# SOLVING SYSTEMS OF EQUATIONS BY ELIMINATION WITH DIFFERENT COEFFICIENT

## About "Solving Systems of Equations by Elimination with Different Coefficient"

Solving Systems of Equations by Elimination with Different Coefficient :

Here we are going to see some example problems of solving linear equations in two variables.

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 fi nd the value of the other unknown.

## Solving Systems of Equations by Elimination with Different Coefficient - Practice questions

(iv) 3(2x + y) = 7xy; 3(x + 3y) = 11xy

Solution :

3(2x + y) = 7xy

6x + 3y  =  7xy

By dividing the entire equation by xy, we get

(6/y) + (3/x)  =  7

Let 1/x  =  a and 1/y  =  b

3a + 6b  =  7   ---------(1)

3(x + 3y) = 11xy

3x + 9y  =  11xy

(3/y) + (9/x)  =  11

9a + 3b  =  11  --------(2)

(1) - 2(2)

3a + 6b  =  7

(2) x 3 ==> 18a + 6b  =  22

(-)   (-)     (-)

------------------

-15a  =  -15

a  =  1

By applying the value of a in (1), we get

3(1) + 6b  =  7

6b  =  7 - 3

6b  =  4

b  =  2/3

By taking reciprocals of a and b, we get the values of x and y respectively.

x  =  1 and y  =  3/2.

(v)  (4/x) + 5y  =  7, (3/x) + 4y  =  5

Solution :

Let 1/x  = a and y  = b

4a + 5b  =  7-------(1)

3a + 4b  =  5 -------(2)

(1) x 4 - (2) x 5

16a + 20b  =  28

15a + 20b  =  25

(-)    (-)        (-)

------------------

a  =  3

By applying the value of a in (1), we get

4(3) + 5b  =  7

5b  =  7 - 12

5b  =  -5

b  =  -1

x  =  1/3 and y  =  -1

Hence the solution is (1/3,  -1).

(vi) 13x + 11y = 70; 11x + 13y = 74

Solution :

13x + 11y = 70 --------(1)

11x + 13y = 74 -------(2)

(1) + (2)

13x + 11y  =  70

11x + 13y  =  74

-----------------

24x + 24y  =  144

x + y  =  6  -----(3)

(1) - (2)

13x + 11y  =  70

11x + 13y  =  74

(-)    (-)       (-)

-----------------

2x - 2y  =  -4

x - y  =  -2  -----(4)

(3) + (4)

x + y  =  6

x - y  =  -2

--------------

2x  =  4

x  =  2

By applying the value of x in (4), we get

2 - y  =  -2

y  =  2 + 2

y  =  4

Hence the solution is (2, 4).

After having gone through the stuff given above, we hope that the students would have understood, "Solving Systems of Equations by Elimination with Different Coefficient"

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