NCERT10th MathematicsExercise 3.4Topic: 3 Pair of Linear Equations in Two Variables
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EXERCISE 3.4
Q1. Solve the following pair of linear equations by the elimination method and the substitution method :
(i) x + y = 5 and 2x – 3y = 4 (ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 (iv) (x/2) + (2y/3) = - 1, x - (y/3) = 3
Explanation:
1) Two equations in two variables are given.
2) Make the coefficients of either x or y the same by multiplying by some non-zero
constant.
3) Eliminate the variable by subtracting/adding one equation from/to the other.
4) This method is known as the elimination method.
Solution:
(i) x + y = 5 and 2x – 3y = 4
a) Elimination method
1) Given equations are
x + y = 5 ---------------equation 1
2x - 3y = 4 ---------------equation 2
2) Here multiply equation 1 by 2 to get the coefficient of x same.
2(x + y) = 2(5)
2x + 2y = 10 ---------------equation 3
3) Subtracting equation 2 from equation 3, we get,
2x + 2y = 102x - 3y = 4
(-) (+) (-)
---------------------------
5y = 6
y = 6/5 ---------------equation 4
4) Put the value of y =6/5 from equation 4 in equation 1, we get,
x + y = 5
x + (6/5) = 5
x = 5 - (6/5)
x = (25 - 6)/5
x = (19)/5
x = 19/5 -------------------------- equation 5.
5) So, x = 19/5 and y = 6/5.
b) Substitution method
6) x + y = 5 ------------ equation 6
7) 2x - 3y = 4 ------------ equation 7
8) Simplify equation 6, and we get
x + y = 5
y = 5 - x ------------ equation 8
9) Substitute the value of y = (5 - x) from equation 8 in equation 7, we get
2x - 3y = 4
2x - 3(5 - x) = 4
2x - 15 + 3x = 4
5x - 15 = 4
5x = 15 + 45x = 19
x = 19/5 ------------ equation 9
10) Put the value of x = 19/5 from equation 9 in equation 8, and we get
y = 5 - xy = 5 - (19/5)
y = (25 - 19)/5
y = 6/5 ------------ equation 10
11) The value of x = 19/5 and the value of y = 6/5.
(ii) 3x + 4y = 10 and 2x – 2y = 2
a) Elimination method
1) Given equations are
3x + 4y = 10 ---------------equation 1
2x - 2y = 2 ---------------equation 2
2) Here multiply equation 2 by 2 to get the coefficient of y same.
2(2x - 2y) = 2(2)
4x - 4y = 4 ---------------equation 3
3) Adding equation 1 and equation 3, we get,
3x + 4y = 104x - 4y = 4
------------------------
7x = 14
x = 14/7
x = 2 ---------------equation 4
4) Put the value of x = 2 from equation 4 in equation 2, we get,
2x - 2y = 2
2(2) - 2y = 2
2y = 4 - 2
2y = 2
y = 1 -------------------------- equation 5.
5) So, x = 2 and y = 1.
b) Substitution method
6) 3x + 4y = 10 ------------ equation 6
7) 2x - 2y = 2 ------------ equation 7
8) Simplify equation 7, we get
2x - 2y = 2
x - y = 1
x = y + 1 ------------ equation 8
9) Substitute the value of x =(y + 1) from equation 8 in equation 6, we get
3x + 4y = 10
3(y + 1) + 4y = 10
3y + 3 + 4y = 10
7y + 3 = 10
7y = 10 - 37y = 7
y = 1 ------------ equation 9
10) Put the value of y = 1 from equation 9 in equation 8, we get
x = y + 1x = 1 + 1
x = 2 ------------ equation 10
11) So, x = 2 and y = 1.
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
a) Elimination method
1) Given equations are
3x - 5y - 4 = 0
3x - 5y = 4 ---------------equation 1
9x = 2y + 7
9x - 2y = 7 ---------------equation 2
2) Here multiply equation 1 by 3 to get the coefficient of x as same.
3(3x - 5y) = 3(4)
9x - 15y = 12 ---------------equation 3
3) Subtracting equation 2 from equation 3, we get,
9x - 15y = 129x - 2y = 7
(-) (+) (-)
-------------------------13 y = 5
y = - (5/13) ---------------equation 4
4) Put the value of y = - (5/13) from equation 4 in equation 1, we get,
3x - 5y = 4
3x - 5(-5/13) = 4
3x + (25/13) = 4
3x = 4 - (25/13)
3x = (52 - 25)/13
3x = (27)/13
x = (27)/(13 x 3)
x = 9/13 -------------------------- equation 5.
5) So, x = 9/13 and y = - (5/13).
b) Substitution method
6) 3x - 5y = 4 ------------ equation 6
7) 9x - 2y = 7 ------------ equation 7
8) Simplify equation 7, we get
3x - 5y = 4
3x = 5y +4
x = (5y + 4)/3 ------------ equation 8
9) Substitute the value of x = (5y + 4)/3 from equation 8 in equation 7, we get
9x - 2y = 7
9(5y + 4)/3 - 2y = 7
3(5y + 4) - 2y = 7
15y + 12 - 2y = 7
13y + 12 = 7
13y = 7 - 12
13y = - 5
y = - (5/13) ------------ equation 9
10) Put the value of y = - (5/13) from equation 9 in equation 8, we get
x = (5(-5/13) + 4)/3
x = (- 25/13) + 4)/3
x = (- 25 + 52)/(13(3))
x = (27)/(13(3))x = (9)/(13)
x = 9/13 ------------ equation 10
11) So, x = 9/13 and y = - (5/13).
(iv) (x/2) + (2y/3) = - 1, x - (y/3) = 3
a) Elimination method
1) Given equations are
(x/2) + (2y/3) = - 1 ---------------equation 1
x – (y/3) = 3 ---------------equation 2
2) Here multiply equation 1 by 2 to get the coefficient of x as same.
2(x/2) + 2(2y/3) = 2(- 1)
x + (4y/3) = - 2 ---------------equation 3
3) Subtracting equation 2 from equation 3, we get,
x + 4y/3 = - 2x - y/3 = 3
(-) (+) (-)
--------------------------5 y/3 = - 5
y/3 = - 1
y = - 3 ---------------equation 4
4) Put the value of y = - 3 from equation 4 in equation 1, we get,
(x/2) + (2y/3) = - 1
(x/2) + (2(- 3)/3) = - 1
(x/2) + (- 2) = - 1
(x/2) = - 1 + 2
(x/2) = 1
x = 2 -------------------------- equation 5.
5) So, x = 2 and y = - 3.
b) Substitution method
6) 3x - 5y = 4 ------------ equation 6
7) 9x - 2y = 7 ------------ equation 7
8) Simplify equation 7, we get
3x - 5y = 4
3x = 5y +4
x = (5y + 4)/3 ------------ equation 8
9) Substitute the value of x = (5y + 4)/3 from equation 8 in equation 7, we get
9x - 2y = 7
9(5y + 4)/3 - 2y = 7
3(5y + 4) - 2y = 7
15y + 12 - 2y = 7
13y + 12 = 7
13y = 7 - 12
13y = - 5
y = - (5/13) ------------ equation 9
10) Put the value of y = - (5/13) from equation 9 in equation 8, we get
x = (5(-5/13) + 4)/3
x = (- 25/13) + 4)/3
x = (- 25 + 52)/(13(3))
x = (27)/(13(3))x = (9)/(13)
x = 9/13 ------------ equation 10
11) So, x = 9/13 and y = - (5/13).
Q2. 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 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 Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 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 Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Explanation:
1) Two equations in two variables are given.
2) Make the coefficients of either x or y the same by multiplying by some non-zero
constant.
3) Eliminate the variable by subtracting/adding one equation from/to the other.
4) This method is known as the elimination method.
Solution:
(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?
1) Let the numerator be x and the denominator be y.
2) According to the first condition, we have,
(x + 1)/(y - 1) = 1
(x + 1) = (y - 1)
x - y = - 2 ------------ equation 1
3) According to the second condition, we have,
(x)/(y + 1) = 1/2
2x = y + 1
2x - y = 1 ------------ equation 2
4) Subtract equation 1 from equation 2, and we get,
2x - y = 1
x - y = - 2
(-) (+) (-)
x = 3
x = 3 ---------------equation 3
5) Put the value of x = 3 from equation 3 in equation 1, we get,
x - y = - 2
3 - y = - 2
y = 3 + 2
y = 5 -------------------------- equation 4.
6) So, here, the numerator is 3 and the denominator is 5. So the fraction is 3/5.
(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?
1) Let Nuri's present age be x and Sonu's present age be y.
2) 5 years ago their ages were (x - 5) and (y - 5).
3) According to the first relation given in the problem,
(x - 5) = 3(y - 5)
(x - 5) = 3y - 15
x - 3y = - 15 + 5
x - 3y = - 10 ---------------------- equation 1
4) 10 years later, their ages will be (x + 10) and (y + 10).
5) According to the relation given in the problem,
(x + 10) = 2(y + 10)
(x + 10) = 2y + 20
x - 2y = 20 - 10
x - 2y = 10 ---------------------- equation 2
6) Subtract equation 1 from equation 2, and we get,
x - 2y = 10
x - 3y = -10
(-) (+) (+)
----------------------------
y = 20
y = 20 ---------------equation 3
7) Put the value of y = 20 from equation 3 in equation 2, we get,
x - 2y = 10
x - 2(20) = 10
x = 10 + 40
x = 50 -------------------------- equation 4.
8) So Nuri's present age is 50 years and Sonu's present age is 20 years.
(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.
1) Let the ones place digit be x and the tens place digit be y.
2) According to the first condition, we have,
x + y = 9
x + y = 9 ------------ equation 1
3) The original number: 10 x + y
4) The number obtained by reversing the digits: 10 y + x
5) According to the second condition, we have,
9(10 x + y) = 2(10 y + x)
90 x + 9 y = 20 y + 2 x
90 x - 2 x + 9 y - 20 y = 0
88 x - 11 y = 0
11(8 x - y) = 0
8 x - y = 0 ------------ equation 2
4) Add equation 1 to equation 2, and we get,
8 x - y = 0
x + y = 9
----------------------------
9x = 9
x = 1 ---------------equation 3
5) Put the value of x = 1 from equation 3 in equation 1, we get,
x + y = 9
1 + y = 9
y = 9 - 1
y = 8 -------------------------- equation 4.
6) So, the unit-placed digit is 1 and the tens-placed digit is 8. So the number is 18.
(iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
1) Let the number of Rs 50 be x and the number of Rs 100 be y.
2) According to the first condition, we have,
x + y = 25
x + y = 25 ------------ equation 1
3) According to the second condition, we have,
50 x + 100 y = 2000
50 (x + 2 y) = 50 (40)
x + 2 y = 40
x + 2 y = 40 ------------ equation 2
4) Subtract equation 1 from equation 2, and we get,
x + 2 y = 40
x + y = 25
(-) (-) (-)
----------------------------
y = 15
y = 15 ---------------equation 3
5) Put the value of y = 15 from equation 3 in equation 1, we get,
x + y = 25
x + 15 = 25
x = 25 - 15
x = 10 -------------------------- equation 4.
6) So, the number of Rs 50 notes is 10 and the number of Rs 100 notes is 15.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for a book she kept for five days. Find the fixed charge and the charge for each extra day.
1) Let the fixed charges for the first 3 days be Rs x.
2) Let the additional charges per extra day be Rs y.
3) Saritha paid Rs 27 for 7 days, so she paid Rs x for the first 3 days and the remaining for
4 days, so we have,
x + 4y = 27
x + 4y = 27 ------------ equation 1
4) Susy paid Rs 21 for 5 days, so she paid Rs x for the first 3 days and the remaining for
2 days, so we have,
x + 2y = 21
x + 2y = 21 ------------ equation 2
5) Subtract equation 2 from equation 1, and we get,
x + 4 y = 27
x + 2 y = 21
(-) (-) (-)
----------------------------
2 y = 6
y = 3 ---------------equation 3
5) Put the value of y = 3 from equation 3 in equation 2, we get,
x + 2 y = 21
x + 2(3) = 21
x + 6 = 21
x = 21 - 6
x = 15 -------------------------- equation 4.
6) So, the fixed charges for the first 3 days are Rs 15 and the additional charges per extra day are Rs 3.
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