System of Linear Equations

System of Linear Equations is a collection of linear equation  Systems of Linear Equations that involve two equations in two variables are simplest to deal.
Suppose there are two linear equation in x and y, then each equation will represent a line in x-y plane. A solution to these equations will be the point where these lines intersect. Thus the solution will be unique value of x and y. If the equations represent parallel lines then there will be no solutions to this system. If the Linear System of Equations contains same coinciding lines then the solutions will be infinite in number.

Any system of linear equation can have following conclusions: no solution, unique solution or infinitely many solutions. A linear system is consistent if it has at least 1 solution and is said to be inconsistent if it has no solution.
Suppose a linear equation is 2x+y=0. Then there will be infinite points satisfying this equation. Like (x, y) ={(1,-2),(0,0),(2,-4)… and many more}. Now suppose there is another line x-y=0, then solutions to this will be (x, y)={(0,0),(1,1),(2,2) and so on}. A common solution to these equations is x=0, y=0. This is hence solution of this system of equations.
We can search solution of more than two equations also by drawing graph of the equations also.

Method of Solving Systems of Linear Equations:
By substitution:
Let two equations are a_1x+b_1y=c_1 and a_2x+b_2y=c_2. Solve first equation for x:
 a_1x+b_1y=c_1  
a_1x=c_1–b_1y or x=(c_1–b_1y)a_1
Substitute this value of x in second equation to get:
a_2((c_1–b_1y)a_1)+b_2y= c_2
Now you get an equation in y. Solve for y. now put the value of y in any of the two equations to solve for x.
You can substitute value of y also from an equation and then substitute it in other equation.

Systems of Linear Equations Word Problems
Q.1) Cost of 2 chairs and 1 table is 1000 while cost of 1 chair and 3 tables is 1500. Find cost of each.
Solution) let cost of chair=x and of table=y.
2x+y=1000………(1)
x+3y=1500………(2)
From second equation: x=1500-3y
Substituting x in first equation = 2(1500-3y)+y=1000
3000-6y+y=1000
3000-1000-5y=0
2000=5y
y=(2000/5)=400
Putting y=400 in first equation we get: 2x+400=1000
2x=(1000-400)=600
x=600/2=300
Linear equation can also be solved by equating coefficients:
Equate coefficients of x by multiplying equation (2) by 2:
2x+2(3y)=2(1500)
Subtract this equation from (1):
   2x+y=1000
-(2x+6y=3000)
     0-5y= -2000
5y=2000
Y=400
Now put value of y in any equation.
You can equate coefficients of y also and then subtract the two equations to get value of x first.