Answer:

Step-by-step explanation:
When working with surds we need to take note of the roots present there.
To expand this equation we can do it the following way noting that √3 X √3 = 3
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<em>Expanding (1-√3)(⅓+√3)</em>
1 X 1/3 = 1/3
1 X √3 = √3
-√3 X 1/3 =-√3/3
√3 X √3 = 3
hence, expanding the equation, we have
1/3 + √3 -√3/3 + 3
We can simply group the like terms and add them up.
[1/3 +3] +[√3-√3/3]
10/3 + 
= 
Notice that
13 - 9 = 4
17 - 13 = 4
so it's likely that each pair of consecutive terms in the sum differ by 4. This means the last term, 149, is equal to 9 plus some multiple of 4 :
149 = 9 + 4k
140 = 4k
k = 140/4
k = 35
This tells you there are 35 + 1 = 36 terms in the sum (since the first term is 9 plus 0 times 4, and the last term is 9 plus 35 times 4). Among the given options, only the first choice contains the same amount of terms.
Put another way, we have

but if we make the sum start at k = 1, we need to replace every instance of k with k - 1, and accordingly adjust the upper limit in the sum.


Answer:
Infinitely many solutions
Step-by-step explanation:
x+y=3
2x+2y=6
x=3-y
2(3-y)+2y=6
6-2y+2y=6
6+0=6
6=6
infinitely many solutions