Solution.

1. The variable we are asked to solve for is \(x\) so our first move is to gather all of the terms involving \(x\) on one side and put the remaining terms on the other. ^[3] (In the margin notes, when we speak of operations, e.g.,‘Subtract \(7x\),’ we mean to subtract \(7x\) from both sides of the equation. The ‘from both sides of the equation’ is omitted in the interest of spacing.)

   \[ \begin{array}{rclr} 3x - 6 & = & 7x + 4 & \\ (3x-6) - 7x + 6 & = & (7x+4) -7x +6 & \text {Subtract $7x$, add $6$} \\ 3x - 7x - 6 + 6 & = & 7x - 7x + 4 + 6 & \text {Rearrange terms} \\ -4x & = & 10 & \text {$3x-7x = (3-7)x = -4x$} \\ \frac {-4x}{-4} & = & \frac {10}{-4} & \text {Divide by the coefficient of $x$} \\ x & = & - \frac {5}{2} & \text {Reduce to lowest terms} \\ \end{array}\]

   To check our answer, we substitute \(x = -\frac {5}{2}\) into each side of the orginial equation to see the equation is satisfied. Sure enough, \(3\left (-\frac {5}{2}\right ) - 6 = -\frac {27}{2}\) and \(7\left (-\frac {5}{2}\right ) + 4 = -\frac {27}{2}\).

2. In our next example, the unknown is \(t\) and we not only have a fraction but also a decimal to wrangle. Fortunately, with equations we can multiply both sides to rid us of these computational obstacles:

   \[ \begin{array}{rclr} 3 - 1.7t & = & \frac {t}{4} & \\ 40(3 - 1.7t) & = & 40 \left ( \frac {t}{4}\right ) & \text {Multiply by $40$} \\ 40(3) - 40(1.7t) & = & \frac {40 t}{4} & \text {Distribute} \\ 120 - 68t & = & 10 t & \\ (120 -68t) + 68 t & = & 10t + 68 t & \text {Add $68t$ to both sides} \\ 120 & = & 78 t & \text {$68t + 10t = (68 + 10)t = 78 t$}\\ \frac {120}{78} & = & \frac {78t}{78} & \text {Divide by the coefficient of $t$}\\ \frac {120}{78} & = & t & \\ \frac {20}{13} & = & t & \text {Reduce to lowest terms} \\ \end{array}\]

   To check, we again substitute \(t = \frac {20}{13}\) into each side of the original equation. We find that \(3 - 1.7 \left (\frac {20}{13}\right ) = 3 - \left (\frac {17}{10}\right )\left (\frac {20}{13}\right ) = \frac {5}{13}\) and \(\frac {(20/13)}{4} = \frac {20}{13} \cdot \frac {1}{4} = \frac {5}{13}\) as well.

3. To solve this next equation, we begin once again by clearing fractions. The least common denominator here is \(36\):

   \[ \begin{array}{rclr} \frac {1}{18}(7 - 4a) + 2 & = & \frac {a}{3} - \frac {4-a}{12} & \\ 36 \left ( \frac {1}{18}(7 - 4a) + 2\right ) & = & 36 \left ( \frac {a}{3} - \frac {4-a}{12}\right ) & \text {Multiply by $36$} \\[13pt] \frac {36}{18} (7-4a) + (36)(2) & = & \frac {36a}{3} - \frac {36(4-a)}{12} & \text {Distribute} \\ 2(7-4a) + 72 & = & 12 a - 3(4-a) & \text {Distribute} \\ 14 - 8a + 72 & = & 12a - 12 + 3a & \\ 86 - 8a & = & 15 a - 12 & \text {$12 a + 3a = (12+3)a = 15a$} \\ (86-8a)+8a+12 & = & (15a-12) + 8a + 12 & \text {Add $8a$ and $12$} \\ 86 + 12 - 8a + 8a & = & 15a + 8a - 12 + 12 & \text {Rearrange terms} \\ 98 & = & 23 a & \text {$15a + 8a = (15+8)a = 23a$} \\ \frac {98}{23} & = & \frac {23a}{23} & \text {Divide by the coefficient of $a$} \\ \frac {98}{23} & = & a & \end{array} \]

   The check, as usual, involves substituting \(a = \frac {98}{23}\) into both sides of the original equation. The reader is encouraged to work through the (admittedly messy) arithmetic. Both sides work out to \(\frac {199}{138}\).

4. The square roots may dishearten you but we treat them just like the real numbers they are. Our strategy is the same: get everything with the variable (in this case \(y\)) on one side, put everything else on the other and divide by the coefficient of the variable. We’ve added a few steps to the narrative that we would ordinarily omit just to help you see that this equation is indeed linear.

   \[ \begin{array}{rclr} 8 y \sqrt {3} + 1 & = & 7 - \sqrt {12}(5 - y) & \\ 8 y \sqrt {3} + 1 & = & 7 - \sqrt {12}(5) + \sqrt {12} y & \text {Distribute} \\ 8 y \sqrt {3} + 1 & = & 7 - (2 \sqrt {3})5 + (2 \sqrt {3})y & \text {$\sqrt {12} = \sqrt {4\cdot 3} = 2 \sqrt {3}$} \\ 8 y \sqrt {3} + 1 & = & 7 - 10 \sqrt {3} + 2y \sqrt {3} & \\ (8 y \sqrt {3} + 1) - 1 - 2y\sqrt {3} & = & (7 - 10 \sqrt {3} + 2y \sqrt {3}) - 1 - 2y\sqrt {3} & \text {Subtract $1$ and $2y\sqrt {3}$} \\ 8 y \sqrt {3} - 2y\sqrt {3} + 1 - 1 & = & 7 - 1 - 10 \sqrt {3} + 2y \sqrt {3} - 2y\sqrt {3} & \text {Rearrange terms} \\ (8\sqrt {3}-2\sqrt {3})y & = & 6 - 10 \sqrt {3} & \\ 6 y \sqrt {3} & = & 6 - 10 \sqrt {3} & \text {See note below} \\ \frac {6 y \sqrt {3}}{6 \sqrt {3}} & = & \frac {6 - 10 \sqrt {3}}{6 \sqrt {3}} & \text {Divide $6\sqrt {3}$} \\ y & = & \frac {2 \cdot \sqrt {3} \cdot \sqrt {3} - 2 \cdot 5 \cdot \sqrt {3}}{2 \cdot 3 \cdot \sqrt {3}} & \\ y & = & \frac {\cancel {2} \cancel {\sqrt {3}}(\sqrt {3} - 5)}{\cancel {2} \cdot 3 \cdot \cancel {\sqrt {3}}} & \text {Factor and cancel} \\ y & = & \frac {\sqrt {3} - 5}{3} & \\ \end{array}\]
   In the list of computations above we marked the row \(6 y \sqrt {3} = 6 - 10 \sqrt {3}\) with a note. That’s because we wanted to draw your attention to this line without breaking the flow of the manipulations. The equation \(6 y \sqrt {3} = 6 - 10 \sqrt {3}\) is in fact linear according to Definition 1: the variable is \(y\), the value of \(A\) is \(6\sqrt {3}\) and \(B = 6 - 10 \sqrt {3}\). Checking the solution, while not trivial, is good mental exercise. Each side works out to be \(\frac {27 - 40 \sqrt {3}}{3}\).

5. Proceeding as before, we simplify radicals and clear denominators. Once we gather all of the terms containing \(x\) on one side and move the other terms to the other, we factor out \(x\) to identify its coefficient then divide to get our answer.

   \[ \begin{array}{rclr} \frac {3x-1}{2} & = & x\sqrt {50} + 4 & \\ \frac {3x-1}{2} & = & 5x\sqrt {2} + 4 & \text {$\sqrt {50} = \sqrt {25 \cdot 2}$} \\ 2\left ( \frac {3x-1}{2}\right ) & = & 2 \left (5x\sqrt {2} + 4 \right ) & \text {Multiply by $2$} \\ \frac {\cancel {2} \cdot (3x-1)}{\cancel {2}} & = & 2 (5x \sqrt {2}) + 2\cdot 4 & \text {Distribute} \\ 3x - 1 & = & 10x \sqrt {2} + 8 & \\ (3x-1) - 10x\sqrt {2} + 1 & = & (10x \sqrt {2} + 8) - 10x\sqrt {2} + 1 & \text {Subtract $10x\sqrt {2}$, add $1$} \\ 3x - 10x\sqrt {2} - 1 + 1 & = & 10x\sqrt {2} - 10x\sqrt {2} + 8 + 1 & \text {Rearrange terms} \\ 3x - 10x\sqrt {2} & = & 9 & \\ (3 - 10\sqrt {2}) x & = & 9 & \text {Factor} \\ \frac {(3 - 10\sqrt {2}) x}{3 - 10\sqrt {2}} & = & \frac {9}{3 - 10\sqrt {2}} & \text {Divide by the coefficient of $x$} \\ x & = & \frac {9}{3 - 10\sqrt {2}} & \\ \end{array} \]
   The reader is encouraged to check this solution - it isn’t as bad as it looks if you’re careful! Each side works out to be \( \frac {12 + 5\sqrt {2}}{3-10\sqrt {2}}\).

6. If we were instructed to solve our last equation for \(x\), we’d be done in one step: divide both sides by \((4-y)\) - assuming \(4-y \neq 0\), that is. Alas, we are instructed to solve for \(y\), which means we have some more work to do.

   \[ \begin{array}{rclr} x(4-y) & = & 8y & \\ 4x - xy & = & 8y & \text {Distribute} \\ (4x - xy) + xy & = & 8y + xy & \text {Add $xy$} \\ 4x & = & (8+x)y & \text {Factor} \\ \end{array}\]
   In order to finish the problem, we need to divide both sides of the equation by the coefficient of \(y\) which in this case is \(8+x\). This expression contains a variable so we need to stipulate that we may perform this division only if \(8 + x \neq 0\), or, in other words, \(x \neq -8\). Hence, we write our solution as:
   \[ y = \frac {4x}{8+x}, \quad \text {provided $x \neq -8$}\]
   What happens if \(x = -8\)? Substituting \(x = -8\) into the original equation gives \((-8)(4-y) = 8y\) or \(-32 + 8y = 8y\). This reduces to \(-32 = 0\), which is a contradiction. This means there is no solution when \(x = -8\), so we’ve covered all the bases. Checking our answer requires some Algebra we haven’t reviewed yet in this text, but the necessary skills should be lurking somewhere in the mathematical mists of your mind. The adventurous reader is invited to plug \(y = \frac {4x}{8 + x}\) into the original equation and show that both sides work out to \(\frac {32x}{x + 8}\). □

1. Solve for \(w\): \(2.1 - 0.01w \leq -3\) or \(2.1-0.01w \geq 3\)

Solution.

1. We begin by clearing denominators and gathering all of the terms containing \(x\) to one side of the inequality and putting the remaining terms on the other.

   \[ \begin{array}{rclr} \frac {7-8x}{2} & \geq & 4x + 1 & \\ 2\left ( \frac {7-8x}{2}\right ) & \geq & 2(4x + 1) & \text {Multiply by $2$} \\ \frac {\cancel {2}(7-8x)}{\cancel {2}} & \geq & 2(4x) + 2(1) & \text {Distribute} \\ 7 - 8x & \geq & 8x + 2 & \\ (7-8x) + 8x-2 & \geq & 8x+2 + 8x -2 & \text {Add $8x$, subtract $2$} \\ 7 - 2 - 8x + 8x & \geq & 8x + 8x + 2 - 2 & \text {Rearrange terms} \\ 5 & \geq & 16x & \text {$8x + 8x = (8+8)x = 16x$} \\ \frac {5}{16} & \geq & \frac {16x}{16} & \text {Divide by the coefficient of $x$} \\ \frac {5}{16} & \geq & x & \\ \end{array} \]

   We get \(\frac {5}{16} \geq x\) or, said differently, \(x \leq \frac {5}{16}\). We express this set ^[4] (Using set-builder notation, our ‘set’ of solutions here is \(\{ x \, | \, x \leq \frac {5}{16} \}\).) of real numbers as \(\left (-\infty , \frac {5}{16}\right ]\). Though not required to do so, we could partially check our answer by substituting \(x = \frac {5}{16}\) and a few other values in our solution set (\(x =0\), for instance) to make sure the inequality holds. (It also isn’t a bad idea to choose an \(x > \frac {5}{16}\), say \(x = 1\), to see that the inequality doesn’t hold there.) The only real way to actually show that our answer works for all values in our solution set is to start with \(x \leq \frac {5}{16}\) and reverse all of the steps in our solution procedure to prove it is equivalent to our original inequality.

2. We have our first example of a ‘compound’ inequality. The solutions to

   \[ \frac {3}{4} \leq \frac {7-y}{2} < 6 \]
   must satisfy
   \[ \frac {3}{4} \leq \frac {7-y}{2} \qquad \text {AND} \qquad \frac {7-y}{2} < 6\]

   One approach is to solve each of these inequalities separately, then intersect their solution sets. While this method works (and will be used later for more complicated problems), our variable \(y\) appears only in the middle expression so we can proceed by working both inequalities at once:

   \[ \begin{array}{rclr} \frac {3}{4} \leq & \frac {7-y}{2} & < 6 & \\ 4\left ( \frac {3}{4} \right ) \leq & 4\left ( \frac {7-y}{2}\right ) & < 4(6) & \text {Multiply by $4$} \\ \frac {\cancel {4} \cdot 3}{\cancel {4}} \leq & \frac {\cancelto {2}{4}(7-y)}{\cancel {2}} & < 24 & \\ 3 \leq & 2(7-y) & < 24 & \\ 3 \leq & 2(7)-2y & < 24 & \text {Distrbute}\\ 3 \leq & 14-2y & < 24 & \\ 3 -14 \leq & (14-2y) - 14 & < 24 - 14 & \text {Subtract $14$}\\ -11 \leq & -2y & < 10 & \\ \frac {-11}{-2} \geq & \frac {-2y}{-2} & > \frac {10}{-2} & \text {Divide by the coefficient of $y$} \\ & & & \text {Reverse inequalities} \\ \frac {11}{2} \geq & y & > -5 & \\ \end{array} \]

   Our final answer is \(\frac {11}{2} \geq y > -5\), or, said differently, \(-5 < y \leq \frac {11}{2}\). In interval notation, this is \(\left ( -5, \frac {11}{2} \right ]\). We could check the reasonableness of our answer as before, and the reader is encouraged to do so.

3. We have another compound inequality and what distinguishes this one from our previous example is that ‘\(t\)’ appears on both sides of both inequalities. In this case, we need to create two separate inequalities and find all of the real numbers \(t\) which satisfy both \(2t-1 \leq 4-t\) and \(4-t < 6t + 1\). The first inequality, \(2t-1 \leq 4-t\), reduces to \(3t \leq 5\) or \(t \leq \frac {5}{3}\). The second inequality, \(4-t < 6t+1\), becomes \(3 < 7t\) which reduces to \(t > \frac {3}{7}\). Thus our solution is all real numbers \(t\) with \(t \leq \frac {5}{3}\) and \(t > \frac {3}{7}\), or, writing this as a compound inequality, \(\frac {3}{7} < t \leq \frac {5}{3}\). Using interval notation, ^[5] (If we intersect the solution sets of the two individual inequalities, we get the answer, too: \(\left (-\infty , \frac {5}{3}\right ] \cap \left (\frac {3}{7}, \infty \right ) = \left ( \frac {3}{7}, \frac {5}{3} \right ]\).) we express our solution as \(\left ( \frac {3}{7}, \frac {5}{3} \right ]\).

4. As before, with this inequality we have no choice but to solve each inequality individually and intersect the solution sets. Starting with the leftmost inequality, we first note that the in the term \(\sqrt {7} x\), the vinculum of the square root extends over the \(7\) only, meaning the \(x\) is not part of the radicand. In order to avoid confusion, we will write \(\sqrt {7} x\) as \(x \sqrt {7}\).

   \[ \begin{array}{rclr} 5 + x\sqrt {7} & \leq & 4x+1 & \\ (5 + x\sqrt {7} ) -4x - 5 & \leq& (4x + 1) - 4x - 5 & \text {Subtract $4x$ and $5$} \\ x\sqrt {7} - 4x + 5 - 5 & \leq & 4x - 4x + 1 - 5 & \text {Rearrange terms} \\ x(\sqrt {7} - 4) & \leq & -4 & \text {Factor} \\ \end{array} \]
   At this point, we need to exercise a bit of caution because the number \(\sqrt {7} - 4\) is negative. ^[6] (Since \(4 < 7 < 9\), it stands to reason that \(\sqrt {4} < \sqrt {7} < \sqrt {9}\) so \(2 < \sqrt {7} < 3\).) When we divide by it the inequality reverses:
   \[ \begin{array}{rclr} x(\sqrt {7} - 4) & \leq & -4 & \\ \frac {x(\sqrt {7}-4)}{\sqrt {7}-4} & \geq & \frac {-4}{\sqrt {7} - 4} & \text {Divide by the coefficient of $x$} \\ [-8pt] & & & \text {Reverse inequalities} \\ x & \geq & \frac {-4}{\sqrt {7} - 4} & \\ x & \geq & \frac {-4}{-(4 - \sqrt {7})} & \\ x & \geq & \frac {4}{4 -\sqrt {7}} & \\ \end{array} \]
   We’re only half done because we still have the rightmost inequality to solve. Fortunately, that one seems rather mundane: \(4x+1 \leq 8\) reduces to \(x \leq \frac {7}{4}\) without too much incident. Our solution is \( x \geq \frac {4}{4-\sqrt {7}}\) and \(x \leq \frac {7}{4}\). We may be tempted to write \(\frac {4}{4-\sqrt {7}} \leq x \leq \frac {7}{4}\) and call it a day but that would be nonsense! To see why, notice that \(\sqrt {7}\) is between \(2\) and \(3\) so \(\frac {4}{4 - \sqrt {7}}\) is between \(\frac {4}{4-2} = 2\) and \(\frac {4}{4-3} = 4\). In particular, we get \(\frac {4}{4 - \sqrt {7}} > 2\). On the other hand, \(\frac {7}{4} < 2\). This means that our ‘solutions’ have to be simultaneously greater than \(2\) AND less than \(2\) which is impossible. Therefore, this compound inequality has no solution, which means we did all that work for nothing. ^[7] (Much like how people walking on treadmills get nowhere. Math is the endurance cardio of the brain, folks!)
5. Our last example is yet another compound inequality but here, instead of the two inequalities being connected with the conjunction ‘and’, they are connected with ‘or’, which indicates that we need to find the union of the results of each. Starting with \(2.1 - 0.01w \leq -3\), we get \(-0.01 w \leq -5.1\), which gives ^[8] (Don’t forget to flip the inequality!) \(w \geq 510\). The second inequality, \(2.1-0.01w \geq 3\), becomes \(-0.01w \geq 0.9\), which reduces to \(w \leq -90\). Our solution set consists of all real numbers \(w\) with \(w \geq 510\) or \(w \leq -90\). In interval notation, this is \((-\infty , -90] \cup [510, \infty )\). □