Question

Find the sum of all positive integral values of a for which every solution to the system of equation $x+ay=3$ and $ax+4y=6$ satisfy the inequalities $x>1,y>0$.

• A. $a>3$
• B. $a>2$
• C. $a>y$
• D. $a>x$

Answer 1

Answer: (B)

$a>2$

0n solving $x+ay=3$ and $ax+4y=6$

$x+a\left(\frac{6-ax}{4}\right)=3$

$x(1-\frac{a^2}{4})=3-\frac{6a}{4}$

$x=\frac{12-6a}{4-a^2}=\frac{6(2-a)}{(2-a)(2+a)}$

for $a\ne 2$

$x=\frac{6}{a+2}>1$ (given)

$\frac{a-4}{a+2}<0$ $aϵ(-2,4)$

positive values of $a$ are $1,3$

sum $=1+3=4$