Question : If $\small x+3y-\frac{2z}{4}=6, \; x+\frac{2}{3}(2y+3z)=33$ and $\frac{1}{7}(x+y+z)+2z=9,$ then what is the value of $46x+131y$?
Option 1: 414
Option 2: 364
Option 3: 384
Option 4: 464
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Correct Answer: 414
Solution :
Given:
$x+3y-\frac{2z}{4} = 6$, $x+\frac{2}{3}(2y+3z) = 33\;$ and $\frac{1}{7}(x+y+z)+2z = 9$
Solution:
$4x+12y-2z = 24......(i)$
$3x+4y+6z = 99.......(ii)$
$x+y+15z = 63........(iii)$
Multiplying equation (i) by 3, we get,
$12x+36y-6z = 72......(iv)$
Now adding equation (ii) and (iv) we get,
$15x+40y = 171.......(v)$
Multiplying equation (ii) by 5 and equation (iii) by 2 and then subtracting equation (iii) from (ii)
$13x+18y = 369$.......(vi)
Multiplying equation (v) by 9 and equation (vi) by 20 and then subtracting equation (v) from (vi), we get,
$125x = 5841$
⇒ $x = 46.728$
Putting the value of $x$ in the above equations, we will get,
$y = –13.248$
Now, $46x+131y = 46×46.728+131×(–13.248) = 2149.488–1735.488 = 414$
Hence, the correct answer is 414.
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