Pair of Linear Equations in Two Variables — Class 10 Mathematics
Mathematics · Chapter 3
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Chapter summary
Key ideas & terms
- Consistent system
- A pair of equations that has at least one solution.
- Inconsistent system
- A pair of equations with no common solution (parallel lines).
- Substitution method
- Solving by expressing one variable in terms of the other.
- Elimination method
- Adding/subtracting equations to remove one variable.
Extra questions & answers
When does a pair of linear equations have no solution?
When
Name three algebraic methods to solve a pair of linear equations.
Substitution, elimination, and cross-multiplication.
What does a pair of linear equations in two variables represent graphically?
Two straight lines in a plane; their solution is the point (or points) common to both lines (refer to the graphs in your NCERT textbook).
Name the algebraic methods of solving a pair of linear equations.
The substitution method, the elimination method and the cross-multiplication method.
When does a pair of linear equations have a unique solution?
When
When does a pair have no solution and when infinitely many?
No solution when
Solve
Adding the equations gives
Solve
Substituting
Is the pair
Here
The sum of two numbers is 20 and their difference is 4. Frame the equations and find them.
Let the numbers be
EXERCISE 3.1
1. Form the pair of linear equations in 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.
Let the numbers of boys and girls be x and y, respectively. Then x+y=10 and y−x=4. The graphs of these equations intersect at (3,7). Therefore, 3 boys and 7 girls 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.
Let the cost of one pencil be ₹x and that of one pen be ₹y. Then 5x+7y=50 and 7x+5y=46. The graphs intersect at (3,5). Hence, one pencil costs ₹3 and one pen costs ₹5.
2. On comparing the ratios
a
2
a
1
,
b
2
b
1
and
c
2
c
1
, 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
Here,
a
2
a
1
=
7
5
and
b
2
b
1
=
6
−4
=−
3
2
. Since
a
2
a
1
=
b
2
b
1
, the lines intersect at one point.
(ii) 9x+3y+12=0
18x+6y+24=0
Here,
a
2
a
1
=
18
9
=
2
1
,
b
2
b
1
=
6
3
=
2
1
, and
c
2
c
1
=
24
12
=
2
1
. Therefore, the lines are coincident.
(iii) 6x−3y+10=0
2x−y+9=0
Here,
a
2
a
1
=
2
6
=3,
b
2
b
1
=
−1
−3
=3, but
c
2
c
1
=
9
10
. Since
a
2
a
1
=
b
2
b
1
=
c
2
c
1
, the lines are parallel.
3. On comparing the ratios
a
2
a
1
,
b
2
b
1
and
c
2
c
1
, find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x+2y=5 ; 2x−3y=7
Writing the equations in standard form gives 3x+2y−5=0 and 2x−3y−7=0. Since
2
3
=
−3
2
, the lines intersect at one point. Therefore, the pair is consistent.
(ii) 2x−3y=8 ; 4x−6y=9
The equations are 2x−3y−8=0 and 4x−6y−9=0. Here,
4
2
=
−6
−3
=
2
1
, while
−9
−8
=
9
8
. Therefore, the lines are parallel and the pair is inconsistent.
(iii)
2
3
x+
3
5
y=7 ; 9x−10y=14
In standard form, the equations are
2
3
x+
3
5
y−7=0 and 9x−10y−14=0. Since
9
2
3
=
6
1
but
−10
3
5
=−
6
1
, the lines intersect. Hence, the pair is consistent.
(iv) 5x−3y=11 ; −10x+6y=−22
The second equation is −2 times the first equation. Thus,
a
2
a
1
=
b
2
b
1
=
c
2
c
1
=−
2
1
. The lines are coincident, so the pair is consistent and has infinitely many solutions.
(v)
3
4
x+2y=8 ; 2x+3y=12
Writing the equations in standard form,
2
3
4
=
3
2
,
3
2
=
3
2
, and
−12
−8
=
3
2
. All three ratios are equal, so the lines are coincident. Therefore, the pair is consistent and has 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
The second equation is twice the first, so both equations represent the same line. The pair is consistent and dependent, with infinitely many solutions. Every point on x+y=5, such as (0,5), (1,4), or (5,0), is a solution.
(ii) x−y=8, 3x−3y=16
In standard form, the equations are x−y−8=0 and 3x−3y−16=0. Here,
3
1
=
−3
−1
=
−16
−8
. The lines are parallel, so the pair is inconsistent and has no solution.
(iii) 2x+y−6=0, 4x−2y−4=0
The lines intersect at (2,2), since 2(2)+2−6=0 and 4(2)−2(2)−4=0. Therefore, the pair is consistent and its unique solution is x=2, y=2.
(iv) 2x−2y−2=0, 4x−4y−5=0
Here,
4
2
=
−4
−2
=
2
1
, but
−5
−2
=
5
2
. Thus, the lines are parallel. The pair is inconsistent and has no solution.
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.
Let the length be l m and the width be w m. Half the perimeter gives l+w=36, and the other condition gives l−w=4. Adding the equations, 2l=40, so l=20. Then w=16. Therefore, the garden is 20 m long and 16 m wide.
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
One suitable equation is x+y−1=0. Its coefficients satisfy
1
2
=
1
3
, so it intersects 2x+3y−8=0 at one point.
(ii) parallel lines
One suitable equation is 4x+6y−10=0. Here,
4
2
=
6
3
=
2
1
, but
−10
−8
=
5
4
. Hence, the lines are parallel.
(iii) coincident lines
One suitable equation is 4x+6y−16=0, which is obtained by multiplying the given equation by 2. Therefore, both equations represent the same line.
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.
The first line meets the x-axis at (−1,0), and the second line meets it at (4,0). Solving x−y+1=0 and 3x+2y−12=0 gives x=2, y=3. Therefore, the vertices of the triangle are (−1,0), (2,3), and (4,0). The region enclosed by these three line segments is the required triangular region.
EXERCISE 3.2
1. Solve the following pair of linear equations by the substitution method.
(i) x+y=14
x−y=4
From x−y=4, x=y+4. Substituting in x+y=14 gives y+4+y=14, so 2y=10 and y=5. Hence, x=9. The solution is x=9, y=5.
(ii) s−t=3
3
s
+
2
t
=6
From s−t=3, s=t+3. Substituting gives
3
t+3
+
2
t
=6. Multiplying by 6, 2t+6+3t=36, so 5t=30 and t=6. Thus, s=9.
(iii) 3x−y=3
9x−3y=9
From the first equation, y=3x−3. Substitution in the second gives 9x−3(3x−3)=9, which simplifies to 9=9. Therefore, the equations represent the same line and have infinitely many solutions, given by y=3x−3.
(iv) 0.2x+0.3y=1.3
0.4x+0.5y=2.3
From 0.2x+0.3y=1.3, 2x+3y=13, so x=
2
13−3y
. Substituting in 4x+5y=23 gives 2(13−3y)+5y=23, so y=3. Hence, x=2.
(v)
2
x+
3
y=0
3
x−
8
y=0
From the first equation, x=−
2
3
y. Substituting in the second gives −
2
3
y−
8
y=0, which implies y=0. Therefore, x=0. The solution is (0,0).
(vi)
2
3x
−
3
5y
=−2
3
x
+
2
y
=
6
13
From the second equation, 2x+3y=13, so x=
2
13−3y
. Substituting in 9x−10y=−12 gives 9(
2
13−3y
)−10y=−12. This yields y=3, and then x=2.
2. Solve 2x+3y=11 and 2x−4y=−24 and hence find the value of ‘m’ for which y=mx+3.
Subtracting the second equation from the first gives 7y=35, so y=5. Substituting in 2x+3y=11 gives 2x+15=11, so x=−2. Now 5=m(−2)+3, giving −2m=2 and m=−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.
Let the larger number be x and the smaller number be y. Then x−y=26 and x=3y. Substitution gives 3y−y=26, so y=13 and x=39. The numbers are 39 and 13.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Let the larger angle be x
∘
and the smaller angle be y
∘
. Then x+y=180 and x−y=18. From x=y+18, substitution gives 2y+18=180, so y=81 and x=99. The angles are 99
∘
and 81
∘
.
(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.
Let one bat cost ₹x and one ball cost ₹y. Then 7x+6y=3800 and 3x+5y=1750. From the second equation, x=
3
1750−5y
. Substitution gives y=50, and hence x=500. Therefore, each bat costs ₹500 and each ball costs ₹50.
(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 charge per km? How much does a person have to pay for travelling a distance of 25 km?
Let the fixed charge be ₹x and the charge per kilometre be ₹y. Then x+10y=105 and x+15y=155. From the first equation, x=105−10y. Substitution in the second gives y=10, so x=5. Thus, the fixed charge is ₹5, the charge per kilometre is ₹10, and the fare for 25 km is ₹(5+25×10)=₹255.
(v) A fraction becomes
11
9
, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes
6
5
. Find the fraction.
Let the fraction be
y
x
. Then
y+2
x+2
=
11
9
and
y+3
x+3
=
6
5
. These give 11x−9y=−4 and 6x−5y=−3. Solving, x=7 and y=9. Therefore, the fraction is
9
7
.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Let Jacob’s present age be x years and his son’s present age be y years. Then x+5=3(y+5) and x−5=7(y−5). From the first equation, x=3y+10. Substitution in the second gives 3y+5=7y−35, so y=10 and x=40. Jacob is 40 years old and his son is 10 years old.
EXERCISE 3.3
1. Solve the following pair of linear equations by the elimination method and the substitution method :
(i) x+y=5 and 2x−3y=4
By elimination, multiply x+y=5 by 2 to obtain 2x+2y=10. Subtracting 2x−3y=4 gives 5y=6, so y=
5
6
and x=
5
19
. By substitution, x=5−y; inserting this into the second equation gives the same values. Thus, x=
5
19
and y=
5
6
.
(ii) 3x+4y=10 and 2x−2y=2
The second equation gives x−y=1, or x=y+1. Substitution in the first gives 3(y+1)+4y=10, so 7y=7, y=1, and x=2. Elimination also gives the same solution. Thus, x=2, y=1.
(iii) 3x−5y−4=0 and 9x=2y+7
The equations are 3x−5y=4 and 9x−2y=7. Multiplying the first by 3 gives 9x−15y=12. Subtracting the second equation yields −13y=5, so y=−
13
5
. Substitution gives x=
13
9
. Thus, x=
13
9
and y=−
13
5
.
(iv)
2
x
+
3
2y
=−1 and x−
3
y
=3
Multiplying the equations by 6 and 3, respectively, gives 3x+4y=−6 and 3x−y=9. Subtracting the second from the first gives 5y=−15, so y=−3. Then x−
3
−3
=3, giving x=2. Hence, x=2, y=−3.
2. Form the pair of linear equations in 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
2
1
if we only add 1 to the denominator. What is the fraction?
Let the fraction be
y
x
. Then
y−1
x+1
=1, giving x−y=−2, and
y+1
x
=
2
1
, giving 2x−y=1. Subtracting the first equation from the second gives x=3, and then y=5. Therefore, the fraction is
5
3
.
(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?
Let their present ages be x years and y years, respectively. Then x−5=3(y−5), so x−3y=−10, and x+10=2(y+10), so x−2y=10. Subtracting gives y=20, and then x=50. Nuri is 50 years old and Sonu is 20 years old.
(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.
Let the tens digit be x and the units digit be y. Then x+y=9. The number is 10x+y, and its reverse is 10y+x. Thus, 9(10x+y)=2(10y+x), which simplifies to 8x−y=0. From y=8x and x+y=9, we get x=1 and y=8. Therefore, the number is 18.
(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.
Let the numbers of ₹50 and ₹100 notes be x and y, respectively. Then x+y=25 and 50x+100y=2000, or x+2y=40. Subtracting x+y=25 gives y=15, and hence x=10. She received 10 notes of ₹50 and 15 notes of ₹100.
(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.
Let the fixed charge be ₹x and the charge for each extra day be ₹y. Saritha paid for 4 extra days, so x+4y=27. Susy paid for 2 extra days, so x+2y=21. Subtracting gives 2y=6, so y=3, and then x=15. The fixed charge is ₹15, and the charge for each extra day is ₹3.