Pair of Linear Equations in Two Variables deals with solving systems of equations involving two variables.
Key Concepts
- Linear Equation in Two Variables: An equation of the form ax + by + c = 0, where a, b, and c are real numbers, and a and b are not both zero.
- Pair of Linear Equations: Two linear equations in the same two variables.
- Solution of a Pair of Linear Equations: The values of the variables that satisfy both equations simultaneously.
- Graphical Method: Solving equations by plotting the graphs of both equations and finding the point of intersection.
- Algebraic Methods:
- Substitution Method: Expressing one variable in terms of the other and substituting it in the second equation.
- Elimination Method: Eliminating one variable by adding or subtracting the equations.
- Cross-multiplication Method: A formula-based method for solving equations.
- Consistent and Inconsistent Systems: A pair of linear equations is consistent if it has at least one solution, and inconsistent if it has no solution.
Applications
Linear equations are widely used in real-life problems involving two unknown quantities. For example, problems related to age, speed, distance, cost, number systems, geometry, etc. can be solved using these equations.
In essence, this chapter provides a comprehensive understanding of how to represent and solve problems involving two unknown quantities using algebraic and graphical methods.
Exercise 3.1
1. Form the pair of linear equations of the following problems and find their solutions graphically:
(i) 10 students of class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.
Ans :
2. On comparing the ratios a1/a2 , b1/b2 and c1/c2 , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 0, 7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0, 2x -y + 9 = 0
Ans :
(i) 5x – 4y + 8 = 0, 7x + 6y – 9 = 0
- a₁/a₂ = 5/7
- b₁/b₂ = -4/6 = -2/3 Since a₁/a₂ ≠ b₁/b₂,
- the lines intersect at a point.
(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0
- a₁/a₂ = 9/18 = 1/2
- b₁/b₂ = 3/6 = 1/2
- c₁/c₂ = 12/24 = 1/2 Since a₁/a₂ = b₁/b₂ = c₁/c₂, the lines are coincident.
(iii) 6x – 3y + 10 = 0, 2x – y + 9 = 0
- a₁/a₂ = 6/2 = 3
- b₁/b₂ = -3/-1 = 3
- c₁/c₂ = 10/9 Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel.
3. On comparing the ratios a1/a2 , b1/b2 and c1/c2 , find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5 ; 2x – 3y = 7
(ii) 2x – 3y = 8 ; 4x – 6y = 9
(iii) 3/2 *x + 5/3* y= 7; 9x – 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v) 4/3 x 2y ;2x + 3y = 12
Ans :
(i) 3x + 2y = 5, 2x – 3y = 7
- a₁/a₂ = 3/2
- b₁/b₂ = 2/-3 Since a₁/a₂ ≠ b₁/b₂,
- the system is consistent with a unique solution.
(ii) 2x – 3y = 8, 4x – 6y = 9
- a₁/a₂ = 2/4 = 1/2
- b₁/b₂ = -3/-6 = 1/2
- c₁/c₂ = 8/9 Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the system is inconsistent with no solution.
(iii) (3/2)x + (5/3)y = 7, 9x – 10y = 14
- a₁/a₂ = (3/2)/9 = 1/6
- b₁/b₂ = (5/3)/(-10) = -1/6 Since a₁/a₂ ≠ b₁/b₂,
- the system is consistent.
(iv) 5x – 3y = 11, -10x + 6y = -22
- a₁/a₂ = 5/-10 = -1/2
- b₁/b₂ = -3/6 = -1/2
- c₁/c₂ = 11/-22 = -1/2 Since a₁/a₂ = b₁/b₂ = c₁/c₂, the system is consistent with infinitely many solutions.
(v) (4/3)x + 2y = 8, 2x + 3y = 12
- a₁/a₂ = (4/3)/2 = 2/3
- b₁/b₂ = 2/3
- c₁/c₂ = 8/12 = 2/3 Since a₁/a₂ = b₁/b₂ = c₁/c₂, the system is consistent with infinitely many solutions.
4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.
(i) x + y = 5, 2x + 2y = 10
(ii) x-y – 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Ans :
5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width is 36 m. Find the dimensions of the garden graphically.
Ans :
6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Ans :
7. Draw the, graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Ans :
Exercise 3.2
1. Solve the following pairs of linear equations by the substitution method:
Ans :
2. Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of’m’ for which y = mx +3.
Ans :
We have the system:
- 2x + 3y = 11 –(1)
- 2x – 4y = -24 –(2)
Let’s use the elimination method. Subtracting equation (2) from equation (1),
- 7y = 35
- y = 5
Substituting y = 5 into equation (1):
- 2x + 3(5) = 11
- 2x + 15 = 11
- 2x = -4
- x = -2
So, the solution to the system is x = -2 and y = 5.
Step 2: Find the value of m
We are given the equation y = mx + 3.
We know that x = -2 and y = 5. Substituting these values, we get:
- 5 = m(-2) + 3
- 2 = -2m
- m = -1
Therefore, the value of m is -1.
3. Form the pair of linear equations for the following problems and find their solution by substitution method:
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charges per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes 5/6. Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five year ago, Jacob’s age was seven times that of his son. What are their present ages?
Ans :
Problem (i)
y – x = 26 (Equation 1)
y = 3x (Equation 2)
Substituting y = 3x in equation (1),
we get:
3x – x = 26
2x = 26
x = 13
Substituting x = 13 in equation (2), we get:
y = 3 * 13
y = 39
So, the smaller number is 13 and the larger number is 39.
Problem (ii)
x + y = 180 (Supplementary angles)
y = x + 18 (Larger angle exceeds smaller by 18)
Substituting y = x + 18 in equation (1), we get:
x + (x + 18) = 180
2x + 18 = 180
2x = 162
x = 81
Substituting x = 81 in equation (2), we get:
y = 81 + 18
y = 99
Problem (iii)
Let the cost of one bat be x and the cost of one ball be y.
7x + 6y = 3800 (Equation 1)
3x + 5y = 1750 (Equation 2)
Multiplying equation (1) by 3 and equation (2) by 7,
we get:
21x + 18y = 11400
21x + 35y = 12250
Subtracting equation (1) from equation (2),
we get:
17y = 850
y = 50
Substituting y = 50 in equation (1),
we get:
7x + 6 * 50 = 3800
7x + 300 = 3800
7x = 3500
x = 500
Problem (iv)
Let the fixed charges be x and the charges per km be y.
x + 10y = 105 (Equation 1)
x + 15y = 155 (Equation 2)
Subtracting equation (1) from equation (2),
we get:
5y = 50
y = 10
Substituting y = 10 in equation (1),
we get:
x + 10 * 10 = 105
x + 100 = 105
x = 5
So, the fixed charges are ₹5 and the charges per km are ₹10.
For travelling a distance of 25 km, the charge would be:
Fixed charges + (Charges per km * Distance)
5 + (10 * 25) = 5 + 250 = ₹255
Problem (v)
(x + 2)/(y + 2) = 9/11 (Equation 1)
(x + 3)/(y + 3) = 5/6 (Equation 2)
From equation (1), we get:
11(x + 2) = 9(y + 2)
11x + 22 = 9y + 18
11x – 9y = -4 (Equation 3)
From equation (2), we get:
6(x + 3) = 5(y + 3)
6x + 18 = 5y + 15
6x – 5y = -3 (Equation 4)
Multiplying equation (3) by 5 and equation (4) by 9, we get:
55x – 45y = -20
54x – 45y = -27
Subtracting equation (2) from equation (1),
we get:
x = 7
Substituting x = 7 in equation (3),
we get:
11 * 7 – 9y = -4
77 – 9y = -4
9y = 81
y = 9
So, the fraction is 7/9.
Problem (vi)
Let Jacob’s present age be x and his son’s present age be y.
(x + 5) = 3(y + 5) (Five years hence)
x – 5 = 7(y – 5) (Five years ago)
From equation (1), we get:
x + 5 = 3y + 15
x – 3y = 10 (Equation 3)
From equation (2), we get:
x – 5 = 7y – 35
x – 7y = -30 (Equation 4)
Subtracting equation (4) from equation (3),
we get:
4y = 40
y = 10
Substituting y = 10 in equation (3),
we get:
x – 3 * 10 = 10
x – 30 = 10
x = 40
Exercise 3.3
1. Solve the following pairs of linear equations by the elimination method and the substitution method:
Ans :
(i) x + y = 5 and 2x – 3y = 4
Elimination Method:
Multiply the first equation by 3: 3x + 3y = 15
Add the equations: 5x = 19
Solve for x: x = 19/5
Substitute x into the first equation: 19/5 + y = 5
Solve for y: y = 6/5
Substitution Method:
Solve the first equation for y:
y = 5 – x
Substitute y in the second equation: 2x – 3(5 – x) = 4
Solve for x: x = 19/5
Substitute x into the first equation: 19/5 + y = 5
Solve for y: y = 6/5
Solution: x = 19/5, y = 6/5
(ii) 3x + 4y = 10 and 2x – 2y = 2
Elimination Method:
Multiply the second equation by 2: 4x – 4y = 4
Add the equations: 7x = 14
Solve for x: x = 2
Substitute x into the first equation: 3(2) + 4y = 10
Solve for y: y = 1
Substitution Method:
Solve the second equation for x: x = y + 1
Substitute x in the first equation: 3(y + 1) + 4y = 10
Solve for y: y = 1
Substitute y into the first equation: 3x + 4(1) = 10
Solve for x: x = 2
Solution: x = 2, y = 1
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
Elimination Method:
Rewrite the second equation: 9x – 2y = 7
Multiply the first equation by 3: 9x – 15y = 12
Subtract the second equation from the first: -13y = 5
Solve for y: y = -5/13
Substitute y into the first equation: 3x – 5(-5/13) – 4 = 0
Solve for x: x = 9/13
Substitution Method:
Solve the second equation for x: x = (2y + 7)/9
Substitute x in the first equation: 3(2y + 7)/9 – 5y – 4 = 0
Solve for y: y = -5/13
Substitute y into the second equation: 9x = 2(-5/13) + 7
Solve for x: x = 9/13
Solution: x = 9/13, y = -5/13
(iv) x/2 + 2y/3 = -1 and x – y/3 = 3
Elimination Method:
Multiply the first equation by 2: x + 4y/3 = -2
Subtract the second equation from the first: 7y/3 = -5
Solve for y: y = -15/7
Substitute y into the first equation: x/2 + 2(-15/7)/3 = -1
Solve for x: x = 2
Substitution Method:
Solve the second equation for x: x = y/3 + 3
Substitute x in the first equation: (y/3 + 3)/2 + 2y/3 = -1
Solve for y: y = -15/7
Substitute y into the second equation: x – (-15/7)/3 = 3
Solve for x: x = 2
Solution: x = 2, y = -15/7
2. Form the pair of linear equations for the following problems and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2, if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Ans :
