In order to solve this addition problem, a common denominator must first be found. 

\(\displaystyle \frac{1}{4}\) should be converted to a fraction in which the denominator is 16.

\(\displaystyle \frac{1}{4}+\frac{3}{16}\)

\(\displaystyle \frac{(1\times4)}{(4\times4)}+\frac{3}{16}\)

\(\displaystyle \frac{4}{16}+\frac{3}{16}\)

Finally, add the fractions.

\(\displaystyle \frac{4+3}{16}\)

\(\displaystyle \frac{7}{16}\)

By the order of operations, the operation within parentheses, which is addition, is carried out first; of the remaining two, exponentiation - squaring here - precedes multiplication.

By the order of operations, any expressions within grouping symbols, such as parentheses and brackets, are carried out from the inside outward. Therefore, the operation in the innermost set of grouping symbols - the subtraction within parentheses - will be carried out first, followed by the remaining operation within the brackets - the multiplication. The remaining operation - the addition - is last.

In base five, each place value is a power of five, starting with 1 at the right, then, going to the left, \(\displaystyle 5 ^{1}=5,5 ^{2}=25,5 ^{3}=125,...\).

\(\displaystyle 2131 _{\textrm{five}}\) can be calculated in base ten as

\(\displaystyle 2 \times 125 + 1 \times 25 + 3 \times 5 + 1 \times 1 = 250 + 25 + 15 + 1 = 291\).

In modulo 15 arithmetic, a number is congruent to the remainder of the divison of that number by 15. Since 

\(\displaystyle 9+ 7+5+ 3 + 1= 25\)

and

\(\displaystyle 25 \div 15 = 1 \textrm{ R }10\),

\(\displaystyle 9+ 7+5+ 3 + 1 \equiv 10 \mod 15\).

This makes 10 the correct choice.

Add the numbers and keep the variable:

\(\displaystyle 7n+3n=10n\)

Answer: \(\displaystyle 10n\)

Add the numbers and keep the variable:\(\displaystyle 14s+3s=17s\)

Answer: \(\displaystyle 17s\)

In order to add variables the terms must be like. In order for terms to be like, the variables must be exactly alike also being raised to the same power by the exponent.

In this case the like terms are \(\displaystyle x^{1}y^{3}\) and \(\displaystyle xy^{3}\). Just because there is a 1 in the exponent for the first term doesnt mean it is different from the second term. With exponents if a variable does not show an exponent, that means it is still to the first power. 

We add the coefficients of the like terms. The coefficient is the number in front of the first variable, in this case it is 1 for both terms because of the identity property of multiplication stating any variable, term, or number multiplied by 1 is itself.

\(\displaystyle x^{1}y^{3}= 1(x^1y^3)\)    \(\displaystyle xy^3= 1(xy^3)\)

Our last term is not like because the \(\displaystyle x\) variable is raised to a different power than the other two. In this case we do not combine it to the like terms, we just add it to the end of the term. 

Remember, for exponent problems, you group together different exponents and different combinations of variables as though each were a different type of variable.  Therefore, you can group your problem as follows:

\(\displaystyle (2x + 15x) + (2x^{2} + 3x^{2}) + (15y + 2y) + (14y^{2} + 3y^{2})\)

Now, just combine like terms:

\(\displaystyle 17x + 5x^{2} + 17y + 17y^{2}\)

You should begin by distributing \(\displaystyle 2\) through the whole group that it precedes:

\(\displaystyle 23x + 22y + 8x + 6y\)

Now, move your like variables next to each other:

\(\displaystyle 23x + 8x + 22y + 6y\)

Finally, combine the like terms:

\(\displaystyle 31x + 28y\)