This is another algebraic method for solving a pair of linear equations. This method is more convenient than the substitution method. Here we eliminate (i.e. remove) one of the two variables in a pair of linear equations, so as to get a linear equation in one variable which can be solved easily.

**Solving by Elimination Method**

This is another
algebraic method for solving a pair of linear equations. This method is more convenient
than the substitution method. Here we eliminate (i.e. remove) one of the two variables
in a pair of linear equations, so as to get a linear equation in one variable which
can be solved easily.

The various
steps involved in the technique are given below:

**Step 1: **Multiply one or both of the equations
by a suitable number(s) so that either the** **coefficients of first variable
or the coefficients of second variable in both the equations become numerically
equal.

**Step 2: **Add both the equations or subtract one
equation from the other, as obtained** **in step 1, so that the terms with equal
numerical coefficients cancel mutually.

**Step 3: **Solve the resulting equation to find
the value of one of the unknowns.

**Step
4: **Substitute this value in any of the two
given equations and find the value of the** **other unknown.

**Example 3.50**

Given
4*a* + 3*b* = 65 and *a* +
2*b* = 35 solve by elimination method.

*Solution*

Thus
the solution is *a* = 5, *b* = 15.

**Example 3.51**

Solve
for *x* and *y*: 8*x* − 3*y* =
5*xy*, 6*x* − 5*y* = −2*xy*
by the method of elimination.

*Solution*

The
given system of equations are

8*x*
−
3*y* = 5*xy *...(1)

6*x*
−
5*y* = −2*xy *...(2)

Observe
that the given system is not linear because of the occurrence of *xy* term.
Also note that if *x* =0, then *y* =0 and vice versa. So, (0,0) is a solution
for the system and any other solution would have both *x* ≠
0 and *y* ≠ 0.

Let
us take up the case where *x* ≠ 0, *y* ≠
0.

Dividing
both sides of each equation *by xy*,

(3)&(4)
respectively become, 8*b* − 3*a* = 5 ………..(5)

6*b* −
5*a* = −2 ………..(5)

which
are linear equations in *a* and *b*.

To eliminate
*a*, we have,

(5) × 5 ⇒ 40b − 15a = 25 .....(7)

(6)×3⇒ 18*b* − 15*a* = −6 .....(8)

Now
proceed as in the previous example to get the solution (11/23 22/31).

Thus,
the system have two solutions (11/23 22/31)
and ( 0, 0) .

Tags : Solving simultaneous linear equations in Two Variables | Example Solved Problems | Algebra | Maths , 9th Maths : UNIT 3 : Algebra

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9th Maths : UNIT 3 : Algebra : Solving by Elimination Method | Solving simultaneous linear equations in Two Variables | Example Solved Problems | Algebra | Maths

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