Question

If $c < 1$ and the system of the equations $x + y - 1 = 0,2x - y - c = 0, - bx + 3ay - c = 0$ is consistent then the possible real values of $b$ are
A. $b \in \left( { - 3,\dfrac{3}{4}} \right)$
B. $b \in \left( { - \dfrac{3}{2},4} \right)$
C. $b \in \left( { - \dfrac{3}{4},3} \right)$
D. $b \in \left( { - \dfrac{3}{2},\dfrac{3}{4}} \right)$

Answer 1

Hint:
If the system of the equations is consistent then it has at least one set of values we can find by equating the determinant of the coefficients of variables as zero. It can also be the case that there are infinitely many solutions.

Complete step by step solution:
Here we are given the set of the equations which are
$
  x + y - 1 = 0 \\
  2x - y - c = 0 \\
  - bx + 3ay - c = 0 \\
 $
And these equations are given to be consistent. First of all we need to know what is the meaning of the consistent? So here there may be cases that the system of equations have an infinite number of roots, finite number of solutions or no solutions. So if the system of equations has no roots then it is said to be inconsistent and if it has at least one solution then it is said to be consistent. As it is said that the system of the equations is consistent if the determinant of the coefficient of the variables is equal to zero.
So we can write it as
$\left| {\begin{array}{*{20}{l}}
  1&1&{ - 1} \\
  2&{ - 1}&{ - c} \\
  { - b}&{3b}&{ - c}
\end{array}} \right| = 0$
Here $1,2, - b$ are all the coefficients of $x$ and similarly $1, - 1,3b$ are the coefficient of $y$ and $ - 1, - c, - c$ are the constant terms respectively.
So on solving we get that
$1( - 1(c)) - 3b( - c)) - 1(2( - c) - ( - b)( - c)) - 1(6b - b)$
$ = $$c + 3bc + 2c + bc - 5b$
$ = 3c + 4bc - 5b$
So we get that
$\left| {\begin{array}{*{20}{l}}
  1&1&{ - 1} \\
  2&{ - 1}&{ - c} \\
  { - b}&{3b}&{ - c}
\end{array}} \right| = 3c + 4bc - 5b$
Equating it to zero we get that
$\Rightarrow 3c + 4bc - 5b = 0$
$\Rightarrow c = \dfrac{{5b}}{{3 + 4b}}$
Now it is given that $c < 1$
So we get that $c = \dfrac{{5b}}{{3 + 4b}} < 1$
$\Rightarrow \dfrac{{5b}}{{3 + 4b}} - 1 < 0$
$\Rightarrow \dfrac{{5b - 3 - 4b}}{{3 + 4b}} < 0$
$\Rightarrow \dfrac{{b - 3}}{{3 + 4b}} < 0$
$\Rightarrow \dfrac{{b - 3}}{{b + \dfrac{3}{4}}} < 0$
Now here we know that $\dfrac{a}{b} < 0$
Case 1 $a > 0,b < 0$
$
\Rightarrow b - 3 > 0 \\
\Rightarrow b > 3 \\
 $ and also we get $
\Rightarrow b + \dfrac{3}{4} < 0 \\
\Rightarrow b < - \dfrac{3}{4} \\
 $
But this cannot be possible
Case 2 $a < 0,b > 0$
$
\Rightarrow b - 3 < 0 \\
\Rightarrow b < 3 \\
 $ and
$
\Rightarrow b + \dfrac{3}{4} > 0 \\
\Rightarrow b > - \dfrac{3}{4} \\
 $

So we get that $b \in \left( { - \dfrac{3}{4},3} \right)$

Note:
We know that the system of equations is consistent then it may have infinite number of solutions and it is possible that when the lines lie over each other or we can say coincide each other and for no solution all lines may be parallel such that it does not intersect.