
x + y = 20 (i)
4x + 2y = 56 (ii)
Solve (i) and (ii)
x = 8 and y = 12

Long Answer type Questions

Question. The sum of the numerator and the denominator of a fraction is 3 less than twice the denominator. If the numerator and the denominator are decreased by one, the numerator becomes half the denominator. Determine the fraction.
Answer : Let the numerator be x and the denominator be y
Fraction is xy According to the first condition x + y = 2y - 3 x + y - 2y = - 3 x - y = - 3 - - - - (1) According to the second condition x - 1 = 12y - 1 2x - 1 = y - 1 2x - 2 = y - 1 2x - y = 1 - - - - - (2) Subtracting eqn (2) from eqn (1) we get - x = - 4 x = 4 Substituting x = 4 in equation 2, we get 2x4 - y = 1 8 - y = 1 y = 7
Fraction is 47

Question. Given the linear equation 7𝑥 − 5𝑦 − 4 = 0. Write another linear equation in two variables such that the geometrical representation of the pair so formed is Intersecting lines, Parallel lines, Coincident lines
Answer : (i) Any linear equation satisfying the condition a1a2 ≠ b1b2
(ii) Any linear equation satisfying the condition a1/a2 = b1/b2 ≠ c1 /c2
(ii) Any linear equation satisfying the condition a1/a2 = b1/b2 = c1/ c2

Question. A number consists of two digits. When the number is divided by the sum of its digits, the quotient is 7. If 27 is subtracted from the number, the digits interchange their places. Find the number.
Answer : Let the digit in ones place be y and the digit in tens place be y.
Two digit number = 10x + y
Given 10x + yx + y = 7
⇒ 10x + y = 7 ( x + y )
∴ 10x + y - 7x - 7y = 0
3x - 6y = 0
x - 2y = 0 - - - - - - - - (1)
According to the second condition.
10x + y - 27 = 10y + x
10x + y - 10y - x = 27
9x - 9y = 27
x - y = 3 - - - - - - - (2)
Equation (1) - (2)
x - 2y - x - y = 0 - 3
x - 2y - x + y = - 3
- y = - 3
y = 3
Substituting y = 3 in equation (2), we get
x - 3 = 3
x = 6
Two - digit number = 10x + y
= 10x6 + 3
= 60 + 3 = 63
20y + y - 27 = 10y + 2y
⇒ 9y = 27
⇒y = 3
Substitute y value in eqn(1)
we get, x = 2× 3
⇒ x = 6
Hence the required number is 63.

Question. The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the fathe
Answer : Let the present age of his two children be “x” years and “y” years.
Present age of father = 2(x + y) - - - - (1)
A.T.Q.
2x + y + 20 = x + 20 + y + 20
2x + 2y + 20 = x + y + 40
2x + 2y - x - y = 40 - 20
x + y = 20 - - - - (2)
Substituting eqn (2) in eqn (1), we get
Present age of father = 2 x 20
= 40 years

Question. A railway half ticket costs half the full fare, but the reservation charges are the same on a half ticket as on a full ticket. One reserved first - class ticket from the station A to B costs ₹2530. Also one reserved first class ticket and one reserved first class half ticket from A to B costs ₹3810.Find the full first class fare from station A to B and also the reservation charges for a ticket.
Answer : Let the cost of full fare be ₹ 𝑥 and the cost of half first class fare be ₹ 𝑥2, respectively and reservation charges be ₹ 𝑦 per ticket.
Case I The cost of one reserved first class ticket from the stations A to B = ₹ 2530
x + y = 2530 ... ( i )
Case II
The cost of one reserved first class ticket and one reserved first class half ticket from stations A to B = ₹ 3810
⇒ 𝑥 + 𝑦 + (𝑥/2) + 𝑦 = 3810
⇒ 𝑥 + (𝑥/2) + 𝑦 + 𝑦 = 3810
⇒3𝑥/2 + 2 𝑦 = 3810
Multiplying throughout by 2, we get
⇒ 3𝑥 + 4𝑦 = 7620 ...(𝑖𝑖)
Now, multiplying Eq. (i) by 4 and then subtracting from Eq. (ii), we get 3 𝑥 + 4 𝑦−4𝑥−4𝑦 = 7620 −10120 − 𝑥 = − 2500 ⇒ 𝑥 = 2500
On putting the value of x in Eq. (i), we get
2500 + y = 2530
⇒ 𝑦 = 30
Hence, full first - class fare from stations A to B is ₹ 2500 and the reservation for a ticket is ₹ 30.

Question. Two numbers are in the ratio 5:6. If 8 is subtracted from each of the numbers, the ratio becomes 4:5. Find the numbers.
Answer : Let the two numbers be x and y
According to the first condition
xy = 56
Cross multiplying, we get
6x = 5y
6x - 5y = 0 - - - - - - (i)
According to the second condition
x - 8y - 8 = 45
Cross multiplying, we get
5x - 8 = 4y - 8
5x - 40 = 4y - 32
5x - 4y = 40 - 32
5x - 4y = 8 - - - - - - - (ii)
Multiplying eqn (i) by 4, we get
24x - 20y = 0 - - - - (iii)
Multiplying eqn (ii) by 5, we get
25x - 20y = 40 - - - - - (iv)
Subtracting eqn (iii) - eqn(iv), we get
24x - 20y - 25x + 20y = - 40 - x = - 40
x = 40
Substituting x = 40 in eqn (i), we get
6 x 40 - 5y = 0
240 - 5y = 0
240 = 5y
y = 48
The numbers are 40 and 48.

Question. A boat takes 4 hours to go 44 km downstream and it can go 20 km upstream in the same time. Find the speed of the stream and that of the boat in still water.
Answer : Let the speed of the stream = y km/hr and
Speed of the boat in still water = x km/hr
Speed of the boat in downstream = (x + y)km/hr
Speed of the boat in upstream = (x - y)km/hr
A.T.Q
44x + y = 4
44 = 4(x + y)
x + y = 11 - - - - - - (1)
Also 20x - y = 4
20 = 4x - y
x - y = 5 - - - - - (2)
Adding eqns 1 and 2 we get, 2x = 16 x = 8
Substituting x = 8 in eqn(1), we get
8 + y = 11
y = 3
Speed of the stream = 3 km/hr and
Speed of the boat in still water = 8 km/hr

Question. There are two examination rooms A and B. If 10 candidates are sent from A to B, the number of candidates in each room is the same. If 20 candidates are sent from B to A, the number of students in A is double the number of students in B. Find the number of students in each room.
Answer : Let the number of students in room A be x and that in room B be y. According to the first condition x - 10 = y + 10 ⟹x−y = 20−−−−−(i) According to the second condition, we get x + 20 = 2(y - 20)
x + 20 = 2y - 40
x - 2y = - 40 - 20 ⟹x−2y = −60−−−−−(ii) Eqn (i) - Eqn (ii) we get, x - y - x + 2y = 20 + 60
y = 80
Substituting y = 80 in eqn (i), we get x - 80 = 20
x = 100
Number of students in room A = 100
Number of students in room B = 80

Question. Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in one hour. What are the speeds of the two cars?
Answer : Let x km/hr be the speed of car from point A and y km/hr be the speed of car from point B.
If the car travels in the same direction,
5x – 5y = 100 x – y = 20 …………………………………(i)
If the car travels in the opposite direction,
x + y = 100………………………………(ii)
Adding equations (i) and (ii), we get
2x = 120
x = 60 km/hr
Substituting this in equation (i), we get,
60 – y = 20
y = 40 km/h
Therefore, the speed of car from point A = 60 km/hr
Speed of car from point B = 40 km/hr.

LINEAR EQUATIONS IN ONE VARIABLE

LINEAR EQUATION
An equation in which the highest power of the unknown variable is one, is a linear equation.

SOME PROPERTIES OF EQUALITY

1. Reflexive Property
Every number is equal to itself.
Example : \(5 = 5\), \(2 = 2\) and so on.

2. Symmetric Property
For any two numbers, if the first number is equal to the second, then the second number is equal to the first.
If \(x\) and \(y\) are two numbers and \(x = y\), then \(y = x\).
Example : \(3 + 4 = 5 + 2\)
\(\Rightarrow 5 + 2 = 3 + 4\)

3. Transitive Property
If \(x\), \(y\) and \(z\) are three numbers such that \(x = y\) and \(y = z\), then \(x = z\).
Example : \(9 + 3 = 12\), \(12 = 3 \times 4\)
\(\therefore 9 + 3 = 3 \times 4\)

4. If \(x\), \(y\) and \(z\) are three numbers such that \(x = y\) and \(x = z\), then \(y = z\).
Example : \(24 = 8 \times 3\), \(24 = 14 + 10\)
\(\Rightarrow 8 \times 3 = 14 + 10\)

SIMULTANEOUS LINEAR EQUATIONS
To solve the equations involving two unknowns, we require two equations.

Solving two Simultaneous Equations
When two equations, each in two variables, are given, they can be solved in four ways.

(a) Elimination by cancellation
Ex. If \(2x + 3y = 10\) and \(x + 2y = 5\), then find the values of \(x + y\)?
Sol. Using this method, the two equations are reduced to a single variable equation by eliminating one of the variables.
Step.1 : Here, we eliminate the \(y\) term and for this, we multiply the first equation with the coefficient of \(y\) in the second equation and multiply the second equation with the coefficient of \(y\) in the first equation so that the coefficient of \(y\) terms in both the equations becomes equal.
\((2x + 3y = 10) \times 2 \Rightarrow 4x + 6y = 20\) --- (3)
\((x + 2y = 5) \times 3 \Rightarrow 3x + 6y = 15\) --- (4)
Step.2 : Subtract equation (4) from equation (3)
\(4x + 6y = 20\)
\(3x + 6y = 15\)
\((-) \quad (-) \quad (-)\)
\(x = 5\)
Step.3 : Substitute the value of \(x\) in equation (1) or equation (2) to find the value of \(y\). Here, substituting the value of \(x\) in the equation (1):
We have -
\(2(5) + 3y = 10\)
\(10 + 3y = 10\)
\(3y = 10 - 10\)
\(3y = 0\)
\(y = 0\)
\(\therefore\) The solution of the given pair of equations is \(x = 5\), \(y = 0\).

(b) Elimination by substitution
Ex. If \(5x - 2y = 17\) and \(3x + y = 8\), then find the value of \(x + y\).
Sol. Using this method, the two equations are reduced to a single variable equation by substituting the value of one variable obtained from one equation, in the other equation.
Step.1 : Using the first equation, find \(x\) in terms of \(y\).
i.e., \(5x - 2y = 17\)
\(\Rightarrow 5x = 17 + 2y\)
\(\Rightarrow x = \frac{17 + 2y}{5}\)
Step.2 : Substitute the value of \(x\) in the equation (2) to find the value of \(y\).
i.e., \(3x + y = 8\)
\(\Rightarrow 3\left(\frac{17 + 2y}{5}\right) + y = 8\)
Step.3 : Simplify the equation in terms of \(y\) and find the value of \(y\).
\(3(17 + 2y) + 5y = 8 \times 5\)
\(\Rightarrow 51 + 6y + 5y = 40\)
\(\Rightarrow 51 + 11y = 40\)
\(\Rightarrow 11y = 40 - 51\)
\(\Rightarrow 11y = -11\)
\(\Rightarrow y = -1\)
Step.4 : Substituting the value of \(y\) obtained in step (3) in equation (1) or (2), we get \(x = 3\).
\(\therefore\) The solution for the given pair of equations is \(x = 3\) and \(y = -1\).

(c) Adding two equations and subtracting one equation from the other.
Ex. \(4x + 3y = 11\) and \(3x + 4y = 10\)
Sol. Adding both equations, we get,
Step.1 : \(7x + 7y = 21\)
\(\Rightarrow 7(x + y) = 7 \times 3\)
\(\Rightarrow x + y = 3\) --- (3)
Step.2 : Subtracting equation (2) from equation (1).
\((4x + 3y) - (3x + 4y) = 11 - 10\)
\(\Rightarrow x - y = 1\) --- (4)
Step.3 : Adding equations (3) and (4), we get.
\((x + y) + (x - y) = 3 + 1\)
\(2x = 4\)
\(\Rightarrow x = 2\)
Substitute \(x = 2\) in equation (3):
\(2 + y = 3\)
\(\Rightarrow y = 1\)
Substituting \(y = 1\) in any of the equations (1), (2), (3) or (4), we get \(x = 2\).
The solution of the pair of equations is \(x = 2\), \(y = 1\).

(A) QUESTIONS

Solve the following linear equations by using the method of Elimination.

Question 1. \(3x + 2y = 11\)
\(2x + 3y = 4\)
Answer: Given equations:
\(3x + 2y = 11\) --- (1)
\(2x + 3y = 4\) --- (2)
Multiply equation (1) by 3 and equation (2) by 2:
\(9x + 6y = 33\) --- (3)
\(4x + 6y = 8\) --- (4)
Subtracting equation (4) from equation (3):
\((9x - 4x) + (6y - 6y) = 33 - 8\)
\(5x = 25\)
\(x = 5\)
Substitute \(x = 5\) in equation (1):
\(3(5) + 2y = 11\)
\(15 + 2y = 11\)
\(2y = 11 - 15\)
\(2y = -4\)
\(y = -2\)
Solution: \(x = 5\), \(y = -2\).
In simple words: Make the coefficients of one variable equal by multiplication, then subtract the equations to find the other variable.

Exam Tip: Always double-check your arithmetic when multiplying coefficients across the entire equation.

Question 2. \(8x + 5y = 9\)
\(3x + 2y = 4\)
Answer: Given equations:
\(8x + 5y = 9\) --- (1)
\(3x + 2y = 4\) --- (2)
Multiply equation (1) by 2 and equation (2) by 5:
\(16x + 10y = 18\) --- (3)
\(15x + 10y = 20\) --- (4)
Subtracting equation (4) from equation (3):
\(16x - 15x = 18 - 20\)
\(x = -2\)
Substitute \(x = -2\) in equation (2):
\(3(-2) + 2y = 4\)
\(-6 + 2y = 4\)
\(2y = 4 + 6\)
\(2y = 10\)
\(y = 5\)
Solution: \(x = -2\), \(y = 5\).
In simple words: Eliminate \(y\) by matching coefficients to 10, solve for \(x\), and substitute back to find \(y\).

Exam Tip: Be careful with negative signs during subtraction and substitution.

Question 3. \(\frac{x}{10} + \frac{y}{5} + 1 = 15\)
\(\frac{x}{8} + \frac{y}{6} = 15\)
Answer: Simplify the first equation:
\(\frac{x}{10} + \frac{y}{5} = 14\) (Multiply by 10)
\(x + 2y = 140\) --- (1)
Simplify the second equation:
\(\frac{x}{8} + \frac{y}{6} = 15\) (Multiply by 24, the LCM of 8 and 6)
\(3x + 4y = 360\) --- (2)
Multiply equation (1) by 2:
\(2x + 4y = 280\) --- (3)
Subtract equation (3) from equation (2):
\((3x - 2x) + (4y - 4y) = 360 - 280\)
\(x = 80\)
Substitute \(x = 80\) into equation (1):
\(80 + 2y = 140\)
\(2y = 60\)
\(y = 30\)
Solution: \(x = 80\), \(y = 30\).
In simple words: Clear the fractions first by multiplying by the LCM, then use standard elimination.

Exam Tip: Simplifying fractions into standard linear form before applying elimination prevents calculation errors.

Question 4. \(11x + 15y + 23 = 0\)
\(7x - 2y - 20 = 0\)
Answer: Rewrite the equations in standard form:
\(11x + 15y = -23\) --- (1)
\(7x - 2y = 20\) --- (2)
Multiply equation (1) by 2 and equation (2) by 15:
\(22x + 30y = -46\) --- (3)
\(105x - 30y = 300\) --- (4)
Add equation (3) and equation (4):
\((22x + 105x) + (30y - 30y) = -46 + 300\)
\(127x = 254\)
\(x = \frac{254}{127} = 2\)
Substitute \(x = 2\) in equation (2):
\(7(2) - 2y = 20\)
\(14 - 2y = 20\)
\(-2y = 20 - 14\)
\(-2y = 6\)
\(y = -3\)
Solution: \(x = 2\), \(y = -3\).
In simple words: Bring constants to the right side, align coefficients of \(y\), and add the equations since signs are opposite.

Exam Tip: When coefficients have opposite signs, adding equations is faster and less prone to sign errors than subtraction.

(B) Solve the following equations by using the method of substitution :-

Question 5. \(2x + 3y = 9\)
\(3x + 4y = 5\)
Answer: Given equations:
\(2x + 3y = 9\) --- (1)
\(3x + 4y = 5\) --- (2)
From equation (1), express \(x\) in terms of \(y\):
\(2x = 9 - 3y\)
\(x = \frac{9 - 3y}{2}\) --- (3)
Substitute \(x\) into equation (2):
\(3\left(\frac{9 - 3y}{2}\right) + 4y = 5\)
Multiply the entire equation by 2 to clear the fraction:
\(3(9 - 3y) + 8y = 10\)
\(27 - 9y + 8y = 10\)
\(27 - y = 10\)
\(-y = 10 - 27\)
\(-y = -17\)
\(y = 17\)
Substitute \(y = 17\) into equation (3):
\(x = \frac{9 - 3(17)}{2}\)
\(x = \frac{9 - 51}{2}\)
\(x = \frac{-42}{2} = -21\)
Solution: \(x = -21\), \(y = 17\).
In simple words: Isolate one variable from one equation and substitute its expression into the other equation.

Exam Tip: Multiplying by the denominator early clears fractions and makes simplification much easier.

Question 6. \(\frac{2x}{a} + \frac{y}{b} = 2\)
\(\frac{x}{a} - \frac{y}{b} = 4\)
Answer: Let \(X = \frac{x}{a}\) and \(Y = \frac{y}{b}\). The equations become:
\(2X + Y = 2\) --- (1)
\(X - Y = 4\) --- (2)
From equation (2), express \(X\) in terms of \(Y\):
\(X = Y + 4\) --- (3)
Substitute \(X\) into equation (1):
\(2(Y + 4) + Y = 2\)
\(2Y + 8 + Y = 2\)
\(3Y + 8 = 2\)
\(3Y = 2 - 8\)
\(3Y = -6\)
\(Y = -2\)
Substitute \(Y = -2\) into equation (3):
\(X = -2 + 4 = 2\)
Now, substitute back to find \(x\) and \(y\):
\(\frac{x}{a} = 2 \Rightarrow x = 2a\)
\(\frac{y}{b} = -2 \Rightarrow y = -2b\)
Solution: \(x = 2a\), \(y = -2b\).
In simple words: Substitute fractional terms with temporary variables to simplify the algebraic structure before solving.

Exam Tip: Variable substitution works exceptionally well for equations containing denominators with constants or parameters like \(a\) and \(b\).

Question 7. \(7(y + 3) - 2(x + 2) = 14\)
\(4(y - 2) + 3(x - 3) = 2\)
Answer: Expand and simplify the first equation:
\(7y + 21 - 2x - 4 = 14\)
\(-2x + 7y + 17 = 14\)
\(-2x + 7y = -3\) --- (1)
Expand and simplify the second equation:
\(4y - 8 + 3x - 9 = 2\)
\(3x + 4y - 17 = 2\)
\(3x + 4y = 19\) --- (2)
From equation (1), express \(2x\):
\(2x = 7y + 3\)
\(x = \frac{7y + 3}{2}\) --- (3)
Substitute \(x\) into equation (2):
\(3\left(\frac{7y + 3}{2}\right) + 4y = 19\)
Multiply by 2 to clear fractions:
\(3(7y + 3) + 8y = 38\)
\(21y + 9 + 8y = 38\)
\(29y + 9 = 38\)
\(29y = 29\)
\(y = 1\)
Substitute \(y = 1\) into equation (3):
\(x = \frac{7(1) + 3}{2} = \frac{10}{2} = 5\)
Solution: \(x = 5\), \(y = 1\).
In simple words: Expand all brackets and simplify into standard linear form before applying substitution.

Exam Tip: Always simplify complex linear equations completely before choosing an elimination or substitution method.

Question 8. Sum of two numbers is 35 and their difference is 13. Find the numbers. (By Adding & Subtracting Rule)
Answer: Let the two numbers be \(x\) and \(y\), where \(x > y\).
According to the question:
\(x + y = 35\) --- (1)
\(x - y = 13\) --- (2)
Using the Adding & Subtracting rule:
Add equation (1) and equation (2):
\((x + y) + (x - y) = 35 + 13\)
\(2x = 48\)
\(x = 24\)
Subtract equation (2) from equation (1):
\((x + y) - (x - y) = 35 - 13\)
\(2y = 22\)
\(y = 11\)
The two numbers are 24 and 11.
In simple words: Add both equations to find the larger number directly, and subtract them to find the smaller number.

Exam Tip: The sum and difference rule (\(x = \frac{\text{sum} + \text{diff}}{2}\) and \(y = \frac{\text{sum} - \text{diff}}{2}\)) is a shortcut for word problems of this type.

Question 9. In a two digit number, the unit's digit is twice the ten's digit. If 27 is added to the number, the digits interchange their places. Find the number. (By any method, out of 3 method)
Answer: Let the ten's digit be \(x\) and the unit's digit be \(y\).
The original two-digit number is \(10x + y\).
From the first condition, the unit's digit is twice the ten's digit:
\(y = 2x\)
\(\Rightarrow 2x - y = 0\) --- (1)
From the second condition, if 27 is added to the number, the digits interchange their places:
\((10x + y) + 27 = 10y + x\)
\(10x - x + y - 10y = -27\)
\(9x - 9y = -27\)
Divide by 9:
\(x - y = -3\) --- (2)
Subtract equation (2) from equation (1):
\((2x - y) - (x - y) = 0 - (-3)\)
\(x = 3\)
Substitute \(x = 3\) into equation (2):
\(3 - y = -3\)
\(-y = -6\)
\(y = 6\)
The ten's digit is 3 and the unit's digit is 6.
Therefore, the number is \(10(3) + 6 = 36\).
In simple words: Represent the number algebraically as \(10x + y\), set up equations based on the digits and condition, then solve.

Exam Tip: For digit problems, always write the number in expanded form \(10x + y\) where \(x\) is tens and \(y\) is units.