System of Linear Equations in two variables

                      Linear equation in two variables 

We know that x, y, z, ... etc are known as variables.  We can model a mathematical problem as an equation that contains two variables. For example, the sum of any two numbers is 45.
In this situation, we don't know the two numbers so we assume one number as x and another number y and the sum of x and y is x+y.
And this sum is equal to 45 therefore, we can write it as $x+y=45$
Hence, we have an equation in two variables. 

Other examples of linear equation in two variables are $3x+2y=15$

$3w+4z=-9\\ 3s-2j=18$ 

etc. 

Note: We can solve an equation in two variables for any variable present in it. 

let's take the above equation $x+y=45$
We can solve it either for x or y.
To solve it for y, we subtract y from both sides, we get
$x+y-y=45-y\\x=45-y$

Similarly, we can solve it for x as-
Subtract x from both sides, we get
$x+y-x=45-x\\y=45-x$ 

Now Try solving the given problems.
1. $ y+3z=25$. Solve it for y

2. $ 2x+3y=15$. Solve for x

Now two equation having the same variables in them are known as a system of equations in two variables. 
Suppose we have two equations  $x-y=15\\x+y=17$
Any system of linear equations containing two variables is called a system of linear equation in two variables.
We can solve a system of equation in two variables using many methods such as
1. Substitution method

2-  Elimination method

3- Cross-multiplication method 

4- Graphing method

5- Matrix method 

6- Using Cramer's rule 
etc.
We will discuss all the methods above by taking an example of the system of linear equations in two variables.
Substitution method:  
Let's take the above example,
$x-y=15$      ...(1)
$x+y=17$      ...(2)

Here, we have two equations. To solve this system of equation using the substitution method, we proceed as follows.
Step I -   We can solve any of the two equations for either x or y.
Step II -  We substitute the solution obtained in Step I in other equation for the variable.
Step III-  We simplify and get the answer for both the variables.

Let's do it.

Step I.  Let's solve equation (1)  for x, we have
$x-y=15$
Add y both sides,

$x=15+y$

Step II-  Now we will put this x=15+y in the equation (2), we have
$x+y=17\\ (15+y)+y=17$
Add like terms,
$2y+15=17$
Subtract 15 from both sides,
$2y=17-15\\ 2y=2$
Divide both sides by 2,
$y=\frac{2}2=1$
Substitute y=1 in x=15+y, we get
$x=15+1=16$
Hence, the solution of the above system of linear eqatuon is x=16 and y=1.
Sometimes, we write the solution in the form of an ordered pair (x,y).
Hence, Solution of the above system of equation is (16,1).

I am sure that you have grasped the concept of the substitution method.
Note: The above concept can be extended to solve a system of equations containing three or more variables.
You can go for the practice the following problems.
Solve the following system of equations using the substitution method-
1.  $ 3x+2y=5\\
        5x-2y=8$
2.  $0.5x+0.3y=14\\
     0.9x+0.14y=28$
3. $\frac{1}x+\frac{1}y=25\\ \frac{1}x-\frac{1}y=10$

Elimination Method:  

In this method, we add or subtract two equations so that one variable is eliminated from the resultant equation because of the opposite sign of the coefficient of the variable.
Let's take an example. Suppose we have 5x and -5x and the sum of these two monomials is
$5x-5x=0$
Similarly, if we have two equations
$5x+3y=12\\
-5x+2y=6$
Add two equation so that left sides are added on the left side and right sides are added on the right side
$5x+3y-5x+2y=12+6\\
  3y+2y=18\\
  5y=18\\
    y=18/5$

We see that in the sum of the left side, the terms containing x have been eliminated. 
This method is known as the elimination method. 

Note: The coefficients of any one variable must be equal in magnitude and opposite in signs if it is not so, we make coefficients equal by multiplying the equations by suitable numbers.

Let's take another example

$3x+2y=7$
$2x+3y=8$ 

In this system of equation, the coefficient of y in the first equation is 2 while the coefficient of y in the second equation is 2. 
Are the coefficients of y in both the equations equal in magnitude?

Well the answer is "No" 

Are the coefficients of x in both the equations equal?

Well Answer is "No" 

Okay, what will we do next?
We will make the coefficients of either variable equal in magnitude in both the equations. 

We also know the LCM is the least common multiple of two or more numbers.
We will make coefficients of y equal to the LCM of 2 and 3. LCM of 2 and 3 is 6, thus We will make coefficients of y in both the equations as 6.
We will multiply the first equation by 3 and the second equation  by 3, we have

$(3x+2y=7 )\times 3 \implies 9x+6y=27 \\
(2x+3y=8)\times 2  \implies 4x+6y=16$ 

Now subtract second equation from the first equation, we get
   $9x+6y=27\\
   4x+6y=16$ 

  $5x     =11 $ 

      $x=\frac{11}5$

Substitute x=11/5 in eqation (1), we get
$3 \times \frac{11}5+2y=7\\
2y=7-\frac{33}5=\frac{35-33}5=\frac{2}5\\
  y=\frac{2}{5\times 2}=\frac{1}{5}\\
  y=\frac{1}5 $

Hence, $x=\frac{11}{5}$  and $y=\frac{1}{5}$ 

Now solve the following systems of equations using the elimination method. 

1.  $3x+2y=5$

        $5x-2y=8$

2.  $0.5x+0.3y=14$

    $0.9x+0.14y=28$

3. $\frac{1}x+\frac{1}y=25\\ \frac{1}x-\frac{1}y=10$