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.

\(\displaystyle x + y = 10\)

By the commutative property, 

\(\displaystyle y + x = 10\) 

By the multiplication property of equality,

\(\displaystyle 5 \left ( y + x \right )= 5\cdot 10\)

By distribution,

\(\displaystyle 5y + 5x = 5 0\)

If \(\displaystyle N > 1\), then \(\displaystyle N ^{3}+ N > N ^{3}= N ^{2} \cdot N > N ^{2} \cdot 1 = N ^{2}\)

If \(\displaystyle N = 1\), then \(\displaystyle N ^{3}+ N = 1 ^{3}+1 = 2 > 1 = 1^{2} = N ^{2}\)

If \(\displaystyle 0 < N < 1\), then \(\displaystyle N ^{3}+ N > N = N \cdot 1 > N \cdot N = N ^{2}\)

Therefore, regardless of the value of \(\displaystyle N\), as long as \(\displaystyle N\) is positive, \(\displaystyle N ^{3} + N > N ^{2}\), and (b) is greater.

\(\displaystyle 100 x + 100 y = \frac{1}{5}\)

From the multiplication property of equality,

\(\displaystyle \frac{1}{100} \cdot \left ( 100 x + 100 y \right ) = \frac{1}{100} \cdot \frac{1}{5}\).

By distribution,

\(\displaystyle x + y = \frac{1}{500} = 0.002 < 0.005\).

\(\displaystyle 5t = 6u\)

\(\displaystyle 5t \div 5= 6u\div 5\)

\(\displaystyle t = \frac{6}{5} u\)

Take the absolute value of both sides:

\(\displaystyle \left |t \right|= \left | \frac{6}{5} u \right |\)

\(\displaystyle \left |t \right|= \frac{6}{5} \left | u \right | > \left | u \right |\)

This makes (a) greater.

(a) \(\displaystyle 4 A + 4 B = \frac{1}{25}\)

\(\displaystyle \frac{1}{4} \cdot \left ( 4 A + 4 B \right ) = \frac{1}{4} \cdot \frac{1}{25}\)

\(\displaystyle A + B = \frac{1}{100 }\)

(b) \(\displaystyle 10 C +10 D = 0.1\)

Rewrite as a fraction:

\(\displaystyle 10 C +10 D = \frac{1}{10 }\)

\(\displaystyle \frac{1}{10 } \cdot \left ( 10 C +10 D \right ) = \frac{1}{10 } \cdot \frac{1}{10 }\)

\(\displaystyle C + D= \frac{1}{100 }\)

\(\displaystyle A + B = C + D\)

(a) \(\displaystyle f(5)\) can be evaluated by using the definition of \(\displaystyle f\) for positive \(\displaystyle x\):

\(\displaystyle f (x) = x - 3\)

\(\displaystyle f (5) =5 - 3 = 2\)

\(\displaystyle f(-5)\) can be evaluated by using the definition of \(\displaystyle f\) for nonpositive \(\displaystyle x\):

\(\displaystyle f (x) = x + 3\)

\(\displaystyle f (-5) =-5 + 3 = -2\)

Add: \(\displaystyle f(5) + f(-5) = 2 + (-2) = 0\)

(b) \(\displaystyle f(0)\) can be evaluated by using the definition of \(\displaystyle f\) for nonpositive \(\displaystyle x\):

\(\displaystyle f (0) = 0 + 3 = 3\)

\(\displaystyle f (0) > f(5) + f(-5)\)