Question

If $c<1$ and the system of eqations $x+y-1=0,2x-y-c=0$ and $-bx+3by-c=0$ is consistent, then the possible real values of $b$ are

• A. $b\in \left(-3\frac{3}{4}\right)$
• B. $b\in \left(\frac{-3}{2},4\right)$
• C. $b\in \left(\frac{-3}{4},3\right)$
• D. $b\in \left(\frac{-3}{2},\frac{3}{4}\right)$

Answer 1

The correct option is C $b\in \left(\frac{-3}{4},3\right)$

Consider the system of equations,

$x+y-1=0,2x-y-c=0$ and $-bx+3by-c=0$

$\left|\begin{array}{ccc}1& 1& -1\\ 2& -1& -c\\ -b& 3b& -c\end{array}\right|=0$

$=1(c+3bc)-1(-2c-bc)-1(6b-b)=0$

$=c+3bc+2c+bc-5b=0$

$=3c+4bc-5b=0$

or $c=\frac{5b}{4b+3}$

Given$:c<1$

$\Rightarrow \frac{5b}{4b+3}<1$

$\Rightarrow \frac{5b}{4b+3}-1<0$

$\Rightarrow \frac{5b-4b-3}{4b+3}<0$

$\Rightarrow \frac{b-3}{b+\frac{3}{4}}<0$

$\therefore b\in (-\frac{3}{4},3)$