Question 22

If $$3x + 6y + 9z = \frac{20}{3}, 6x + 9y + 3z = \frac{17}{3}$$ and $$18x + 27y - z = \frac{113}{9}$$, then what is the value of $$75x + 113y$$?

Solution

$$3x+6y+9z=\frac{20}{3}.$$

or, $$x+2y+3z=\frac{20}{9}...............\left(1\right)$$

$$6x+9y+3z=\frac{17}{3}.$$

or, $$2x+3y+z=\frac{17}{9}................\left(2\right)$$

$$18x+27y-z=\frac{113}{9}.$$

or, $$72x+108y-4z=\frac{452}{9}.......................\left(3\right)$$

By adding (1),(2) & (3) :

$$75x+113y=\frac{452}{9}+\frac{20}{9}+\frac{17}{9}=\frac{489}{9}=\frac{163}{3}.$$

A is correct choice.

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