\(\displaystyle |4y - 27| = 80\), so either

\(\displaystyle 4y - 27 = - 80\) or \(\displaystyle 4y - 27 = 80\)

We solve both equations separately:

\(\displaystyle 4y - 27 = - 80\)

\(\displaystyle 4y - 27 + 27 = - 80 + 27\)

\(\displaystyle 4y = -53\)

\(\displaystyle \frac{4y}{4} = \frac{-53}{4}\)

\(\displaystyle y = -13 \frac{1}{4}\)

\(\displaystyle 4y - 27 = 80\)

\(\displaystyle 4y - 27 + 27 = 80 + 27\)

\(\displaystyle 4y = 107\)

\(\displaystyle \frac{4y}{4} = \frac{107}{4}\)

\(\displaystyle y = 26 \frac{3}{4}\)

Since the negative solution is being requested, we choose \(\displaystyle y = -13 \frac{1}{4}\).

\(\displaystyle |3t + 12| - 15 = 34\)

\(\displaystyle |3t + 12| - 15 + 15 = 34 + 15\)

\(\displaystyle |3t + 12| =49\)

Either

\(\displaystyle 3t + 12 = -49\) or \(\displaystyle 3t + 12 = 49\),

so we solve the equations separately:

\(\displaystyle 3t + 12 = -49\)

\(\displaystyle 3t + 12 - 12 = -49 - 12\)

\(\displaystyle 3t = -61\)

\(\displaystyle \frac{3t}{3} =\frac{ -61}{3}\)

\(\displaystyle t = -20 \frac{1}{3}\)

or

\(\displaystyle 3t + 12 = 49\)

\(\displaystyle 3t + 12 - 12 = 49 - 12\)

\(\displaystyle 3t = 37\)

\(\displaystyle \frac{3t}{3} =\frac{37}{3}\)

\(\displaystyle t = 12\frac{1}{3}\)

The solution set is \(\displaystyle \left \{-20 \frac{1}{3},12\frac{1}{3} \right \}\)

Since the absolute value of a nonnegative number is the number itself, and the absolute value of a negative number is its (positive) opposite, we have to examine up to three cases:  \(\displaystyle x \ge 0\), \(\displaystyle -5 \le x < 0\), and \(\displaystyle x < -5\).

However, let us examine that third case. 

This makes \(\displaystyle x\) and \(\displaystyle x+5\) negative, so the equation can be rewritten:

\(\displaystyle |x + 5| = |x | - 5\)

\(\displaystyle -(x+5) = -x - 5\)

\(\displaystyle -x-5= -x - 5\)

This statement is identically true. Therefore, all values of \(\displaystyle x\) less than \(\displaystyle -5\) work, and we have already proved that there are infinitely many solutions. We do not need to go further.

\(\displaystyle t+ 13 = u + 8\),

so 

\(\displaystyle t+ 13- 13 = u + 8 - 13\)

\(\displaystyle t = u - 5\)

\(\displaystyle u + 4 = v - 7\)

\(\displaystyle u + 4 - 4 = v - 7 - 4\)

\(\displaystyle u = v-11\)

\(\displaystyle v- 11 = w - 6\)

\(\displaystyle v- 11 + 11= w - 6 + 11\)

\(\displaystyle v = w+5\)

Using two substitutions:

\(\displaystyle t = u - 5\)

\(\displaystyle t = v-11 - 5 = v-16\)

\(\displaystyle t = v-16 = w+5 -16 = w -11\)

The correct choice is \(\displaystyle t = w -11\).

\(\displaystyle 3t = 7u\)

\(\displaystyle \frac{3t}{3} =\frac{ 7u}{3}\)

\(\displaystyle t =\frac{ 7}{3}u\)

Similarly,

\(\displaystyle 4u = 7v\)

\(\displaystyle u = \frac{7}{4}v\)

\(\displaystyle 9v = 2w\)

\(\displaystyle v = \frac{2}{9}w\)

By substitution:

\(\displaystyle t =\frac{ 7}{3}u =\frac{ 7}{3} \cdot \frac{7}{4}v =\frac{ 7}{3} \cdot \frac{7}{4} \cdot \frac{2}{9}w\)

\(\displaystyle t = \frac{49}{54}w\)

\(\displaystyle 54 \cdot t =54 \cdot \frac{49}{54}w\)

\(\displaystyle 54t = 49 w\)

\(\displaystyle 8 ^{3x- 8} = 4 ^{y}\)

\(\displaystyle \left (2^{3} \right )^{3x- 8} = (2^{2}) ^{y}\)

\(\displaystyle 2^{3(3x- 8)} = 2^{2y}\)

\(\displaystyle 2y = 3(3x-8)\)

\(\displaystyle 2y = 9x-24\)

\(\displaystyle \frac{2y }{2}= \frac{9x-24}{2}\)

\(\displaystyle y = \frac{9}{2}x- 12\)

\(\displaystyle 6 ^{x} \cdot 36 ^{y} = 216 ^{t}\)

\(\displaystyle 6 ^{x} \cdot\left ( 6 ^{2} \right )^{y} = \left (6^{3} \right )^{t}\)

\(\displaystyle 6 ^{x} \cdot 6 ^{2 y} = 6^{3 t}\)

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

\(\displaystyle 3t = x+2y\)

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

\(\displaystyle t = \frac{1}{3}x+ \frac{2}{3}y\)

\(\displaystyle \frac{4 ^{x} }{8 ^{y} } = 16^{t}\)

\(\displaystyle \frac{\left (2^{2} \right )^{x} }{\left (2^{3} \right ) ^{y} } =\left ( 2^{4} \right ) ^{t}\)

\(\displaystyle \frac{ 2^{2x} }{ 2^{3y} } = 2^{4t}\)

\(\displaystyle 2^{2x -3y} = 2^{4t}\)

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

\(\displaystyle \frac{4t }{4}= \frac{2x-3y}{4}\)

\(\displaystyle t = \frac{1}{2}x- \frac{3}{4}y\)

\(\displaystyle (g \circ f) (-3) = g [f(-3)]\)

From the diagram below, it can be seen that \(\displaystyle f(-3) = 2\).

\(\displaystyle (g \circ f) (-3) = g [f(-3)] = g(2)\)

\(\displaystyle g(x) = 4x+8\), so

\(\displaystyle g(x) = 4 (2)+8 = 8 + 8 = 16\)

Therefore, 

\(\displaystyle (g \circ f) (-3) = 16\).

\(\displaystyle \left (f \circ g \right ) (-3) = f \left [ g (-3)\right ]\)

\(\displaystyle g(x) = 4x+8\)

\(\displaystyle g(-3) = 4 (-3)+8 = -12 + 8 = -4\)

\(\displaystyle f \left [ g (-3)\right ] = f(-4)\).

Examine the diagram below.

As can be seen, \(\displaystyle f(-4)= 4\). Therefore, \(\displaystyle (f \circ g) (-3) = 4\).