The two equations are actually equivalent, as is proved here:

\(\displaystyle b =- a+\frac{1}{2 }\)

\(\displaystyle a + b =a - a+\frac{1}{2 }\)

\(\displaystyle a + b = \frac{1}{2 }\)

\(\displaystyle 2 \cdot \left (a + b \right )= 2 \cdot \frac{1}{2 }\)

\(\displaystyle 2a + 2b = 1\)

Therefore, we need only test the first equation. However, it can be shown that it is possible for either of the two to be greater or both to be equal; as can be determined from that third equation \(\displaystyle a + b = \frac{1}{2 }\) , any two values of \(\displaystyle a\) and \(\displaystyle b\) that add up to \(\displaystyle \frac{1}{2}\) will solve the system, such as \(\displaystyle \left ( 0, \frac{1}{2}\right )\), \(\displaystyle \left ( \frac{1}{4}, \frac{1}{4} \right )\), or \(\displaystyle \left ( \frac{1}{2}, 0\right )\).

\(\displaystyle -\frac{3}{4} M =- \frac{6}{7}\)

\(\displaystyle \left ( - \frac{4}{3}\right ) \cdot\left ( -\frac{3}{4} M \right )=\left ( - \frac{4}{3}\right ) \cdot\left ( - \frac{6}{7} \right )\)

\(\displaystyle M = \frac{24}{21} = \frac{8}{7} = 1 \frac{1}{7}\)

\(\displaystyle -\frac{4}{5} N =- \frac{5}{6}\)

\(\displaystyle \left (-\frac{5} {4} \right ) \cdot \left (-\frac{4}{5} N \right )= \left (-\frac{5} {4} \right ) \cdot \left (- \frac{5}{6} \right )\)

\(\displaystyle N = \frac{25}{24} = 1\frac{1}{24}\)

\(\displaystyle 1 \frac{1}{7} > 1 \frac{1}{24}\), so \(\displaystyle M > N\), making (A) greater.

The point \(\displaystyle \left ( - \frac{1}{4}, y\right )\) is on the graph of the equation \(\displaystyle \frac{3}{x} + \frac{2}{y} = 5\). Finding the \(\displaystyle y\)-coordinate of this point is the same as evaluating \(\displaystyle y\) for \(\displaystyle x= -\frac{1}{4}\). Substitute, and we get: 

\(\displaystyle \frac{3}{x} + \frac{2}{y} = 5\)

\(\displaystyle \frac{3}{ -\frac{1}{4}} + \frac{2}{y} = 5\)

\(\displaystyle -12+ \frac{2}{y} = 5\)

\(\displaystyle -12+ \frac{2}{y} + 12 = 5 + 12\)

\(\displaystyle \frac{2}{y} = 17\)

\(\displaystyle \frac{2}{y}\cdot \frac{y}{17} = 17 \cdot \frac{y}{17}\)

\(\displaystyle \frac{2} {17} = y\)

The \(\displaystyle y\)-coordinate is therefore \(\displaystyle \frac{2} {17}\).

The point \(\displaystyle \left ( 64, y\right )\) is on the graph of the equation \(\displaystyle 2\sqrt{x} + \sqrt{y} = 25\). Finding the \(\displaystyle y\)-coordinate of this point is the same as evaluating \(\displaystyle y\) for \(\displaystyle x=64\). Substitute, and we get: 

\(\displaystyle 2\sqrt{x} + \sqrt{y} = 25\)

\(\displaystyle 2\sqrt{64} + \sqrt{y} = 25\)

\(\displaystyle 2 \cdot 8 + \sqrt{y} = 25\)

\(\displaystyle 16+ \sqrt{y} = 25\)

\(\displaystyle 16+ \sqrt{y} - 16 = 25 - 16\)

\(\displaystyle \sqrt{y} = 9\)

\(\displaystyle \left (\sqrt{y} \right )^{2} = 9 ^{2}\)

\(\displaystyle y =81\)

The point \(\displaystyle \left ( 64, y\right )\) is on the graph of the equation \(\displaystyle 3\sqrt{x} +2 \sqrt{y} = 23\). Finding the \(\displaystyle y\)-coordinate of this point is the same as evaluating \(\displaystyle y\) for \(\displaystyle x=64\). Substitute, and we get: 

\(\displaystyle 3\sqrt{x} +2 \sqrt{y} = 23\)

\(\displaystyle 3\sqrt{64} +2 \sqrt{y} = 23\)

\(\displaystyle 3 \cdot8 +2 \sqrt{y} = 23\)

\(\displaystyle 24 +2 \sqrt{y} = 23\)

\(\displaystyle 24 +2 \sqrt{y} -24 = 23 -24\)

\(\displaystyle 2 \sqrt{y} = -1\)

\(\displaystyle 2 \sqrt{y} \div 2 = -1 \div 2\)

\(\displaystyle \sqrt{y} = - \frac{1}{2}\)

Since the square root of a number must be positive, there is no solution. Therefore, there is no point on this graph with \(\displaystyle x\)-coordinate 64.

The point \(\displaystyle \left ( \frac{1}{3}, y\right )\) is on the graph of the equation \(\displaystyle \frac{4}{x} + \frac{3}{y} = 12\). Finding the \(\displaystyle y\)-coordinate of this point is the same as evaluating \(\displaystyle y\) for \(\displaystyle x= \frac{1}{3}\). Substitute, and we get: 

\(\displaystyle \frac{4}{x} + \frac{3}{y} = 12\)

\(\displaystyle \frac{4}{\frac{1}{3}} + \frac{3}{y} = 12\)

\(\displaystyle 12+ \frac{3}{y} = 12\)

\(\displaystyle 12+ \frac{3}{y}- 12 = 12 - 12\)

\(\displaystyle \frac{3}{y} = 0\)

However, there is no number that can be divided into 3 to yield a quotient of 0, so there is no solution. Therefore, there is no point on this graph with \(\displaystyle x\)-coordinate \(\displaystyle \frac{1}{3}\).

Substitute \(\displaystyle \frac{1}{3} x + \frac{4}{3}\) for \(\displaystyle y\) in the second equation:

\(\displaystyle 2x - 3y = 4\)

\(\displaystyle 2x - 3 \left ( \frac{1}{3} x + \frac{4}{3} \right ) = 4\)

\(\displaystyle 2x - x -4 = 4\)

\(\displaystyle x -4 = 4\)

\(\displaystyle x -4+ 4 = 4 + 4\)

\(\displaystyle x = 8\)

Solve for \(\displaystyle x\) in the top equation:

\(\displaystyle y = \frac{1}{3} x + \frac{4}{3}\)

\(\displaystyle 3 \cdot y = 3 \cdot \left ( \frac{1}{3} x + \frac{4}{3} \right )\)

\(\displaystyle 3 y = x +4\)

\(\displaystyle 3 y-4 = x +4 -4\)

\(\displaystyle x = 3 y-4\)

Substitute \(\displaystyle 3 y-4\) for \(\displaystyle x\) in the second equation:

\(\displaystyle 2x - 3y = 4\)

\(\displaystyle 2 \left ( 3 y-4 \right )- 3y = 4\)

\(\displaystyle 6 y-8- 3y = 4\)

\(\displaystyle 3 y-8 = 4\)

\(\displaystyle 3 y-8 + 8 = 4 + 8\)

\(\displaystyle 3 y = 12\)

\(\displaystyle 3 y \div 3 = 12 \div 3\)

\(\displaystyle y = 4\)

To solve this problem, isolate for x. First, move the y term over to the left side. This gives you \(\displaystyle \frac{x}{4}=16-2y\). Then, multiply both sides by 4. This gives you \(\displaystyle x=(16-2y)4\). Then, distribute the four to the terms inside the parantheses. This gives you a final answer of \(\displaystyle x=64-8y\).

Substitute \(\displaystyle 0.5x + 1.4\) for \(\displaystyle y\) in the second equation as follows:

\(\displaystyle 1.5x-0.5y = 9.8\)

\(\displaystyle 1.5x-0.5 \left ( 0.5x + 1.4 \right ) = 9.8\)

\(\displaystyle 1.5x-0.25 x - 0.7 = 9.8\)

\(\displaystyle 1.25 x - 0.7 = 9.8\)

\(\displaystyle 1.25 x - 0.7 + 0.7= 9.8 + 0.7\)

\(\displaystyle 1.25 x =10.5\)

\(\displaystyle 1.25 x\div 1.25 =10.5 \div 1.25\)

\(\displaystyle x = 8.4\)