Question 522095
if x + 3y + 2z = 48 and 2x + 3y + 4z = 69, then the value of 3x + 3y + 6z is ?
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The trick here is to observe that: 

1. the left side of the first equation contains x + 2z
2. the left side of the second equation contains 2x + 4z, which is 2(x + 2z).
3. the expression to find the value of contains 3x + 6z, which is 3(x + 2z)

So we rewrite the problem by rearranging the terms
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if x + 2z + 3y = 48 and 2x + 4z + 3y = 69, then the value of 3x + 6z + 3y is ?
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Then again by factoring the expressions 2x + 4z and 3x + 6z and writing x + 2z
in parentheses:
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if (x + 2z) + 3y = 48 and 2(x + 2z) + 3y = 69, then the value of 3(x + 2z) + y is ?
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Now let w = x + 2z and we can rewrite the problem again as
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if w + 3y = 48 and 2w + 3y = 69, then the value of 3w + 3y is ?
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So we solve this system of equations:

 w + 3y = 48
2w + 3y = 69

and get w = 21, y = 9

And therefore 3w + 3y = 3(w + y) = 3(21 + 9) = 3(30) = 90

Answer = 90

Edwin</pre>