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This question belongs to Arithmetic Ability Algebra
Algebra
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If $$x + 3y - \frac{{2z}}{4} = 6,\,x + \frac{2}{3}\left( {2y + 3z} \right) = 33$$        and $$\frac{1}{7}$$(x + y + z) + 2z = 9 then what is the value of 46x + 131y?

Answer & Solution
Correct Answer: Option A
$$\eqalign{ & x + 3y - \frac{{2z}}{4} = 6 \cr & 2x + 6y - z = 12\,........\,\left( {\text{i}} \right) \cr & x + \frac{2}{3}\left( {2y + 3z} \right) = 33 \cr & 3x + 4y + 6z = 99\,........\,\left( {{\text{ii}}} \right) \cr & \frac{1}{7}\left( {x + y + z} \right) + 2z = 9 \cr & x + y + z + 14z = 63 \cr & x + y + 15z = 63\,........\,\left( {{\text{iii}}} \right) \cr & {\text{Add}}\,\left[ {21 \times \left( {\text{i}} \right)} \right] + \left( {{\text{ii}}} \right) + \left( {{\text{iii}}} \right){\text{ }}\left( {{\text{remove }}z} \right) \cr & 21\left( {2x + 6y + z} \right) = 12 \times 21 \cr & 3x + 4y + 6z = 99 \cr & x + y + 15z = 63 \cr & 42x + 126y - 21z = 252 \cr & 3x + 4y + 6z = 99 \cr & \underline {x + y + 15z = 63\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \cr & 46x + 131y = 414 \cr} $$
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