If two equations $$x-4y=0$$ and $$4x+3y=19$$ have a solution $$(a,b)$$. Find the value of $$\frac{ab}{a+4b}$$
Equations : $$x-4y=0$$ and $$4x+3y=19$$
Substituting $$x=4y$$ from equation (i) in the second equation
=> $$4(4y)+3y=19$$
=> $$16y+3y=19y=19$$
=> $$y=\frac{19}{19}=1$$
Thus, $$x=4(1)=4$$
Now, $$(a,b)=(4,1)$$
$$\therefore$$Â $$\frac{ab}{a+4b}$$
= $$\frac{4\times1}{4+4(1)}=\frac{4}{8}=\frac{1}{2}$$
=> Ans - (B)
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