The answer is 30. You just multiply 51 by (10/17)
The correct answer is the first option.
If you want to use elimination, you can sum the two equations for example, so that the x's simplify:
Plug this value for y in one of the equations to derive the value of x:
So, the solution is 
You're looking for the largest number <em>x</em> such that
<em>x</em> ≡ 1 (mod 451)
<em>x</em> ≡ 4 (mod 328)
<em>x</em> ≡ 1 (mod 673)
Recall that
<em>x</em> ≡ <em>a</em> (mod <em>m</em>)
<em>x</em> ≡ <em>b</em> (mod <em>n</em>)
is solvable only when <em>a</em> ≡ <em>b</em> (mod gcd(<em>m</em>, <em>n</em>)). But this is not the case here; with <em>m</em> = 451 and <em>n</em> = 328, we have gcd(<em>m</em>, <em>n</em>) = 41, and clearly
1 ≡ 4 (mod 41)
is not true.
So there is no such number.
Nanometers are a really small measurement, one billionth of a meter, I believe.
So nanometers is used to describe tiny stuff, like atoms of a water molecule, and waves of light.
The answer is 64 .............
