Question

If $2a=b$, the pair of equations $ax+by=2a^2-3b^2,x+2y=2a-6b$ posses:

• A. no solution
• B. only one solution
• C. only two solutions
• D. an infinite number of solutions

Answer 1

The correct option is D an infinite number of solutions

We replace $b$ by $2a$, write the equations in the form $ax+by+c=0$.

Check for $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ ......(i) for infinite solutions,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$ .......(ii) for no solution and

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$ .........(iii) for a unique solution.

The equations are

$ax+by=2a^2-3b^2⟹ax+2ay-2a^2+3{\left(2a\right)}^2=0⟹ax+2ay+10a^2=\mathrm{0......}\left(1\right)x+2y=2a-6b⟹x+2y-2a+6\times 2a=0⟹x+2y+10a=\mathrm{0.......}\left(2\right)$

Here $a_1=a,a_2=1,b_1=2a,b_2=2,c_1=10a^2,c_2=10a$.

$\therefore \frac{a_1}{a_2}=\frac{a}{1}=a,\frac{b_1}{b_2}=\frac{2a}{2}=a\frac{c_1}{c_2}=\frac{10a^2}{10a}=a\therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Hence, there are infinite number of solutions as the lines are coincident.