Exercise 10.3.3

1. Solve the following pair of linear equations by the elimination method and the substitution method.

(i) x + y = 5 and 2x – 3y = 4

Elimination Method:

Given Equation : 1.x + y = 5 , 2. 2x - 3y = 4.

To eliminate one of the variables, multiply equation (1) by 2 to align the coefficients of x with equation (2)

Multiply equation (1) by 2: ⇒ 2(x+y)=2⋅5 ⇒ 2x+2y=10

Now, write down equation (2): 2x−3y=4

Now, subtract equation (1) from equation (2): ⇒ (2x−3y)−(2x+2y)=4−10 ⇒ 2x−3y−2x−2y=−6 ⇒ −5y=−6.

Divide both sides by -5 to solve for y : y = ⇒ y = .

Now substitute y=1.2 back into equation (1) to find x: x + 1.2 = 5 ⇒ x = 3.8.

Substitution Method:

From equation (1), solve for x : x=5−y

Substitute x=5−y into equation (2): 25−y−3y = 4 ⇒ 10−5y=4.

Subtract 10 from both sides: −5y = −6.

Divide both sides by -5 to solve for y = ⇒ y = .

Now substitute y=1.2 back into x=5−y to find x: x = 5 - 1.2 ⇒ x = .

(ii)x2+2y3 = -1 and x-y3 = 3

Elimination Method:

Given Equation : x2+2y3 = -1 and x-y3 = 3

Multiply equation (2) by 3 to eliminate fractions:3(x− y3) = 3⋅3 ⇒ 3x−y=9

Now, equation (1) remains the same: x2 + 2y3 = -1.

Now, multiply equation (1) by 3 to align the coefficients of y: 3(x2 + 2y3) = 3(-1) ⇒ 3x2 + 2y = -3.

Multiply Equation 2 by 2 to eliminate fractions: 3x+4y=−6.

Now, subtract Equation 3 from Equation 4: (3x+4y)−(3x−y)=−6−9 ⇒ 5y=−15.

Divide both sides by 5 to solve for y = 155 ⇒ y = .

Now substitute y=−3 back into Equation 3 to find x: 3x−(−3)=9 ⇒ x = 2.

Substitution Method:

From equation (2), solve for x: x = 3 + y3.

Substitute x = 3 + y3 into equation (1): 3+y32 + 2y3= -1.

Simplify and solve for y: 36 + y6+ 2y3 = -1.

Multiply through by 6 to eliminate fractions: 9+y+4y=−6 ⇒ 5y=−15.

Divide both sides by 5 to solve for y = −155 ⇒ y = .

Now substitute y=−3 back into x=3+y3to find x = 3 + −33 ⇒ x = .

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 12 if we only add 1 to the denominator. What is the fraction?

If we add 1 to the numerator and subtract 1 from the denominator, the fraction reduces to 1:

x+1y−1 = 1 ⇒ x - y = -2.

It becomes 12if we only add 1 to the denominator:

xy+1 = 12 ⇒ 2x= y + 1.

Now, solve these equations using the elimination method: From Equation 2:y=2x−1

Substitute y=2x−1 into Equation 1: x−(2x−1)=−2 ⇒ x = .

Now, substitute x=3 back into Equation 2 to find y:

y=2(3)−1 ⇒ y = .

So,the fraction is 35.

(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?

Five years ago, Nuri was thrice as old as Sonu: x−5=3(y−5) ⇒ x−3y=−10.

Ten years later, Nuri will be twice as old as Sonu:x+10=2(y+10) ⇒ x−2y=10.

Now, solve these equations using the elimination method:

Subtract Equation 2 from Equation 1: (x−3y)−(x−2y)=−10−10 ⇒ y = .

Now, substitute y=20 back into Equation 2 to find x:

x−2(20)=10 ⇒ x−40=10 ⇒ x=50.

So, 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.

The sum of the digits is 9:x+y=9

Nine times this number is twice the number obtained by reversing the digits:9(10x+y)=2(10y+x) ⇒ 8x−y=0.

Now, solve these equations using the elimination method:

From Equation 2:y=8x

Substitute y=8x into Equation 1:x+8x=9 ⇒ x =

Now, substitute x=1 back into Equation 2 to find y=8(1) y=8

So, the number is : 10x+y = 10(1)+8 = .

(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.

The total number of notes is 25:x+y=25

The total amount is ₹2000:50x+100y=2000 ⇒ x+2y=40.

Now, solve these equations using the elimination method:

Subtract Equation 1 from Equation 2:(x+2y)−(x+y)=40−25 x+2y−x−y=15 y=15

Now, substitute y=15 back into Equation 1 to find x:x+15=25 ⇒ x=10.

So, Meena 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.

Saritha paid ₹27 for a book kept for seven days:x+4y=27

Susy paid ₹21 for the book she kept for five days:x+2y=21

Now, solve these equations using the elimination method:

Subtract Equation 2 from Equation 1:(x+4y)−(x+2y)=27−21 ⇒ x+4y−x−2y=6 ⇒2y=6 ⇒y=.

Now, substitute y=3 back into Equation 2 to find x:x+2(3)=21 ⇒x+6=21 ⇒x=.

So, the fixed charge is ₹15 and the charge for each extra day is ₹3

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Exercise 10.3.3