SOLVING PAIR OF LINEAR EQUATION USING SUBSTITUTION:
Algorithm to find the solution:
A system of a pair of linear equations in two variables:
Step 1: Find the value of one variable, say y in terms of the other variable, i.e.,x from either equation, whichever is convenient consider this equation as equation (3).
Step 2: Subtitute this value of y in the other equation, and reduce it to an equation in ne variable, i.e., in terms of x, which can be solved. Smetimes, you get statements with no variable. If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions. If the statement is false, then the pair of linear equations is inconsistent and hence no solution.
Step 3: Subtitute the value of x (or y) obtained in Step 2 in the equation (1) , (2) or (3) to obtan the value of the other variable.
Example : Fiind the solution of the following system by subtitution method:
4x + y = 11 , x + 2y = 8
SOLUTION:
It is easy to solve for either y in the first equation or x in the secon equation. Let us solve for y in the first equation. The result is y = 11 - 4x
Since equals may be subtituted for equals, we may subtitute this value of y wherever y apperas in the second equation. THus,
x + 2 ( 11 - 4x ) = 8
We now have one equation that is linear in x, that is, the equation contains only the variable x.
Removing the parentheses and solving for x, we find that
x + 22 - 8x = 8 or -7x = 8 - 22 or -7x = -14 or x = 2
To get the corresponding value of y, we subtitute x = 2 in y = 11 - 4x. THe result is
y = 11 - 4 ( 2) = 11 - 8 = 3
Thus, the solution for the two original equations are x = 2 and y = 3.
If x + y = 6 and 3x - y = 4 then x - y is equal to:
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Right Option : A | |||
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