Question 4

If two equations $$x-4y=0$$ and $$4x+3y=19$$ have a solution $$(a,b)$$. Find the value of $$\frac{ab}{a+4b}$$

Solution

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, $$(a,b)=(4,1)$$

$$\therefore$$ $$\frac{ab}{a+4b}$$

= $$\frac{4\times1}{4+4(1)}=\frac{4}{8}=\frac{1}{2}$$

=> Ans - (B)

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