- 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

$ax+b=0$

a and b are constant

$ax+by+c=0$

a,b,c are constants

So, $ax^2 +b=0$ is not a linear equation

Similary $ \frac {a}{x}+b=0$ is not a linear equation

Linear equations are straight lines when plotted on Cartesian plane

S.no |
Type of equation |
Mathematical representation |
Solutions |

1 |
Linear equation in one Variable |
$ax+b=0 \; , \; a \neq 0 $ a and b are real number |
One solution |

2 |
Linear equation in two Variables |
$ax+by+c=0 \; , \; a \neq 0 \; and \; b \neq 0$ a, b and c are real number |
Infinite solution possible |

3 |
Linear equation in three Variables |
$ax+by+cz+d=0 \; , \; a \neq 0 \;, \; b \neq 0 \; and \; c \neq 0$ a, b, c, d are real number |
Infinite solution possible |

- Linear equation in two variables is represented by straight line the Cartesian plane.
- Every point on the line is the solution of the equation.
- Infact Linear equation in one variable can also be represented on Cartesian plane, it will be a straight line either parallel to x –axis or y –axis

$x-2=0$ , (straight line parallel to y axis). It means ( 2,<any value on y axis ) will satisfy this line

$y-2=0$, ( straight line parallel to x axis ). It means ( <any value on x-axis ),2 ) will satisfy this line

- Suppose the equation given is

$ax+by+c=0 \; , \; a \neq 0 \; and \; b \neq 0$ - Find the value of y for x=0

$y=\frac {-c}{b}$

This point will lie on Y –axis. And the coordinates will be $(0,\frac{-c}{b})$ - Find the value of x for y=0

$x=\frac {-c}{a}$

This point will lie on X –axis. And the coordinates will be $(\frac {-c}{a}, 0)$ - Now we can draw the line joining these two points

$a_1x+b_1y+c_1=0$

$a_2x+b_2y+c_2=0$

Graphically it is represented by two straight lines on Cartesian plane.

Simultaneous pair of Linear equation |
Condition |
Graphical representation |
Algebraic interpretation |

$a_1x+b_1y+c_1=0$ $a_2x+b_2y+c_2=0$ Example $x-4y+14=0$ $3x+2y-14=0$ |
$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$ |
Intersecting lines. The intersecting point coordinate is the only solution |
One unique solution only. |

$a_1x+b_1y+c_1=0$ $a_2x+b_2y+c_2=0$ Example $2x+4y=16$ $3x+6y=24$ |
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ |
Coincident lines. The any coordinate on the line is the solution. |
Infinite solution. |

$a_1x+b_1y+c_1=0$ $a_2x+b_2y+c_2=0$ Example $2x+4y=6$ $4x+8y=18$ |
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$ |
Parallel Lines |
No solution |

The graphical solution can be obtained by drawing the lines on the Cartesian plane.

S.no |
Type of method |
Working of method |

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 |

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 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 |

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 |

- First frame the two linear equation in two variable
- We can solve them either using elimination , substitution or cross multiplication method as explained above
- We can also solve them in graphical manner. We need ton draw the lines on the graph using the technique given above and then finding the intersection
- Always validate the solutions with both the equations at the end

$a_1x + b_1y+c_1z+d_1=0$

$a_2x + b_2y+c_2z+d_2=0$

$a_3x + b_3y+c_3z+d_3=0$

- Find the value of variable z in term of x and y in First equation
- Substitute the value of z in Second and third equation.
- Now the equation obtained from 2 and 3 are linear equation in two variables. Solve them with any algebraic method
- Substitute the value x and y in equation first and get the value of variable z

$3(x+3)=2(x+1)$

$3x+9=2x+2$

or x+7=0

or x=-7

$x+y=6$

$2x+y=12$

We will go with elimination method

Step 1 ) Choose one equation

$x+y=6$

$x=6-y$

Step 2) Substitute this value of x in second equation

$2x+y=12$

$2(6-y)+y=12$

$12-2y+y=12$

or y=0

Step 3) Substitute of value of y in any of these equation to find the value of x

$x+y=6$

x=6

So x=6 and y=0 Satisfy both the equations

$x+2y =10 \; , \; 4x+8y=40$

As per algebraic condition

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 |

- Read the problem carefully and note what is given and what is required and what is given. If it is not clear in first time,read it again.
- Denote the unknown by the variables as x, y, …….
- Translate the problem to the language of mathematics or mathematical statements.
- Form the linear equation in two variable using the conditions given in the problems.
- Solve the equation for the unknown either by substitution, elimination or cross multiplication methods
- Verify to be sure whether the answer satisfies the conditions of the problem.

Problem type |
Steps to be followed |

Problem based on Distance, speed and time. |
We need to remember the Speed formula in these problem to derive the correct mathematical statement Speed = Distance /Time Or Distance = Speed X time Or Time = Distance /Speed |

Problem Based on moving upstream and downstream in the river |
We need to generally calculate the speed of Boat in still water and Speed of the stream. Let’s us assume them as x and y, then For downstream movement, speed of the boat would = x+y For upstream movement, speed of the boat would =x-y |

Problem based on Geometry and Mensuration |
We need to remember the Angle sum property for Triangle, quadrilaterals, cyclic quadrilateral, parallelogram For Triangle, Sum of angles =180 ^{0}For quadrilateral, Sum of angles=360 ^{0}Similarly In case of parallelogram which is a special case of quadrilateral, the opposite angles are equal In case of cyclic quadrilaterals, the opposite angles are supplementary |

Problem based on Numbers and digits. |
A two-digit number (xy y -in unit place and x in tenth place) can be expressed as 10x+y If the number are interchanged, it will be expressed as $10y+x$ So you can formulate the mathematical expression in the unknown using the above formula For three-digit number(xyz), the expression would $(100x+10y+z)$ |

Problem based on Work Rate |
The standard equation that will be needed for these problems is Portion of Job done in given time= (Work rate of the person) X (Time taken working) |

Age Problems |
It is good to take present age of the person as x and y. Then k year ago, there age would be (x-k) and (y-k) respectively Similarly, k year after, their age would be (x+k) and (y+k) respectively So you can formulate the mathematical expression based on this and then solve the problem |

Fractions problem |
For Fraction problem, if the numerator is x and denominator is y, then the fraction is x/y. We can use the above expression to formulate the mathematical expressions |

Mixing solution problems |
In this problem type, we will be looking at mixing solutions of different percentages to get a new percentage. The solution will consist of a secondary liquid (acid or alcohol) mixed in with water. Amount of secondary liquid will be given by Amount of secondary liquid= (percentage of secondary liquid) X (Volume of the solution) |

Commercial mathematics problems |
Commercial mathematics problems include interest rate, cost price, selling price problem. Selling price or Marked price = Cost price + Profit So x be the selling price and y be the cost price Profit = y-x In case selling at loss, loss would be =x-y Similarly, simple interest problem could be formulated Simple interest = (% Rate of interest) X (number of year) X (Principle amount) |

Let present age of Asha be x years and present age of Rita be y years

Therefore, x = y + 5

or x – y = 5 ...(1)

5 years ago, Asha was x – 5 years and Rita was (y – 5) years old.

Therefore, x – 5 = 2(y – 5)

or x – 2y = – 5 ...(2)

Solving (1) and (2), we get y = 10 and x = 15

Let the length of garden = x m

and width of garden = y m

Therefore x = y + 4 ...(1)

Also, perimeter is 20 m, therefore

2(x + y) = 20

or x + y = 10 ...(2)

Solving (1) and (2), we get x = 7, y = 3

Hence, length = 7 m and breadth = 3m

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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