Question

Consider the system of simultaneous linear equation:
$y+ax=b$
$x+3y=10$
Where $a$ and $b$ are constants.
Which of the following possible values of $a$ and $b$ makes the equation having no solution.

• A. $a=\frac{1}{3}$ and $b=2$
• B. $a=\frac{1}{3}$ and $b=\frac{10}{3}$
• C. $a=\frac{1}{3}$ and $b=6$
• D. None of the above.

Answer 1

Answer: (A), (C)

$a=\frac{1}{3}$ and $b=6$

Concept: For simultaneous equations having no solution, Slope of both the lines must be same that means both lines should be parallel to each other having different intercepts on $x$ and $y$ axes.

Slope of line $x+3y=10$ will be calculated by arranging equation in slope-intercept form.
$\Rightarrow 3y=-x+10\Rightarrow y=\frac{-1}{3}x+\frac{10}{3}$
$\therefore \text{slope}=\frac{-1}{3}$.

Slope of line $y+ax=b$ must be $\frac{-1}{3}$ for being parallel with $x+3y=10$.

$\Rightarrow y=-ax+b$
$\Rightarrow -a=\frac{-1}{3}\Rightarrow a=\frac{1}{3}.$

Now, For lines to be parallel their intercept must be different.
$\therefore b\ne \frac{10}{3}$

So, the possible values of $a$ and $b$ will be option (a) and (c).