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Class 10 Maths notes for Linear Equation


What is Linear equations


A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable. Linear equations can have one or more variables
Example
ax+b=0
a and b are constant
ax+by+c=0
a,b,c are constants
In Linear equation ,No variable in a linear equation is raised to a power greater than 1 or used as the denominator of a fraction.
So  ax2 +b=0   is not a linear equation
Similary  (a/x)+b=0  is not a linear equation

Linear equations  are  straight lines when plotted on Cartesian plane

Linear equations Solutions


S.no
Type of equation
Mathematical representation
Solutions
1
Linear equation in one Variable
ax+b=0  ,a≠0
a and b are real number
One solution
2
Linear equation in two Variable
ax+by+c=0 , a≠0 and b≠0
a, b and c are real number
Infinite solution possible
3
Linear equation in three Variable
ax+by+cz+d=0 , a≠0 ,b≠0 and c≠0
a, b, c, d  are real number
Infinite solution possible

Graphical Representation of Linear equation in one and two variable

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

Steps to Draw the Given line on Cartesian plane| how to graph the linear equations

1) Suppose the equation given is
ax+by+c=0 , a≠0 and b≠0
2) Find the value of y for x=0
y=-c/b
This point will lie on Y –axis. And the coordinates will be (0,-c/b)
3) Find the value of x for y=0
x=-c/a
This point will lie on X –axis. And the coordinates will be (-c/a, 0)
4) Now we can draw the line joining these two points

Simultaneous pair of Linear equation:

A pair of Linear equation in two variables
a1x+b1y+c1=0
a2x +b2y+c2=0
Graphically it is represented by two straight lines on Cartesian plane.
Simultaneous pair of Linear equation
Condition
Graphical representation
Algebraic interpretation
a1x+b1y+c1=0
a2x +b2y+c2=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
intersecting lines
  One unique solution only.
a1x+b1y+c1=0
a2x +b2y+c2=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.
same lines
Infinite solution.
a1x+b1y+c1=0
a2x +b2y+c2=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
parallel lines
No solution
The graphical solution can be obtained by drawing the lines on the Cartesian plane.

Algebraic Solution of system of Linear equation in two variable

S.no
Type of method
Working of method
1
Method of elimination by substitution
1) Suppose the equation are
a1x+b1y+c1=0
a2x +b2y+c2=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
a1x+b1y+c1=0
a2x +b2y+c2=0
2) Find the LCM of a1 and a2 .Let it k.
3) Multiple the first equation by the value k/a1
4) Multiple the first equation by the value k/a2
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
a1x+b1y+c1=0
a2x +b2y+c2=0
2) This can be written as
Cross multipication 3) This can be written as
Cross multipication 4) Value of x and y can be find using the
x => first and last expression
y=> second and last expression

How to solve the Linear equations in Two variables

1) First frame the two linear equation in two variable
2) We can solve them either using elimination , substitution or cross multiplication method as explained above
3) 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
4) Always validate the solutions with both the equations at the end

 Simultaneous pair of Linear equation in Three Variable

  Three Linear equation in three variables
a1x + b1y+c1z+d1=0
a2x + b2y+c2z+d2=0
a3x + b3y+c3z+d3=0

Steps to solve the Linear equations in Three variables

1) Find the value of variable z in term of x and y in First equation
2) Substitute the value of z in Second and third equation.
3) Now the equation obtained from 2 and 3 are linear equation in two variables. Solve them with any algebraic method
4) Substitute the value x and y in equation first and get the value of variable z
Solved Examples
1)   Solve the linear equation
3(x+3)=2(x+1)
Solution
3x+9=2x+2
or x+7=0
or x=-7
2)  Solve the Simultanous linear pair of equations
x+y=6
2x+y=12
Solution
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
3) Solve the equations
x+2y =10 , 4x+8y=40
Solution
As per algebric 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
So here infinite solutions are possible
Steps involved in solving a linear equation word problem:
1) 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.
2) Denote the unknown by the variables as x, y, …….
3) Translate the problem to the language of mathematics or mathematical statements.
4) Form the linear equation in two variable using the conditions given in the problems.
5) Solve the equation for the unknown either by sunstitution, elimination or cross multiplication methods
6) Verify to be sure whether the answer satisfies the conditions of the problem.
Here are some of the Tips for specific word problems type
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 =1800
For quadrilateral, Sum of angles=3600
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)
Example
1) Asha is five years older than Rita. Five years ago, Asha was twice as old as Rita was then. Find their present ages.
Solution: 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
2) The perimeter of a rectangular garden is 20 m. If the length is 4 m more than the breadth, find the length and breadth of the garden
Solution:
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

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