# Which of the following pairs of linear equations has unique solutions, no solution, or infinitely solution ? In case there is a unique solution, find it by using cross multiplication method.

Question : Which of the following pairs of linear equations has unique solutions, no solution, or infinitely solution ? In case there is a unique solution, find it by using cross multiplication method.

(i)  x – 3y – 3 = 0       and      3x – 9y – 2 = 0

(ii)  2x + y = 5      and       3x + 2y = 8

(iii)  3x – 5y = 20      and      6x – 10y = 40

(iv)   x – 3y – 7 = 0     and      3x – 3y – 15 = 0

## Solution :

(i)  The given equations are

x – 3y – 3 = 0

and  3x – 9y – 2 = 0

These equations are of the form   $$a_1x + b_1y + c_1$$ = 0

and  $$a_2x + b_2y + c_2$$ = 0

where $$a_1$$ = 1, $$b_1$$ = -3, $$c_1$$ = -3   and   $$a_2$$ = 3, $$b_2$$ = -9,  $$c_2$$ = -2

We have :  $$a_1\over a_2$$ = $$1\over 3$$,  $$b_1\over b_2$$ = $$-3\over -9$$ = $$1\over 3$$

and  $$c_1\over c_2$$  =  $$-3\over -2$$ = $$3\over 2$$

Clearly we see that, $$a_1\over a_2$$ = $$b_1\over b_2$$ $$\ne$$  $$c_1\over c_2$$

So, the given linear system of equations has no solution (i.e. system of equations is inconsistent).

(ii) The given equations are

2x + y – 5 = 0

and  3x + 2y – 8 = 0

These equations are of the form   $$a_1x + b_1y + c_1$$ = 0

and  $$a_2x + b_2y + c_2$$ = 0

where $$a_1$$ = 2, $$b_1$$ = 1, $$c_1$$ = -5   and   $$a_2$$ = 3, $$b_2$$ = 2,  $$c_2$$ = -8

We have :  $$a_1\over a_2$$ = $$2\over 3$$,  $$b_1\over b_2$$ = $$1\over 2$$

Clearly we see that, $$a_1\over a_2$$ $$\ne$$ $$b_1\over b_2$$

So, the given linear system of equations has unique solution.

To find the solution, we will be using cross-multiplication method here. By cross-multiplication method,

$$x\over {1\times (-8) – 2\times (-5)}$$ = $$y\over {-5 \times 3 – (-8) \times 2}$$ = $$1\over {2\times 2 – 3\times 1}$$

$$\implies$$   $$x\over -8 + 10$$ = $$y\over -15 + 16$$ = $$1\over 4 – 3$$

$$\implies$$  $$x\over 2$$ = $$y\over 1$$ = $$1\over 1$$     or    x = 2,  y = 1

Hence, the given linear system of equation has unique solution given by x = 2, y = 1.

(iii)  The given equations are

3x – 5y – 20 = 0

and  6x – 10y – 40 = 0

These equations are of the form   $$a_1x + b_1y + c_1$$ = 0

and  $$a_2x + b_2y + c_2$$ = 0

where $$a_1$$ = 3, $$b_1$$ = -5, $$c_1$$ = -20   and   $$a_2$$ = 6, $$b_2$$ = -10,  $$c_2$$ = -40

We have :  $$a_1\over a_2$$ = $$3\over 6$$ = $$1\over 2$$,  $$b_1\over b_2$$ = $$-5\over -10$$ = $$1\over 2$$

and  $$c_1\over c_2$$  =  $$-20\over -40$$ = $$1\over 2$$

Clearly we see that, $$a_1\over a_2$$ = $$b_1\over b_2$$ = $$c_1\over c_2$$

So, the given linear system of equations has infinitely many solutions.

(iv) The given equations are

x – 3y – 7 = 0

and  3x – 3y – 15 = 0

These equations are of the form   $$a_1x + b_1y + c_1$$ = 0

and  $$a_2x + b_2y + c_2$$ = 0

where $$a_1$$ = 1, $$b_1$$ = -3, $$c_1$$ = -7   and   $$a_2$$ = 3, $$b_2$$ = -3,  $$c_2$$ = -15

We have :  $$a_1\over a_2$$ = $$1\over 3$$,  $$b_1\over b_2$$ = 1

Clearly we see that, $$a_1\over a_2$$ $$\ne$$ $$b_1\over b_2$$

So, the given linear system of equations has unique solution.

To find the solution, we will be using cross-multiplication method here. By cross-multiplication method,

$$x\over 45 – 21$$ = $$y\over -21 + 15$$ = $$1\over -3 + 9$$

$$\implies$$  $$x\over 24$$ = $$y\over -6$$ = $$1\over 6$$     or    x = 4,  y = -1

Hence, the given linear system of equation has unique solution given by x = 4, y = -1.