Introduction:
We have already discussed the steps for solving linear equations of one variable and two variables. Let us discuss the various methods of representing a linear equation in two variables.Writing linear equations in standard form:
We are very much aware that the graph of the linear equation of two variables is a straight line on the two dimensional graph.
The standard form is very much useful to find the x and y intercepts of the graph of the equation.
Let us discuss an example:
Write the standard form of the linear equation, 2x + 3y = 10 = 0. Hence find the x and y intercepts.
Solution:
We have 2x + 3y 12 = 12To write this in the standard form, ax + by = c, let us eliminate the constant term 10 from the left hand side.
2x + 3y  12 + 12 = 0 + 12
=> 2x + 3y = 12
To find the xintercept : The xintercept is the point where the line meets the xaxis. (i.e) on the xaxis, y = 0
Therefore substituting y = 0 in the above standard form, we get2x + 3 (0) = 12
=> 2x + 0 = 12
=> 2x = 12
=> x = 12/2 = 6
Therefore the coordinate of xintercept is ( 6,0)
To find the yintercept : The yintercept is the point where the line meets the yaxis. (i.e) on the the yaxis, x = 0.
Therefore substituting x = 0, in the above standard form,2 (0) + 3y = 12
=> 0 + 3y = 12
=> 3y = 12
=> y = 12/3
= 4
Therefore the coordinate of yintercept is ( 0,4)
Writing linear equations in slope intercept form:
The slope intercept form of the linear equation of two variables is,
y = mx + c, where 'm' is the slope and 'c' the yintercept of the line in the graph.
Writing slopeintercept form from the standard form:
We have the standard form as ax + by = c
by = ax + c
$\frac{by}{b}$ =  $\frac{ax}{b}$ + $\frac{c}{b}$ [ dividing both each term by a on both sides ]
y =  $\frac{a}{b}$ x + $\frac{c}{b} $, which is the slopeintercept form.
Comparing this with the slope intercept form,
y = mx + c,
we get the slope m =  $ \frac{a}{b}$ and
the yintercept is c = $\frac{c}{b}$
Example: Express the equation 3x + 5y = 15 in slope intercept form. Also find the x and yintercepts of the equation.
Solution:
We have 3x + 5y = 155y =  3x + 15
y =  $\frac{3}{5}$ + $\frac{15}{5}$
y =  $\frac{3}{5}$ + 3 , which is of the form y = mx + c, the slope  intercept form.
Comparing we get, the slope m = $\frac{3}{5}$ , and the yintercept c = 3
To find the x intercept:
We have 3x + 5y = 15Substituting y = 0, we get 3x + 5 ( 0) = 15
=> 3x + 0 = 15
=> 3x = 15
=> x = 15/3
= 5
The coordinate of the xintercept is ( 5,0)
To find the yintercept :
Substituting x = 0 in 3x + 5y = 15, we get,3 (0 ) + 5y = 15
=> 0 + 5y = 15
=> 5y = 15
=> y = 15/5
= 3
The coordinate of the yintercept is ( 0, 3)
Writing linear equations from graphs:
If a and b are the x and y  intercepts of a graph of a line, then the equation of a line is given by,
$\frac{x}{a}$ + $ \frac{y}{b} $ = 1
For example, if a line meets the xaxis at x = 4 and yaxis at y = 3, then the equation of the line is $\frac{x}{4}$ + $ \frac{y}{3}$ = 1
Hence the above equation can be written as, $\frac{x}{4}$  $ \frac{y}{3}$ = 1
Multiplying both sides by the LCM of the denominations = 12,
12 ( $\frac{x}{4}$)  12 $(\frac{y}{3})$ = 12(1)
=> 3x  4y = 12, which is the standard form of the equation.
Writing linear equations word problems:
Step 1: Read the problem carefully and identify the unknown quantities. Assume these quantities a variable name x, y.
Step 2: Read the problem carefully and list the given quantities / table the given quantities.
Step 3: Read the problem again and form the two equations.
Step 4: Solve the two simultaneous equations for the variables x and y.
Note: Elimination method or cross multiplication method is the most suitable method for solving for the variables.
Example : Ten years ago, father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be. Find their present ages.
Solution:
Let us assume the present age of father as x years and that of the son's y years.Let us table the given data.
Father's age 
Son's Age  Relation (or) Equation 

2 yrs ago  x  10  y  10 
Father's age = 12 (Son's age) x  10 = 12( y  10 ) 
Present age 
x 
y 

10 yrs hence  x + 10 
y + 10 
Father's age = 2(Son's age) x + 10 = 2 ( y + 10) 
Hence from the above table, we have the two equations,
x  10 = 12 ( y  10) (1)
x + 10 = 2 ( y + 10) (2)
Let us simplify and write the equations (1) and (2) in standard form,
x  10 = 12 y  120
=> x  12y = 120 + 10 [ by transposition ]
=> x  12y = 110 (1)
x + 10 = 2 ( y + 10 )
x + 10 = 2y + 20
x  2y = 20  10
x  2y = 10 (2)
x  12 y =  110
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Subtracting (1) from (2), we get
10 y = 120
=> y = 120/10
= 12
Substituting y = 12 in Equation (1), we get,
x  12 y = 110
=> x  12 ( 12) = 110
=> x  144 = 110
=> x = 144  110
= 34
=> x = 34
Therefore, Father's Present age = 34 yrs and Son's Present age = 12 yrs
Practice Questions:
2. Points A and B are 90 km apart from each other on a highway. A car starts from A and another from B at the same time. If they go in the same direction they meet in 9 hrs and if they go in the opposite direction they meet in 9/7 hrs. Find their speeds.
3. Study the given graph and write the equation in
a. Intercept form.
b. slopeintercept form
c. standard form
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