Create New Account
Create New Account

Sign in to Innings2

or

New Account     Reset Password     

    Exercise 10.3.2

Powered by Innings 2

Reset Progress

This will delete your progress and chat data for all chapters in this course, and cannot be undone!

Glossary

Select one of the keywords on the left…

6th class > > Exercise 10.3.2

Exercise 10.3.2

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

(i) x + y = 14 ,x – y = 4

From Equation 1.x = 14 - y ,substitute the value in equation.2 ⇒ (14 - y)- y = 4.
⇒ 14 - 2y = 4 ⇒ y = .
Substitute y = 5 back into the first equation to find x.
= x + y = 14 ⇒ x + = 14 ⇒ x = .

(ii) s - t = 3, s3 + t2 = 6

Solve the first equation for s : s - t = 3 : s = t + 3.
Substitute this experssion for s into the second equation ⇒ s3 + t2 = 6 ⇒ t+33 + t2 = 6.
Simplify and solve for t ⇒ t+33 + t2 = 6 ⇒ 2t+36 + 3t6 = 6 ⇒ 5t+66 = 6.
Multiply both sides by 6 ⇒ 5t + 6 = 36 ⇒ 5t = 30 ⇒ t = .
Substitute t=6 back into the first equation to find s ⇒ s = t + 3 ⇒ s = .

(iii) 3x−y=3, 9x−3y=9

Solve the first equation for y:3x−y=3 ⇒ y = 3x - 3.
Substitute this expression for y into the second equation ⇒ 9x−3y=9 ⇒ 9x−33x−3 = 9 ⇒ 9 = 9.
Analyze the result ⇒ 9 = 9.
Since the equations are dependent, there are infinitely many solutions. The solution set can be expressed in terms of x as follows: x,y = x,3x−3.
⇒ (x,y) = (x, 3x - 3).

(iv) 0.2x+0.3y=1.3, 0.4x+0.5y=2.3

`0.2x + 0.3y = 1.3 Multiply both side by 10 ⇒ (0.2x + 0.3y) x 10 = 1.3 x 10.
⇒ 2x + 3y = 13 ⇒ 2x = 13-3y ⇒ x = 13−3y2.
putting value of x in equation 2 ⇒ 0.4x+0.5y=2.3 ⇒ 0.413−3y2+0.5y = 2.3.
Multiplying both sides by 10 ⇒ +10 x 0.5y = 10 x 2.3 ⇒ 413−y2+5y = 23 ⇒ y = .
Putting y = 3 in equation 1 ⇒ 0.2x+0.3y=13 ⇒ 0.2x+0.33=13 ⇒ x = .

(v) ,.

From Equation 1 ⇒ = ⇒ x = .
Substituting x in equation 2 ⇒ ⇒ y = .
Putting the value of y in equation 1 ⇒ ⇒ x = .

(iv) 3x2-5y3= -2,x3+y2=136.

Solve x from equation 2 ⇒ x3+y2=136 ⇒ Multiply through by 6 to eliminate the fractions: 2x+3y=13.
Solve Equation 1 for x ⇒ 3x2-5y3= -2 ⇒ Multiply through by 6 to eliminate the fractions: 9x+10y=−12.
Solve equation 2 for x ⇒ 2x+3y=13 ⇒ x = 13−3y2.
Substitute x into equation 1 ⇒ 313−3y22-5y3= -2 ⇒ y = .
Substitute y=3 back unto x ⇒ x = 13−332 ⇒ x = .

2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

From both Equations,notice that 2x is common ⇒ Subtract Equation 2 from Equation 1 to eliminate x: (2x+3y)−(2x−4y)=11−(−24) ⇒ 7y=35.
Divide both sides by 7 to solve for y ⇒ y= 357 ⇒ y = .
Substitute y = 5 Back into equation 1 ⇒ 2x+3(5)=11 ⇒ x=.
So the solution of the system of equation is x=-2 and y=5 .
Find the value of m for which y = mx + 3 ⇒ 5=m(-2)+3 ⇒ m = .

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.

The difference between two numbers is 26: x−y=26

One number is three times the other: x=3y

From equation (2), substitute x=3y into equation (1):

3y-y=26 ⇒ y = .

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

The larger of two supplementary angles exceeds the smaller by 18 degrees:x−y=18

Supplementary angles sum up to 180 degrees: x+y=180

From equation (1), solve for x: x=y+18

Substitute x=y+18 into equation (2):(y+18)+y=180 ⇒ y = .

Now, substitute y=81 back into x=y+18 to find x: x = 81 + 18 ⇒ x = .

(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.

First equation from the first purchase: 7b+6c=3800

Second equation from the second purchase:3b+5c=1750

Substitute b into equation (2):33800−6c7+5c = 1750.

Multiply through by 7 to eliminate the fraction:3(3800−6c)+35c=12250

11400−18c+35c=12250

C= .

Now, substitute c=50 back into 𝑏 = 3800−6c/7.

b= 3800−6.5/7 ⇒ b = . ​

(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?

First equation from the first journey:F+10C=105

Second equation from the second journey:F+15C=155

From equation (1), solve for F=105−10C

Substitute F into equation (2): 105-10c+15c=155 ⇒ C = .

So, the fixed charge is ₹5 and the charge per km is ₹10.

F+25C=255

(v) A fraction becomes 911 , if 2 is added to both the numerator and the denominator.If, 3 is added to both the numerator and the denominator it becomes 56 Find the fraction.

From the first condition: a+2b+2 = 911.

Cross-multiply: 11a−9b = −4

From the second condition: a+3b+3 = 56

Cross-multiply: 6a−5b = −3

Multiply Equation (1) by 6 and Equation (2) by 11 to align the coefficients of a: 66a−54b=−24, 66a−55b=−33

Subtract Equation (2) from Equation (1):−54b+55b=−24+33 ⇒ b = .

Now substitute b=9 into Equation (1): a = .