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 explain  elimination  method

In elimination method, we eliminate one of the variables from the given equations. To do this, we multiply the given equations with suitable numbers so that when we add or subtract the given equations, then one of the variables cancel out. We then find the value of the other variable and then substitute this value in one of the given equations to find the value of the second variable. Go through few of the solved examples given in our study material to understand it in a better way. 

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Instructions

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      Clear all fractions and decimals and put the equations in standard format. For fractions, multiply both sides of an equation by the lowest common denominator of the fractions in the equation or by all the denominators. For decimals, multiply both sides of an equation by 10 for one place behind the decimal, 100 for two places and so on. After completing the clearing of fractions and decimals, all the equations will be in the form Ax + By + Cz = D for three equations with three unknowns and A, B, C and D representing whole numbers.

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      Choose two equations and a variable to eliminate. Multiply both sides of each of the two equations by a factor so that the whole number in front of the chosen variable becomes equal in both equations. Subtract one equation from the other by subtracting the x terms, the y terms and so on. The chosen variable is eliminated because the two equations had the same term for that variable. If there were only two equations, you can now solve for the remaining variable. If there were more, repeat the procedure with two other equations and eliminate the same variable. You will eventually be left with two equations with two variables that can be solved in a final step.

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      As an example, take the equations 0.2x + 5 = 0.3y and y - x = 1/2. Clearing the fractions and decimals by multiplying the first equation by 10 and the second equation by 2 gives 2x + 50 = 3y and 2y - 2x = 1. Putting the equations in standard form Ax + By = C gives 2x - 3y = -50 and -2x + 2y = 1. Multiplying the first equation by -2 and the second equation by 3 gives -4x + 6y = 100 and -6x + 6y = 3. Subtracting the second equation from the first gives 2x = 97. Then x = 48.5 and y - 48.5 = 1/2. Solving for y gives 49. You can verify that the answers are correct by substituting them into the original equations.

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This method makes use of the fact that the solution of an equation is not changed if we

  • multiply both sides of the equation by the same factor. 
     
  • subtract equal quantities from both sides of an equation.

This means that we can take one equation and subtract a multiple of another equation from it without changing the solution of the equations.

The elimination method uses this fact to solve a system of linear equations. Suppose we start with a system of n equations in n unknowns. Pick the first equation and subtract suitable multiples of it from the other n − 1 equations. In each case the multiple is chosen so that the subtraction cancels or eliminates the same variable, say x. The result is that the n − 1 equations contain only n − 1 unknowns (x no longer appears).

We repeat this elimination process until we get 1 equation in 1 unknown, which is then easily solved.

The final step is to back-substitute the solution already obtained for the 1 unknown into the previous equations to find the values of all the other unknowns. 



Example: Solve this system of equations by elimination:

Solution: Let’s take twice the first equation, namely:

2 x + 2 y = 8

and subtract it from the second equation, like this:

The result is one equation in the one unknown, y. The other unknown, x, has been eliminated. Solving this equation yields y = 0.4.

It remains to find x. If we back-substitute y = 0.4 into either of the original equations we get x = 3.6. Thus the solution is:

{ x = 3.6, y = 0.4 }.

(Note that we could have found x without back-substitution if we had subtracted 3 times the first equation from the second equation, since this eliminates y.)

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