- 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

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- 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

a) Concepts questions

b) Calculation problems

c) Multiple choice questions

d) Long answer questions

e) Fill in the blank's

- 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)

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 _{1}x+b_{1}y+c_{1}=0a _{2}x +b_{2}y+c_{2}=02) 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) ( 1,0)

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 _{1}x+b_{1}y+c_{1}=0a _{2}x +b_{2}y+c_{2}=02) Find the LCM of a _{1} and a_{2} .Let it k.3) Multiple the first equation by the value k/a _{1}4) Multiple the first equation by the value 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)

a) (a-b)x + (a=b) y=a

b) x+y=5 ,2x-3y=4

3 |
Cross Multiplication method |
1) Suppose the equation are a _{1}x+b_{1}y+c_{1}=0a _{2}x +b_{2}y+c_{2}=02) 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

(a+b)x+(a+b)y-(a

By cross multiplication we have

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

b) (19/5,6/5)

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

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 12/36=14/p => p=42

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

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

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

x+37y=123

21x-41y=125

a) No solution

b) One solution

c) infinite many

d) None of these

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

Distance =speed X 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=50X12=600 Km

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