# CLASS-9PROBLEM & SOLUTION OF CROSS MULTIPLICATION - SIMULTANEOUS LINEAR EQUATIONS

PROBLEM & SOLUTION OF CROSS MULTIPLICATION -

Example.1) Solve the system of equations, 2x – 3y + 4 = 0, 3x + y + 5 = 0 by the method of cross multiplications.

Ans.) The given equations are –

2x – 3y + 4 = 0 ……………………..(i)

3x + y + 5 = 0  ……………………..(ii)

By cross multiplication, we have – x                       y                          1

=> ---------------- = ----------------- = -----------------

(2 X 1) – {3 X (-3)}      {(-3) X 5} – (1 X 4)       (4 X 3) – (5 X 2)

x                       y                      1

=> --------------- = --------------- = ----------------

2 + 9                -15 – 4                12 – 10

x                  y                1

=> ----------- = ------------ = ----------

11               -19                2

Hence,  x = 11/2, and y = -19/2 is the required solution.  (Ans.)

Example.2) Solve the system of equations 2x – 3y = 12, 3x + 2y = - 5 by the method of cross multiplications.

Ans.)  The given equations are –

2x – 3y = 12

=>   2x – 3y – 12 = 0 ………………………..(i)

And,   3x + 2y = - 5

=>   3x + 2y + 5 = 0  …………………..(ii)

By cross multiplication, we have – x                         y                        1

=> ----------------- = ------------------ = ----------------

(2 X 2) – {3 X (-3)}       {(-3) X 5} – {(-12) X 2}    {(-12) X 3} – (5 X 2)}

x                    y                    1

=> ------------ = -------------- = --------------

4 + 9            - 15 + 24           - 36 – 10

x              y              1

=> --------- = --------- = ----------

13             9            - 46

Hence, x = - 13/46, and y = - 9/46 is the required solution.  (Ans.)

5           2

Example.3) solve the system of equations ------- - ------ + 1 = 0,

x + y       x – y

15             7

--------- + -------- = 10 ( xy )

x + y         x – y

Ans.)   Equations given –

5              2

--------- - --------- + 1 = 0 ........................ (i

x + y          x – y

15              7

---------- + --------- = 10  ........................ (ii)

x + y          x – y

1                          1

Let,  u  = ----------,  and  v = ---------- we get -

x + y                     x – y

5u – 2v + 1 = 0 ...................(iii)

15u + 7v – 10 = 0.....................(iv)

By cross multiplication, we have – u                        v                          1

=> ---------------- = ----------------- = -----------------

(5 X 7) – {15 X (-2)}      {(-2) X (-10)} – (1 X 7)      (1 X 15) – {(-10) X 5)}

u                     v                      1

=> ------------- = --------------- = ---------------

(35 + 30)             (20 – 7)              (15 + 50)

u                v               1

=> ---------- = ---------- = ----------

65              13              65

Or,   u = 1

Replacing, x + y = 1

So,   x + y – 1 = 0 ………………….. (v)

And,   v  = 1/5

Replacing,  x – y = 5

So,  x – y – 5 = 0 …………………….(vi)

By cross multiplication, we have – x                        y                        1

=> ----------------- = ----------------- = -----------------

{1 X (-1)} – (1 X 1)     {1 X (-5)} – {(-1) X (-1)}   {(-1) X 1} – {1 X (-5)}

x                     y                      1

=> ------------- = --------------- = ----------------

(-1 – 1)              (-5 – 1)               (-1 + 5)

x                 y                 1

=> ----------- = ------------ = -----------

- 2               - 6                 4

So, x = (-2)/4  = - 1/2

And, y = (-6)/4 = - 3/2

Hence, x = - 1/2, and y = - 3/2 is the required solution.   (Ans.)