Let \(\displaystyle (b,0),(0,a)\) be the \(\displaystyle x\)- and \(\displaystyle y\)-intercepts, respectively. We know that the line does not pass through the origin - so \(\displaystyle a,b \neq 0\) .

Then the slope is:

\(\displaystyle m = \frac{a-0}{0-b} = \frac{a}{-b} = - \frac{a}{b}\)

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

\(\displaystyle a = -2b\)

Either \(\displaystyle a\) or \(\displaystyle b\) can be the greater. For example, if \(\displaystyle b = 1\), then \(\displaystyle a = -2\), and if \(\displaystyle b =- 1\), then \(\displaystyle a = 2\). 

\(\displaystyle 36 = 6^{2}\), so substitute and use the power of a power rule.

\(\displaystyle 6^{3x}= 36 ^{x+ 3}\)

\(\displaystyle 6^{3x}=\left ( 6^{2} \right )^{x+ 3}\)

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

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

\(\displaystyle 3x = 2x + 6\)

\(\displaystyle 3x -2x= 2x-2x + 6\)

\(\displaystyle x = 6\)

This makes (a) and (b) equal.

Both equations are in slope-intercept form, so compare the coefficients of \(\displaystyle x\). The coefficients in (a) and (b) are 5 and 4, respectively, so these are the slopes of the lines. The line in (a) has the greater slope.

Using two different cases, we show that it is impossible to tell which is greater.

Case 1: \(\displaystyle x = 1\). Then \(\displaystyle y = 2x + 1 = 2 \cdot 1+ 1 = 2 + 1 = 3\), and \(\displaystyle y > x\).

Case 2: \(\displaystyle x = -3\). Then \(\displaystyle y = 2x + 1 = 2 \cdot (-3)+ 1 = -6+ 1 = -5\), and \(\displaystyle y < x\).

To solve the system of equations, add the left and right sides of the equation separately:

  \(\displaystyle x + y = 10\)

\(\displaystyle \underline{2x - y = 14}\)

\(\displaystyle 3x\; \; \; \; \; \; = 24\)

Divide:

\(\displaystyle 3x \div 3 = 24\div 3\)

\(\displaystyle x=8\)

Substitute to get \(\displaystyle y\):

\(\displaystyle x+y = 10\)

\(\displaystyle 8+y = 10\)

\(\displaystyle 8-8+y = 10-8\)

\(\displaystyle y = 2\)

\(\displaystyle x\) is greater.

We show that it is possible for either \(\displaystyle t\) or \(\displaystyle u\) to be the greater by giving one of each case.

Case 1: \(\displaystyle u = 1\). Then \(\displaystyle t + 1 = \left | 1\right | = 1\), so \(\displaystyle t = 0\)

Case 2: \(\displaystyle u = -1\). Then \(\displaystyle t + 1 = \left | -1\right | = 1\), so \(\displaystyle t = 0\)

In Case 1, \(\displaystyle t < u\); in Case 2, \(\displaystyle t > u\)

Multiply both sides of the top equation by 3, and subtract both sides of the second equation.

\(\displaystyle x + 3y = 28\)

\(\displaystyle 3\left ( x + 3y \right )= 3 \cdot 28\)

\(\displaystyle 3 x + 9y \right )= 84\)

\(\displaystyle \underline{3x + 2y = 35}\)

          \(\displaystyle 7y = 49\)

\(\displaystyle 7y \div 7= 49 \div 7\)

\(\displaystyle y = 7\)

Now substitute to find \(\displaystyle x\):

\(\displaystyle x + 3y = 28\)

\(\displaystyle x + 3 \cdot 7 = 28\)

\(\displaystyle x + 21= 28\)

\(\displaystyle x + 21-21= 28-21\)

\(\displaystyle x = 7\)

The two are equal.

Triple both sides of the top equation, and subtract both sides of the bottom equation:

\(\displaystyle 4x +y = 20\)

\(\displaystyle 3 \left ( 4x +y \right ) = 3\left ( 20\right )\)

\(\displaystyle 12x +3y = 60\)

  \(\displaystyle \underline{2x + 3y = 30}\)

\(\displaystyle 10x\)           \(\displaystyle =30\)

\(\displaystyle 10x \div 10 = 30 \div 10\)

\(\displaystyle x = 3\)

Now substitute to find \(\displaystyle y\):

\(\displaystyle 4x +y = 20\)

\(\displaystyle 4 \cdot 3 +y = 20\)

\(\displaystyle 12 +y = 20\)

\(\displaystyle 12 +y -12= 20 -12\)

\(\displaystyle y = 8\)

This makes \(\displaystyle y > x\)

We show that it cannot be determined which of \(\displaystyle t\) and \(\displaystyle u\), if either, is greater, by showing one case in which \(\displaystyle t > u\) and one case in which \(\displaystyle t < u\).

Case 1: \(\displaystyle u = 0\). Then 

\(\displaystyle t = |u - 1| = |0 - 1| = | - 1| = 1\)

and \(\displaystyle t > u\).

Case 2: \(\displaystyle u = 1\). Then 

\(\displaystyle t = |u - 1| = |1 - 1| = | 0| = 0\)

and \(\displaystyle t < u\).

If \(\displaystyle u < 0\), then, since \(\displaystyle t\) is nonnegative, \(\displaystyle t > u\).

If \(\displaystyle u \geq 0\), then \(\displaystyle u + 1 \geq 1 > 0\), so the equation becomes \(\displaystyle t = u + 1\). Therefore, \(\displaystyle t > u\).

Either way, (a) is greater.