Set up an equation for the sentence: 15 – 3 = 2x and solve for x.  X equals 6. If you plug in 6 for x in the expression 24/x, you get 24/6 = 4. 4 is only choice greater than a. 

Notice that - (a-b) = b-a, so statement I & III are true after substituting the expression. Substitute the expression for statement II gives ((a+b)/(a-b))((a+b)/(b-a))=((a+b)(b+a))/((-1)(a-b)(a-b))=-1 〖(a+b)〗²/〖(a-b)〗² =-((a+b)/(a-b))² = -(a♦b)² ≠ (a♦b)²

The best way to solve this problem is to plug in an appropriate value for a.  For example, plug-in 11 for a because 11 divided by 7 will give us a remainder of 4. 

Then 3a + 5, where a = 11, gives us 38.  Then 38 divided by 3 gives a remainder of 2.

The algebra method is as follows:

a divided by 7 gives us some positive integer b, with a remainder of 4.

Thus,

a / 7 = b  4/7

a / 7 = (7b + 4) / 7

a = (7b + 4)

then 3a + 5 = 3 (7b + 4) + 5

(3a+5)/3 = [3(7b + 4) + 5] / 3

= (7b + 4) + 5/3

The first half of this expression (7b + 4) is a positive integer, but the second half of this expression (5/3) gives us a remainder of 2.

Cancel by subtracting the exponents of like terms:

\(\displaystyle \frac{6x^7y^3z^9}{3x^6y^3z} = 2x^{7-6}y^{3-3}z^{9-1}=2xy^0z^8=2xz^8\)

It is not necessary to work a long division if you recognize \(\displaystyle x^{3} + 125y ^{3}\) as the sum of two perfect cube expressions:

\(\displaystyle x^{3} + 125y ^{3} = x^{3} + 5^{3} y ^{3} = x^{3} + (5y )^{3}\)

A sum of cubes can be factored according to the pattern

\(\displaystyle A^{3}+ B^{3} = (A+B) (A^{2}+ AB +B^{2})\),

so, setting \(\displaystyle A = x, B = 5y\),

\(\displaystyle x^{3} + (5y )^{3} =(x+5y) [x ^{2}+ x \cdot 5y + (5y)^{2}]\)

\(\displaystyle =(x+5y)(x ^{2}+ 5xy + 25y^{2})\)

Therefore, 

\(\displaystyle \frac{x^{3} + 125y ^{3}}{x+ 5y} = \frac{(x+5y)(x ^{2}+ 5xy + 25y^{2})}{x+ 5y} = x ^{2}+ 5xy + 25y^{2}\)

Divide \(\displaystyle 6x^{3} +20x^{2} +5x-21\) by \(\displaystyle 3x+7\) by setting up a long division. 

Divide the lead term of the dividend, \(\displaystyle 6x^{3}\), by that of the divisor, \(\displaystyle 3x\); the result is \(\displaystyle \frac{6x^{3}}{3x} = 2 x^{2}\)

Enter that as the first term of the quotient. Multiply this by the divisor:

\(\displaystyle 2 x^{2} (3x+7) = 2 x^{2} (3x)+2 x^{2} ( 7) = 2x^{3} + 14x^{2}\)

Subtract this from the dividend. This is shown in the figure below.

Repeat the process with the new difference:

\(\displaystyle \frac{6x^{2}}{3x} = 2 x\)

\(\displaystyle 2 x (3x+7) = 2 x (3x)+2 x ( 7) = 2x^{2} + 14x\)

Repeating:

\(\displaystyle \frac{-9x}{3x} = -3\)

\(\displaystyle -3 (3x+7) = -3 (3x)+ (-3) ( 7) = -9x+ 21\)

The quotient - and the correct response - is \(\displaystyle 2x^{2 }+ 2x - 3\).