- 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 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

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.ax+b=0

a and b are constant

ax+by+c=0

a,b,c are constants

So ax

Similary (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≠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 |

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

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

a

a

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

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

a_{1}x+b_{1}y+c_{1}=0a _{2}x +b_{2}y+c_{2}=0Example 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_{1}x+b_{1}y+c_{1}=0a _{2}x +b_{2}y+c_{2}=0Example 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_{1}x+b_{1}y+c_{1}=0a _{2}x +b_{2}y+c_{2}=0Example 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 _{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 |

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 |

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 |

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

a

a

a

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

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 |

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.

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) |

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

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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