7  Solving by operations

In this section we introduce the operation framework, which is a method for converting an equation into an equivalent equation.

7.1 Equal operations preserve equality

The key idea is this: if two things are equal, and we do the same operation to both them, then the results are also equal.

For example:

  • Start with \(2\cdot 3\) and \(6\), which are equal.
  • Perform the operation \(+5\) to both.
  • The result is \(2\cdot 3 + 5\) and \(6+5\), which are also equal.

Here is one way to write this process in mathematical symbols \[ \begin{gathered} 2\cdot 3 = 6 \\ +5\qquad +5 \\ 2\cdot 3 +5 = 6+5 \end{gathered} \]

The previous example is a bit silly, so let’s look at another example that shows how this concept helps us solve equations.

Here is the process in words:

  • We start with the equation \(x+3 = 22\).
  • We perform the operation \(-3\) to both \(x+3\) and to \(22\).
  • The result is that \(x+3-3\) is equal to \(22-3\).
  • But \(+3-3\) cancels itself out, and we are left with \(x = 22-3\).
  • This means that \(x=19\).

Here is one way to write this in math symbols. \[ \begin{gathered} x+3 = 22 \\ -3\qquad -3 \\ x+3-3 = 22-3 \\ x = 22-3 \\ x = 19 \end{gathered} \]

7.2 Practice with operations

Example 7.1 For each equation, decide what operation would improve the equation. Apply that operation and then simplify to find the value of \(x\).

  1. \(5x = 35\)

  2. \(x + 9 = 24\)

  3. \(-3x = 21\)

  4. \(x - 7 = 15\)

  5. \(8x = 72\)

  6. \(x + 12 = 30\)

  7. \(-6x = 42\)

  8. \(4x = -28\)

  9. \(x - 11 = 19\)

  10. \(7x = 91\)

  1. Divide both sides by 5: \(x = 35 \div 5 \Rightarrow x = 7\)

  2. Subtract 9 from both sides: \(x = 24 - 9 \Rightarrow x = 15\)

  3. Divide both sides by -3: \(x = 21 \div (-3) \Rightarrow x = -7\)

  4. Add 7 to both sides: \(x = 15 + 7 \Rightarrow x = 22\)

  5. Divide both sides by 8: \(x = 72 \div 8 \Rightarrow x = 9\)

  6. Subtract 12 from both sides: \(x = 30 - 12 \Rightarrow x = 18\)

  7. Divide both sides by -6: \(x = 42 \div (-6) \Rightarrow x = -7\)

  8. Divide both sides by 4: \(x = -28 \div 4 \Rightarrow x = -7\)

  9. Add 11 to both sides: \(x = 19 + 11 \Rightarrow x = 30\)

  10. Divide both sides by 7: \(x = 91 \div 7 \Rightarrow x = 13\)

7.3 Operations with fractions

In order to solve the equation \[ \frac23x = 6 \] we can apply the operation “multiply by \(\frac{3}{2}\)” to both sides. If we do this, the left side cleans up nicely: \[ \begin{gathered} \frac{3}{2}\cdot\frac{2}{3}x = \frac{3}{2}\cdot 6 \\ x = 9 \end{gathered} \]

Example 7.2 (Practice: operations with fractions)  

  1. \(\dfrac{2}{3}x = 18\)

  2. \(\dfrac{5}{4}x = \dfrac{15}{8}\)

  3. \(-\dfrac{3}{7}x = 12\)

  4. \(\dfrac{4}{5}x = 20\)

  5. \(\dfrac{3}{8}x = \dfrac{9}{16}\)

  6. \(-\dfrac{2}{5}x = \dfrac{6}{7}\)

  7. \(\dfrac{7}{9}x = 14\)

  8. \(-\dfrac{5}{6}x = -\dfrac{10}{3}\)

  1. Multiply both sides by \(\dfrac{3}{2}\) to obtain \(x = 18 \cdot \dfrac{3}{2} \Rightarrow x = 27\)

  2. Multiply both sides by \(\dfrac{4}{5}\) to obtain \(x = \dfrac{15}{8} \cdot \dfrac{4}{5} \Rightarrow x = \dfrac{60}{40} \Rightarrow x = \dfrac{3}{2}\)

  3. Multiply both sides by \(-\dfrac{7}{3}\) to obtain \(x = 12 \cdot \left(-\dfrac{7}{3}\right) \Rightarrow x = -28\)

  4. Multiply both sides by \(\dfrac{5}{4}\) to obtain \(x = 20 \cdot \dfrac{5}{4} \Rightarrow x = 25\)

  5. Multiply both sides by \(\dfrac{8}{3}\) to obtain \(x = \dfrac{9}{16} \cdot \dfrac{8}{3} \Rightarrow x = \dfrac{72}{48} \Rightarrow x = \dfrac{3}{2}\)

  6. Multiply both sides by \(-\dfrac{5}{2}\) to obtain \(x = \dfrac{6}{7} \cdot \left(-\dfrac{5}{2}\right) \Rightarrow x = -\dfrac{30}{14} \Rightarrow x = -\dfrac{15}{7}\)

  7. Multiply both sides by \(\dfrac{9}{7}\) to obtain \(x = 14 \cdot \dfrac{9}{7} \Rightarrow x = 18\)

  8. Multiply both sides by \(-\dfrac{6}{5}\) to obtain \(x = -\dfrac{10}{3} \cdot \left(-\dfrac{6}{5}\right) \Rightarrow x = \dfrac{60}{15} \Rightarrow x = 4\)

7.4 Two-operation solutions

Some equations require two operations in order to solve for the unknown. For example, consider the equation \[ 5x - 3 = 17. \] There are two operations on the left side that we must un-do:

  • The variable \(x\) is multiplied by \(5\), which can be un-done with \(\div 5\).
  • The quantity \(5x\) has \(3\) subtracted, which can be un-donw with \(+3\).

Which operation to do first? The key to deciding is to keep in mind that the operation applies to the entirety of the two sides.

  • If we apply the operation \(\div 5\), then the result is \[ (5x-3)\div 5 = 17\div 5 \] which we can also write as \[ \frac{5x-3}{5} = \frac{17}{5}. \]

  • If we apply the operation \(+3\), then the result is \[ (5x-3)+3 = 17 +3 \] which we can also write as \[ 5x -3 +3 = 17 +3. \]

Which of these operations makes the left side of the equation look simpler?

We choose the second option. Thus our plan is to first add \(3\) and then divide by \(5\). Here is the full solution: \[ \begin{aligned} 5x -3 &= 17 & \qquad &\text{add $3$} \\[10pt] 5x -3 +3 &= 17 +3 & &\text{tidy both sides} \\[10pt] 5x &= 20 & &\text{divide by 5} \\[10pt] \frac{5x}{5} &= \frac{20}{5} & &\text{more cleaning!} \\[10pt] x &= 4 \end{aligned} \]

Which operation to do first?

When deciding which operation to do first, it is important to keep in mind the order of operations. The operations we choose are designed to reverse the operations being done to the variable, and so must happen in the reverse order.

In the previous example, the left side \(5x -3\) means first multiply by \(5\), then subtract \(3\). And so to un-do that, we must first add \(3\) and then divide by \(5\).

7.5 Practice with two operation solutions

Example 7.3 First decide which operations you will do, and in what order. Then use the operations to solve for \(x\).

  1. \(3x + 7 = 25\)

  2. \(4(x + 5) = 32\)

  3. \(\dfrac{x + 3}{2} = 9\)

  4. \(6x - 11 = 31\)

  5. \(5(x - 4) = 35\)

  6. \(\dfrac{x - 7}{3} = 5\)

  1. \(-2x + 9 = 3\)

  2. \(7(x + 2) = 49\)

  3. \(\dfrac{x + 8}{4} = 6\)

  4. \(8x - 15 = 41\)

  5. \(3(x - 6) = 21\)

  6. \(\dfrac{x - 5}{6} = 4\)

Problem 1: \[ \begin{aligned} 3x + 7 &= 25 & \qquad &\text{subtract $7$} \\[10pt] 3x + 7 - 7 &= 25 - 7 & &\text{tidy both sides} \\[10pt] 3x &= 18 & &\text{divide by 3} \\[10pt] \frac{3x}{3} &= \frac{18}{3} & &\text{more cleaning!} \\[10pt] x &= 6 \end{aligned} \]

Problem 2: \[ \begin{aligned} 4(x + 5) &= 32 & \qquad &\text{divide by 4} \\[10pt] \frac{4(x + 5)}{4} &= \frac{32}{4} & &\text{tidy both sides} \\[10pt] x + 5 &= 8 & &\text{subtract 5} \\[10pt] x + 5 - 5 &= 8 - 5 & &\text{more cleaning!} \\[10pt] x &= 3 \end{aligned} \]

Problem 3: \[ \begin{aligned} \frac{x + 3}{2} &= 9 & \qquad &\text{multiply by 2} \\[10pt] 2 \cdot \frac{x + 3}{2} &= 9 \cdot 2 & &\text{tidy both sides} \\[10pt] x + 3 &= 18 & &\text{subtract 3} \\[10pt] x + 3 - 3 &= 18 - 3 & &\text{more cleaning!} \\[10pt] x &= 15 \end{aligned} \]

Problem 4: \[ \begin{aligned} 6x - 11 &= 31 & \qquad &\text{add $11$} \\[10pt] 6x - 11 + 11 &= 31 + 11 & &\text{tidy both sides} \\[10pt] 6x &= 42 & &\text{divide by 6} \\[10pt] \frac{6x}{6} &= \frac{42}{6} & &\text{more cleaning!} \\[10pt] x &= 7 \end{aligned} \]

Problem 5: \[ \begin{aligned} 5(x - 4) &= 35 & \qquad &\text{divide by 5} \\[10pt] \frac{5(x - 4)}{5} &= \frac{35}{5} & &\text{tidy both sides} \\[10pt] x - 4 &= 7 & &\text{add 4} \\[10pt] x - 4 + 4 &= 7 + 4 & &\text{more cleaning!} \\[10pt] x &= 11 \end{aligned} \]

Problem 6: \[ \begin{aligned} \frac{x - 7}{3} &= 5 & \qquad &\text{multiply by 3} \\[10pt] 3 \cdot \frac{x - 7}{3} &= 5 \cdot 3 & &\text{tidy both sides} \\[10pt] x - 7 &= 15 & &\text{add 7} \\[10pt] x - 7 + 7 &= 15 + 7 & &\text{more cleaning!} \\[10pt] x &= 22 \end{aligned} \]

Problem 7: \[ \begin{aligned} -2x + 9 &= 3 & \qquad &\text{subtract $9$} \\[10pt] -2x + 9 - 9 &= 3 - 9 & &\text{tidy both sides} \\[10pt] -2x &= -6 & &\text{divide by $-2$} \\[10pt] \frac{-2x}{-2} &= \frac{-6}{-2} & &\text{more cleaning!} \\[10pt] x &= 3 \end{aligned} \]

Problem 8: \[ \begin{aligned} 7(x + 2) &= 49 & \qquad &\text{divide by 7} \\[10pt] \frac{7(x + 2)}{7} &= \frac{49}{7} & &\text{tidy both sides} \\[10pt] x + 2 &= 7 & &\text{subtract 2} \\[10pt] x + 2 - 2 &= 7 - 2 & &\text{more cleaning!} \\[10pt] x &= 5 \end{aligned} \]

Problem 9: \[ \begin{aligned} \frac{x + 8}{4} &= 6 & \qquad &\text{multiply by 4} \\[10pt] 4 \cdot \frac{x + 8}{4} &= 6 \cdot 4 & &\text{tidy both sides} \\[10pt] x + 8 &= 24 & &\text{subtract 8} \\[10pt] x + 8 - 8 &= 24 - 8 & &\text{more cleaning!} \\[10pt] x &= 16 \end{aligned} \]

Problem 10: \[ \begin{aligned} 8x - 15 &= 41 & \qquad &\text{add $15$} \\[10pt] 8x - 15 + 15 &= 41 + 15 & &\text{tidy both sides} \\[10pt] 8x &= 56 & &\text{divide by 8} \\[10pt] \frac{8x}{8} &= \frac{56}{8} & &\text{more cleaning!} \\[10pt] x &= 7 \end{aligned} \]

Problem 11: \[ \begin{aligned} 3(x - 6) &= 21 & \qquad &\text{divide by 3} \\[10pt] \frac{3(x - 6)}{3} &= \frac{21}{3} & &\text{tidy both sides} \\[10pt] x - 6 &= 7 & &\text{add 6} \\[10pt] x - 6 + 6 &= 7 + 6 & &\text{more cleaning!} \\[10pt] x &= 13 \end{aligned} \]

Problem 12: \[ \begin{aligned} \frac{x - 5}{6} &= 4 & \qquad &\text{multiply by 6} \\[10pt] 6 \cdot \frac{x - 5}{6} &= 4 \cdot 6 & &\text{tidy both sides} \\[10pt] x - 5 &= 24 & &\text{add 5} \\[10pt] x - 5 + 5 &= 24 + 5 & &\text{more cleaning!} \\[10pt] x &= 29 \end{aligned} \]

7.6 Homework exercises

Exercise 7.1 For each equation, decide what operation would improve the equation. Apply that operation to both sides and then simplify to find the value of \(x\).

  1. \(6x = 48\)

  2. \(x + 11 = 35\)

  3. \(-4x = 32\)

  4. \(x - 9 = 18\)

  5. \(9x = 81\)

  6. \(x + 15 = 42\)

  7. \(-7x = 56\)

  8. \(5x = -40\)

  9. \(x - 13 = 27\)

  10. \(8x = 96\)

  1. Divide both sides by 6 to obtain \(x = 48 \div 6 \Rightarrow x = 8\)

  2. Subtract 11 from both sides to obtain \(x = 35 - 11 \Rightarrow x = 24\)

  3. Divide both sides by -4 to obtain \(x = 32 \div (-4) \Rightarrow x = -8\)

  4. Add 9 to both sides to obtain \(x = 18 + 9 \Rightarrow x = 27\)

  5. Divide both sides by 9 to obtain \(x = 81 \div 9 \Rightarrow x = 9\)

  6. Subtract 15 from both sides to obtain \(x = 42 - 15 \Rightarrow x = 27\)

  7. Divide both sides by -7 to obtain \(x = 56 \div (-7) \Rightarrow x = -8\)

  8. Divide both sides by 5 to obtain \(x = -40 \div 5 \Rightarrow x = -8\)

  9. Add 13 to both sides to obtain \(x = 27 + 13 \Rightarrow x = 40\)

  10. Divide both sides by 8 to obtain \(x = 96 \div 8 \Rightarrow x = 12\)

Exercise 7.2 For each equation, decide what operation would improve the equation. Apply that operation to both sides and then simplify to find the value of \(x\).

  1. \(\dfrac{3}{5}x = 15\)

  2. \(\dfrac{7}{4}x = \dfrac{21}{8}\)

  3. \(-\dfrac{4}{9}x = 16\)

  4. \(\dfrac{x}{5} = 8\)

  5. \(\dfrac{5}{6}x = 25\)

  6. \(x + \dfrac{2}{3} = \dfrac{5}{6}\)

  7. \(-\dfrac{3}{8}x = \dfrac{9}{4}\)

  8. \(\dfrac{x}{7} = 12\)

  9. \(-\dfrac{7}{10}x = -\dfrac{14}{5}\)

  10. \(x + \dfrac{3}{4} = \dfrac{7}{8}\)

  1. Multiply both sides by \(\dfrac{5}{3}\) to obtain \(x = 15 \cdot \dfrac{5}{3} \Rightarrow x = 25\)

  2. Multiply both sides by \(\dfrac{4}{7}\) to obtain \(x = \dfrac{21}{8} \cdot \dfrac{4}{7} \Rightarrow x = \dfrac{84}{56} \Rightarrow x = \dfrac{3}{2}\)

  3. Multiply both sides by \(-\dfrac{9}{4}\) to obtain \(x = 16 \cdot \left(-\dfrac{9}{4}\right) \Rightarrow x = -36\)

  4. Multiply both sides by 5 to obtain \(x = 8 \cdot 5 \Rightarrow x = 40\)

  5. Multiply both sides by \(\dfrac{6}{5}\) to obtain \(x = 25 \cdot \dfrac{6}{5} \Rightarrow x = 30\)

  6. Subtract \(\dfrac{2}{3}\) from both sides to obtain \(x = \dfrac{5}{6} - \dfrac{2}{3} \Rightarrow x = \dfrac{5}{6} - \dfrac{4}{6} \Rightarrow x = \dfrac{1}{6}\)

  7. Multiply both sides by \(-\dfrac{8}{3}\) to obtain \(x = \dfrac{9}{4} \cdot \left(-\dfrac{8}{3}\right) \Rightarrow x = -\dfrac{72}{12} \Rightarrow x = -6\)

  8. Multiply both sides by 7 to obtain \(x = 12 \cdot 7 \Rightarrow x = 84\)

  9. Multiply both sides by \(-\dfrac{10}{7}\) to obtain \(x = -\dfrac{14}{5} \cdot \left(-\dfrac{10}{7}\right) \Rightarrow x = \dfrac{140}{35} \Rightarrow x = 4\)

  10. Subtract \(\dfrac{3}{4}\) from both sides to obtain \(x = \dfrac{7}{8} - \dfrac{3}{4} \Rightarrow x = \dfrac{7}{8} - \dfrac{6}{8} \Rightarrow x = \dfrac{1}{8}\)

Exercise 7.3 For each equation, decide which operations to apply and in what order. Show all steps to find the value of \(x\).

  1. \(4x + 9 = 37\)

  2. \(5(x + 3) = 40\)

  3. \(\dfrac{x + 6}{3} = 8\)

  4. \(7x - 12 = 23\)

  5. \(6(x - 5) = 42\)

  6. \(\dfrac{x - 4}{5} = 7\)

  7. \(-3x + 14 = 5\)

  8. \(8(x + 4) = 64\)

  9. \(\dfrac{x + 10}{6} = 5\)

  10. \(9x - 18 = 54\)

Problem 1: \[ \begin{aligned} 4x + 9 &= 37 & \qquad &\text{subtract $9$} \\[10pt] 4x + 9 - 9 &= 37 - 9 & &\text{tidy both sides} \\[10pt] 4x &= 28 & &\text{divide by 4} \\[10pt] \frac{4x}{4} &= \frac{28}{4} & &\text{more cleaning!} \\[10pt] x &= 7 \end{aligned} \]

Problem 2: \[ \begin{aligned} 5(x + 3) &= 40 & \qquad &\text{divide by 5} \\[10pt] \frac{5(x + 3)}{5} &= \frac{40}{5} & &\text{tidy both sides} \\[10pt] x + 3 &= 8 & &\text{subtract 3} \\[10pt] x + 3 - 3 &= 8 - 3 & &\text{more cleaning!} \\[10pt] x &= 5 \end{aligned} \]

Problem 3: \[ \begin{aligned} \frac{x + 6}{3} &= 8 & \qquad &\text{multiply by 3} \\[10pt] 3 \cdot \frac{x + 6}{3} &= 8 \cdot 3 & &\text{tidy both sides} \\[10pt] x + 6 &= 24 & &\text{subtract 6} \\[10pt] x + 6 - 6 &= 24 - 6 & &\text{more cleaning!} \\[10pt] x &= 18 \end{aligned} \]

Problem 4: \[ \begin{aligned} 7x - 12 &= 23 & \qquad &\text{add $12$} \\[10pt] 7x - 12 + 12 &= 23 + 12 & &\text{tidy both sides} \\[10pt] 7x &= 35 & &\text{divide by 7} \\[10pt] \frac{7x}{7} &= \frac{35}{7} & &\text{more cleaning!} \\[10pt] x &= 5 \end{aligned} \]

Problem 5: \[ \begin{aligned} 6(x - 5) &= 42 & \qquad &\text{divide by 6} \\[10pt] \frac{6(x - 5)}{6} &= \frac{42}{6} & &\text{tidy both sides} \\[10pt] x - 5 &= 7 & &\text{add 5} \\[10pt] x - 5 + 5 &= 7 + 5 & &\text{more cleaning!} \\[10pt] x &= 12 \end{aligned} \]

Problem 6: \[ \begin{aligned} \frac{x - 4}{5} &= 7 & \qquad &\text{multiply by 5} \\[10pt] 5 \cdot \frac{x - 4}{5} &= 7 \cdot 5 & &\text{tidy both sides} \\[10pt] x - 4 &= 35 & &\text{add 4} \\[10pt] x - 4 + 4 &= 35 + 4 & &\text{more cleaning!} \\[10pt] x &= 39 \end{aligned} \]

Problem 7: \[ \begin{aligned} -3x + 14 &= 5 & \qquad &\text{subtract $14$} \\[10pt] -3x + 14 - 14 &= 5 - 14 & &\text{tidy both sides} \\[10pt] -3x &= -9 & &\text{divide by $-3$} \\[10pt] \frac{-3x}{-3} &= \frac{-9}{-3} & &\text{more cleaning!} \\[10pt] x &= 3 \end{aligned} \]

Problem 8: \[ \begin{aligned} 8(x + 4) &= 64 & \qquad &\text{divide by 8} \\[10pt] \frac{8(x + 4)}{8} &= \frac{64}{8} & &\text{tidy both sides} \\[10pt] x + 4 &= 8 & &\text{subtract 4} \\[10pt] x + 4 - 4 &= 8 - 4 & &\text{more cleaning!} \\[10pt] x &= 4 \end{aligned} \]

Problem 9: \[ \begin{aligned} \frac{x + 10}{6} &= 5 & \qquad &\text{multiply by 6} \\[10pt] 6 \cdot \frac{x + 10}{6} &= 5 \cdot 6 & &\text{tidy both sides} \\[10pt] x + 10 &= 30 & &\text{subtract 10} \\[10pt] x + 10 - 10 &= 30 - 10 & &\text{more cleaning!} \\[10pt] x &= 20 \end{aligned} \]

Problem 10: \[ \begin{aligned} 9x - 18 &= 54 & \qquad &\text{add $18$} \\[10pt] 9x - 18 + 18 &= 54 + 18 & &\text{tidy both sides} \\[10pt] 9x &= 72 & &\text{divide by 9} \\[10pt] \frac{9x}{9} &= \frac{72}{9} & &\text{more cleaning!} \\[10pt] x &= 8 \end{aligned} \]