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.
