- What is Linear equations
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- Linear equations Solutions
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- Graphical Representation of Linear equation in one and two variable
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- Steps to Draw the Given line on Cartesian plane
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- Simultaneous pair of Linear equation
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- Algebraic Solution of system of Linear equation
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- Simultaneous pair of Linear equation in Three Variable
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- Steps to solve the Linear equations

- NCERT Solutions Exercise 3.1
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- NCERT Solutions Exercise 3.2
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- NCERT Solutions Exercise 3.3,3.4,3.5
- NCERT Solutions Exercise 3.6
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- NCERT Solutions Exercise 3.7

- Problem and Solutions
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- Linear equations Problems
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- Linear equation worksheet
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- Linear equation word problems
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- Linear equations graphical problems

Given below are the **Class 10 Maths** Important Questions for Linear equation

a. Concepts questions

b. Calculation problems

c. Multiple choice questions

d. Long answer questions

e. Fill in the blank's

a. Concepts questions

b. Calculation problems

c. Multiple choice questions

d. Long answer questions

e. Fill in the blank's

Which of these equation have i) Unique solution ii) Infinite solutions iii) no solutions

- $152x-378y=-74$ , $-378x +152y=-604$
- $x+2y =10$ , $3x+6y=30$
- $3x+4y=6$ , $12x+16x=30$
- $7x-11y=53$ ,$19x -17y=456$
- $x=7$,$y=-2$
- $ax+by=a-b$ , $bx-ay=a+b$
- $2x+3y=0$, $124x+13y=0$
- $y=11$ ,$y =-11$

Condition |
Algebraic interpretation |

$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$ |
One unique solution only. |

$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ |
Infinite solution. |

$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$ |
No solution |

Infinite solution :( b)

No solution: (c), (h)

Using substitution method solve the below equation

a. $x-2y+300=0$, $6x-y-70=0$

b. $5x-y=5, 3x-y=3$

1 |
Method of elimination by substitution |
1. Suppose the equation are $a_1x+b_1y+ c_1=0$ $a_2x +b_2y+c_2=0$ 2. Find the value of variable of either x or y in other variable term in first equation 3. Substitute the value of that variable in second equation 4. Now this is a linear equation in one variable. Find the value of the variable 5. Substitute this value in first equation and get the second variable |

$x=2y-300$

Substituting this in second equation

$6(2y-300)- y-70=0$ => $y=170$

Putting this Ist equation

$x=40$

b. Similarly,this can be solved. Answer is ( 1,0)

Using elimination method ,solve the following

a. $x+y-40=0$ ,$7x+3y=180$

b. $x+10y =68$ , $x+15y=98$

Solution

2 |
Method of elimination by equating the coefficients |
1. Suppose the equation are $a_1x+b_1y+ c_1=0$ $a_2x +b_2y+c_2=0$ 2. Find the LCM of $a_1$ and $a_2$ .Let it k. 3. Multiple the first equation by the value $ \frac {k}{a_1}$ 4. Multiple the first equation by the value $ \frac {k}{a_2}$ 4. Subtract the equation obtained. This way one variable will be eliminated and we can solve to get the value of variable y 5. Substitute this value in first equation and get the second variable |

$7x+3y=180$ ---(2)

Multiplying equation (1) by 7

$7x+7y-280=0$ ---(3)

Subtracting equation 2 from equation 7

$7x+7y-280=0$

$7x+3y=180$

We get

$4y=100$ => $y=25$

Substituting this in (1) ,we get $x=15$

b. ( 8,6)

Solve the below linear equation using cross-multiplication method

a. $(a-b)x + (a=b) y=a^2-2ab-b^2$ , $(a+b)(x+y)=a^2+ b^2$

b. $x+y=5$ ,$2x-3y=4$

3 |
Cross Multiplication method |
1. Suppose the equation are $a_1x+b_1y+ c_1=0$ $a_2x +b_2y+c_2=0$ 2. This can be written as 3. This can be written as 4. Value of x and y can be find using the x => first and last expression y=> second and last expression |

$(a-b)x + (a=b) y=a^2-2ab-b^2$

$(a+b)(x+y)=a^2+ b^2$

By cross multiplication we have

$x=a+b$, $y=-2ab(a+b)^{-1}$

b. (19/5,6/5)

True or False statement

a. Line $4x+5y=0$ and $11x+17y=0$ both passes through origin

b. Pair of lines $117x+14y=30$ , $65x+11y=19$ are consistent and have a unique solution

c. There are infinite solution for equation $17x+12y=30$

d. $x=0$ ,$y=0$ has one unique solution

e. Lines represented by $x-y=0$ and $x+y=0$ are perpendicular to each other

f. $2x+6y=12$ and $8x+24y=65$ are consistent pair of equation

g. $x+6y=12$ and $4x+24y=64$ are inconsistent pair of equation

- True
- True
- True
- True
- True
- False
- True

find the value of p for which the linear pair has infinite solution

$12x+14y=0$

$36x+py=0$

a. 14

b. 28

c. 56

d. 42

For infinite solution

$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}$

So $ \frac {12}{36}=\frac {14}{p}$ => p=42

There are 10 students in XII class. Some are maths and some bio student. The no of bio students are 4 more then math’s students. Find the no of math’s and bio students

a. 1,9

b. 4,6

c. 2,8

d. 3,7

Let x be math’s students

y be bio students

Then

$x+y=10$

$y=x+4$

Solving these linear pair through any method we get

x=3 and y=7

which of the below pair are consistent pair?

a. $x-3y=3$ , $3x-9y=2$

b. $51x +68y=110$ ,$3x+4y=99$

c. $2x+3y=10$ , $9x+11y=12$

d. None of these

For consistent pair

$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}$

Or

$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

Analyzing all of them ,we get c as the answer

There are two numbers. Two conditions are there for them

(i) Sum of these two numbers are 100

(ii) One number is four time another number.

What are these numbers?

a. 20,80

b. 30,70

c. 40,60

d. 25,75

Let x and y are the number

$x+y=100$

$y=4x$

Solving them we get x=20 and y=80

The number of solution of the linear pair

$x+37y=123$

$21x-41y=125$

a. No solution

b. One solution

c. infinite many

d. None of these

The sum of a 2 digit number and number obtained by reversing the order of the digits is 99. If the digits of the number differ by 3. Find the number

Rajdhani train covered the distance between Lucknow and Delhi at a uniform speed. It is observed that if rajdhani would have run slower by 10 km/hr,it would have taken 3 hours more to reach the destination and if rajdhani would have run faster by 10 km/hr,it would have taken 2 hours less. Find the distance Lucknow and Delhi?

Let x be the speed and t be the original timing ,then distance between Lucknow and Delhi

$Distance =speed \times time =xy$

Now from first observation

$xy=(x-10)(y+3)$ => $3x-10y-30=0$

From second observation

$xy=(x+10)(y-2)$ => $2x-10y+20=0$

Solving both we get

x=50km/hr

y=12hours

So distance between Lucknow and Delhi=$50 \times 12=600$ Km

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.

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