Question
Write a pair of linear equations which has the unique solution $x = -1, y = 3$. How many such pairs can you write?
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Answer
Condition for the pair of system to have unique solution
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
Let the equations are,
$a_1x + b_1y + c_1 = 0$
and $a_2x + b_2y + c_2 = 0$
Since, x = -1 and y = 3 is the unique solution of these two equations, then
$a_1(-1) + b_1(3) + c_1 = 0$
$\Rightarrow -a_1 + 3b_1 + c_1 = 0$
and $a_2(-1) + b_2(3) + c2 = 0$
$\Rightarrow -a_2 + 3b_2 + c_2 = 0$
So, the different valume of $a_1, a_2, b_1, b_2, c_1$ and $c_2$ satisfy the eqs. (i) and (ii).
Hence, infinitely many pairs of linear equations are possible.
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
Let the equations are,
$a_1x + b_1y + c_1 = 0$
and $a_2x + b_2y + c_2 = 0$
Since, x = -1 and y = 3 is the unique solution of these two equations, then
$a_1(-1) + b_1(3) + c_1 = 0$
$\Rightarrow -a_1 + 3b_1 + c_1 = 0$
and $a_2(-1) + b_2(3) + c2 = 0$
$\Rightarrow -a_2 + 3b_2 + c_2 = 0$
So, the different valume of $a_1, a_2, b_1, b_2, c_1$ and $c_2$ satisfy the eqs. (i) and (ii).
Hence, infinitely many pairs of linear equations are possible.
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