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.

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