6  Equivalent equations

6.1 Equivalent statements

Vocabulary: equivalent statements

We say that the statements \[ 12\div 4 = 3\quad\text{ and }\quad 12 = 4\cdot 3 \] are equivalent statements because they express the same basic fact, but with opposite operations.

Example 6.1 (Practice: equivalent statements) For each, write an equivalent statement using the opposite operation. There might be more than one possible answer.

  1. \(15 + 7 = 22\)

  2. \(8 \cdot 5 = 40\)

  3. \(30 - 12 = 18\)

  4. \(56 \div 8 = 7\)

  5. \(9 + 14 = 23\)

  6. \(45 \div 5 = 9\)

  7. \(7 \cdot 6 = 42\)

  8. \(25 - 11 = 14\)

  1. \(22 - 7 = 15\) or \(22 - 15 = 7\)

  2. \(40 \div 5 = 8\) or \(40 \div 8 = 5\)

  3. \(18 + 12 = 30\) or \(30 = 18 + 12\)

  4. \(56 = 7 \cdot 8\) or \(56 = 8 \cdot 7\)

  5. \(23 - 14 = 9\) or \(23 - 9 = 14\)

  6. \(45 = 9 \cdot 5\) or \(45 = 5 \cdot 9\)

  7. \(42 \div 6 = 7\) or \(42 \div 7 = 6\)

  8. \(14 + 11 = 25\) or \(25 = 14 + 11\)

6.2 Equivalent scenarios

If Jacob drives 120 miles in two hours, then we can compute his average speed by \[ \frac{120\text{ miles}}{2\text{ hours}} = 60\text{ miles/hour}. \]

An equivalent scenario is that Jordan averages 60 miles per hour over the course of two hours, with the result that the distance traveled is \[ (60\text{ miles/hour})\cdot (2\text{ hours}) = 120\text{ miles}. \]

Example 6.2 (Practice: equivalent scenarios) For each scenario, write an equivalent scenario using the opposite operation.

  1. Maria runs 15 miles in 3 hours. Her average speed is \(\dfrac{15\text{ miles}}{3\text{ hours}} = 5\text{ miles/hour}\).

  2. A worker earns $25 per hour and works for 8 hours.

    The total earnings are \((25\text{ dollars/hour}) \cdot (8\text{ hours}) = 200\text{ dollars}\).

  3. A recipe uses 12 cups of flour to make 4 batches of cookies. The amount of flour per batch is \(\dfrac{12\text{ cups}}{4\text{ batches}} = 3\text{ cups/batch}\).

  4. Water flows from a hose at 6 gallons per minute for 10 minutes. The total amount of water is \((6\text{ gallons/minute}) \cdot (10\text{ minutes}) = 60\text{ gallons}\).

  5. Rachel travels by bicycle 80 kilometers in 4 hours.

    Her average speed is \(\dfrac{80\text{ km}}{4\text{ hours}} = 20\text{ km/hour}\).

  1. Equivalent scenario: Maria runs at an average speed of 5 miles per hour for 3 hours. The total distance traveled is \((5\text{ miles/hour}) \cdot (3\text{ hours}) = 15\text{ miles}\).

  2. Equivalent scenario: A worker earns $200 total over 8 hours. The hourly wage is \(\dfrac{200\text{ dollars}}{8\text{ hours}} = 25\text{ dollars/hour}\).

  3. Equivalent scenario: A recipe uses 3 cups of flour per batch to make 4 batches. The total amount of flour is \((3\text{ cups/batch}) \cdot (4\text{ batches}) = 12\text{ cups}\).

  4. Equivalent scenario: A total of 60 gallons of water flows from a hose over 10 minutes. The flow rate is \(\dfrac{60\text{ gallons}}{10\text{ minutes}} = 6\text{ gallons/minute}\).

  5. Equivalent scenario: Rachel travels by bicycle at an average speed of 20 km/hour for 4 hours. The total distance traveled is \((20\text{ km/hour}) \cdot (4\text{ hours}) = 80\text{ km}\).

6.3 Equivalent equations

It is also possible to have equivalent equations. For simplicity, we focus on equations with only one variable.

The equation \(x + 5 = 23\) has two different equivalent equations:

  • One option is \(5 = 23-x\).
  • Another option is \(x = 23- 5\).

Both of the options are mathematically valid, but the second one is more useful because it helps us deduce the value \(x=18\).

Example 6.3 (Practice: equivalent equations) For each equation find an equivalent equation that allows us to deduce the value of the variable.

  1. \(x + 7 = 20\)

  2. \(3y = 45\)

  3. \(m - 8 = 15\)

  4. \(12 = 4n\)

  5. \(18 = p + 9\)

  6. \(56 = 7k\)

  7. \(w - 12 = 30\)

  8. \(5t = 65\)

  1. \(x = 20 - 7\), so \(x = 13\)

  2. \(y = 45 \div 3\), so \(y = 15\)

  3. \(m = 15 + 8\), so \(m = 23\)

  4. \(n = 12 \div 4\), so \(n = 3\)

  5. \(p = 18 - 9\), so \(p = 9\)

  6. \(k = 56 \div 7\), so \(k = 8\)

  7. \(w = 30 + 12\), so \(w = 42\)

  8. \(t = 65 \div 5\), so \(t = 13\)

6.4 Equivalent scenarios with variables

Example 6.4 Currently, Hail Creek is selling regular gasoline at $3.85 per gallon. I have $20 to spend on gas, but I don’t know how many gallons of gasoline that will be.

I let \(x\) represent the number of gallons. I know that \[ \text{total price} = \$3.85\cdot\text{ number of gallons} \] and so my equation is \[ 20 = 3.85 x. \]

This is equivalent to the equation \[ \frac{20}{3.85} = x. \] Using my calculator, I see that \(x \approx 5.19\), which means I can purchase 5.19 gallons of gas.

Example 6.5 (Practice: equivalent scenarios with variables)  

  1. A painter charges $45 per hour. Maria has a budget of $270 for painting her living room. Let \(h\) represent the number of hours the painter can work. Write an equation and solve for \(h\).

  2. Ground beef costs $6.50 per pound at the grocery store. Kevin wants to spend $32.50 on ground beef. Let \(p\) represent the number of pounds he can buy. Write an equation and solve for \(p\).

  1. A plumber charges $85 per hour. A homeowner has $425 available to pay for plumbing repairs. Let \(h\) represent the number of hours the plumber can work. Write an equation and solve for \(h\).

  2. Maria has some money saved. After adding $35 to her savings, she has $127 total. Let \(s\) represent her original savings amount. Write an equation and solve for \(s\).

  3. Firewood is sold for $225 per cord. A family has $675 budgeted for firewood. Let \(c\) represent the number of cords they can buy. Write an equation and solve for \(c\).

  4. After spending some money on groceries, Tom has $43 left from his original $150. Let \(g\) represent the amount he spent on groceries. Write an equation and solve for \(g\).

  5. A landscaper charges $55 per hour. A client has a budget of $330 for landscaping work. Let \(h\) represent the number of hours the landscaper can work. Write an equation and solve for \(h\).

  1. Total cost = $45 per hour × number of hours, so \(270 = 45h\).

    Equivalent equation: \(h = 270 \div 45 = 6\) hours

  2. Total cost = $6.50 per pound × number of pounds, so \(32.50 = 6.50p\).

    Equivalent equation: \(p = 32.50 \div 6.50 = 5\) pounds

  3. Total cost = $85 per hour × number of hours, so \(425 = 85h\).

    Equivalent equation: \(h = 425 \div 85 = 5\) hours

  4. Original savings + $35 = total, so \(s + 35 = 127\).

    Equivalent equation: \(s = 127 - 35 = 92\) dollars

  5. Total cost = $225 per cord × number of cords, so \(675 = 225c\).

    Equivalent equation: \(c = 675 \div 225 = 3\) cords

  6. Original amount - groceries = amount left, so \(150 - g = 43\).

    Equivalent equation: \(g = 150 - 43 = 107\) dollars

  7. Total cost = $55 per hour × number of hours, so \(330 = 55h\).

    Equivalent equation: \(h = 330 \div 55 = 6\) hours

6.5 Equivalence with fractions

Suppose we want to create an equivalent equation for \(\frac23x = 8\). Since the original equation involves multiplication of \(\frac23\) and \(x\), the equivalent equaiton will involve division: \[ x = 8\div\frac{2}{3}. \]

But we can convert division by a fraction into multiplication! So another version of our equivalent equation is \[ x = 8\cdot\frac{3}{2}. \]

Example 6.6 (Practice: equivalence with fractions) For each equation below, construct an equivalent equation that involves multiplication.

  1. \(\dfrac{3}{4}x = 12\)

  2. \(\dfrac{2}{5}y = 10\)

  3. \(18 = \dfrac{5}{6}m\)

  4. \(\dfrac{4}{7}n = 20\)

  5. \(15 = \dfrac{3}{8}p\)

  6. \(\dfrac{5}{9}w = 25\)

  1. \(x = 12 \div \dfrac{3}{4} \Rightarrow x = 12 \cdot \dfrac{4}{3} \Rightarrow x = 16\)

  2. \(y = 10 \div \dfrac{2}{5} \Rightarrow y = 10 \cdot \dfrac{5}{2} \Rightarrow y = 25\)

  3. \(m = 18 \div \dfrac{5}{6} \Rightarrow m = 18 \cdot \dfrac{6}{5} \Rightarrow m = 21.6\)

  4. \(n = 20 \div \dfrac{4}{7} \Rightarrow n = 20 \cdot \dfrac{7}{4} \Rightarrow n = 35\)

  5. \(p = 15 \div \dfrac{3}{8} \Rightarrow p = 15 \cdot \dfrac{8}{3} \Rightarrow p = 40\)

  6. \(w = 25 \div \dfrac{5}{9} \Rightarrow w = 25 \cdot \dfrac{9}{5} \Rightarrow w = 45\)

6.6 Homework exercises

Exercise 6.1 Write an equivalent statement using the opposite operation.

  1. \(18 + 9 = 27\)

  2. \(6 \cdot 7 = 42\)

  3. \(35 - 14 = 21\)

  4. \(72 \div 9 = 8\)

  5. \(13 + 19 = 32\)

  6. \(63 \div 7 = 9\)

  7. \(8 \cdot 9 = 72\)

  8. \(50 - 23 = 27\)

  9. \(12 \cdot 5 = 60\)

  10. \(81 \div 9 = 9\)

  1. \(27 - 9 = 18\) or \(27 - 18 = 9\)

  2. \(42 \div 7 = 6\) or \(42 \div 6 = 7\)

  3. \(21 + 14 = 35\) or \(35 = 21 + 14\)

  4. \(72 = 8 \cdot 9\) or \(72 = 9 \cdot 8\)

  5. \(32 - 19 = 13\) or \(32 - 13 = 19\)

  6. \(63 = 9 \cdot 7\) or \(63 = 7 \cdot 9\)

  7. \(72 \div 9 = 8\) or \(72 \div 8 = 9\)

  8. \(27 + 23 = 50\) or \(50 = 27 + 23\)

  9. \(60 \div 5 = 12\) or \(60 \div 12 = 5\)

  10. \(81 = 9 \cdot 9\)

Exercise 6.2 For each scenario, write an equivalent scenario using the opposite operation.

  1. A train travels 240 miles in 4 hours. The average speed is \(\dfrac{240\text{ miles}}{4\text{ hours}} = 60\text{ miles/hour}.\)

  2. A baker makes cookies at a rate of 36 cookies per hour for 5 hours. The total number of cookies is \((36\text{ cookies/hour}) \cdot (5\text{ hours}) = 180\text{ cookies}.\)

  3. A garden hose fills a pool with 150 gallons of water in 6 hours. The flow rate is \(\dfrac{150\text{ gallons}}{6\text{ hours}} = 25\text{ gallons/hour}.\)

  4. A construction worker earns $28 per hour and works for 7 hours. The total earnings are \((28\text{ dollars/hour}) \cdot (7\text{ hours}) = 196\text{ dollars}.\)

  5. A student reads 180 pages in 6 hours. The reading rate is \(\dfrac{180\text{ pages}}{6\text{ hours}} = 30\text{ pages/hour}.\)

  6. A factory produces 75 items per hour for 8 hours. The total production is \((75\text{ items/hour}) \cdot (8\text{ hours}) = 600\text{ items}.\)

  7. A car uses 56 gallons of gas to travel 896 miles. The fuel efficiency is \(\dfrac{896\text{ miles}}{56\text{ gallons}} = 16\text{ miles/gallon}.\)

  8. A runner maintains a pace of 8 minutes per mile for 5 miles. The total time is \((8\text{ minutes/mile}) \cdot (5\text{ miles}) = 40\text{ minutes}.\)

  1. Equivalent scenario: A train travels at an average speed of 60 miles per hour for 4 hours. The total distance traveled is \[(60\text{ miles/hour}) \cdot (4\text{ hours}) = 240\text{ miles}.\]

  2. Equivalent scenario: A baker makes a total of 180 cookies over 5 hours. The production rate is \[\dfrac{180\text{ cookies}}{5\text{ hours}} = 36\text{ cookies/hour}.\]

  3. Equivalent scenario: A garden hose flows at a rate of 25 gallons per hour for 6 hours. The total amount of water is \[(25\text{ gallons/hour}) \cdot (6\text{ hours}) = 150\text{ gallons}.\]

  4. Equivalent scenario: A construction worker earns $196 total over 7 hours. The hourly wage is \[\dfrac{196\text{ dollars}}{7\text{ hours}} = 28\text{ dollars/hour}.\]

  5. Equivalent scenario: A student reads at a rate of 30 pages per hour for 6 hours. The total number of pages read is \[(30\text{ pages/hour}) \cdot (6\text{ hours}) = 180\text{ pages}.\]

  6. Equivalent scenario: A factory produces a total of 600 items over 8 hours. The production rate is \[\dfrac{600\text{ items}}{8\text{ hours}} = 75\text{ items/hour}.\]

  7. Equivalent scenario: A car has a fuel efficiency of 16 miles per gallon and uses 56 gallons of gas. The total distance traveled is \[(16\text{ miles/gallon}) \cdot (56\text{ gallons}) = 896\text{ miles}.\]

  8. Equivalent scenario: A runner completes 5 miles in a total time of 40 minutes. The pace is \[\dfrac{40\text{ minutes}}{5\text{ miles}} = 8\text{ minutes/mile}.\]

Exercise 6.3 For each equation find an equivalent equation that allows us to deduce the value of the variable.

  1. \(x + 9 = 25\)

  2. \(4y = 52\)

  3. \(m - 6 = 19\)

  4. \(18 = 3n\)

  5. \(27 = p + 11\)

  6. \(72 = 8k\)

  7. \(w - 15 = 40\)

  8. \(6t = 84\)

  9. \(35 = r + 14\)

  10. \(9s = 108\)

  1. \(x = 25 - 9\), so \(x = 16\)

  2. \(y = 52 \div 4\), so \(y = 13\)

  3. \(m = 19 + 6\), so \(m = 25\)

  4. \(n = 18 \div 3\), so \(n = 6\)

  5. \(p = 27 - 11\), so \(p = 16\)

  6. \(k = 72 \div 8\), so \(k = 9\)

  7. \(w = 40 + 15\), so \(w = 55\)

  8. \(t = 84 \div 6\), so \(t = 14\)

  9. \(r = 35 - 14\), so \(r = 21\)

  10. \(s = 108 \div 9\), so \(s = 12\)

Exercise 6.4 For each scenario, define the variable, write an equation, and solve for the variable.

  1. A mechanic charges $75 per hour. A customer has $450 available for car repairs. Let \(h\) represent the number of hours the mechanic can work. Write an equation and solve for \(h\).

  2. Fresh salmon costs $14.50 per pound at the fish market. A chef wants to spend $87 on salmon. Let \(p\) represent the number of pounds they can buy. Write an equation and solve for \(p\).

  3. A tree removal service charges $180 per tree. A homeowner has a budget of $900 for tree removal. Let \(t\) represent the number of trees they can have removed. Write an equation and solve for \(t\).

  4. Sarah has some money in her checking account. After depositing $250, she has $815 total. Let \(b\) represent her original balance. Write an equation and solve for \(b\).

  1. Premium coffee beans cost $16.75 per pound. A coffee shop has $134 to spend on beans. Let \(p\) represent the number of pounds they can buy. Write an equation and solve for \(p\).

  2. A personal trainer charges $65 per session. A client has $390 budgeted for training. Let \(s\) represent the number of sessions they can afford. Write an equation and solve for \(s\).

  3. Ceramic tiles cost $4.25 per tile. A homeowner has $255 to spend on tiles. Let \(t\) represent the number of tiles they can purchase. Write an equation and solve for \(t\).

  4. After paying some bills, Carlos has $185 remaining from his original $520. Let \(b\) represent the amount he paid in bills. Write an equation and solve for \(b\).

  5. Organic strawberries cost $5.50 per basket. A customer has $27.50 to spend. Let \(b\) represent the number of baskets they can buy. Write an equation and solve for \(b\).

  6. An electrician charges $95 per hour. A business has $570 budgeted for electrical work. Let \(h\) represent the number of hours the electrician can work. Write an equation and solve for \(h\).

  1. Total cost = $75 per hour × number of hours, so \(450 = 75h\).

    Equivalent equation: \(h = 450 \div 75 = 6\) hours

  2. Total cost = $14.50 per pound × number of pounds, so \(87 = 14.50p\).

    Equivalent equation: \(p = 87 \div 14.50 = 6\) pounds

  3. Total cost = $180 per tree × number of trees, so \(900 = 180t\).

    Equivalent equation: \(t = 900 \div 180 = 5\) trees

  4. Original balance + $250 = total, so \(b + 250 = 815\).

    Equivalent equation: \(b = 815 - 250 = 565\) dollars

  5. Total cost = $16.75 per pound × number of pounds, so \(134 = 16.75p\).

    Equivalent equation: \(p = 134 \div 16.75 = 8\) pounds

  6. Total cost = $65 per session × number of sessions, so \(390 = 65s\).

    Equivalent equation: \(s = 390 \div 65 = 6\) sessions

  7. Total cost = $4.25 per tile × number of tiles, so \(255 = 4.25t\).

    Equivalent equation: \(t = 255 \div 4.25 = 60\) tiles

  8. Original amount - bills = amount remaining, so \(520 - b = 185\).

    Equivalent equation: \(b = 520 - 185 = 335\) dollars

  9. Total cost = $5.50 per basket × number of baskets, so \(27.50 = 5.50b\).

    Equivalent equation: \(b = 27.50 \div 5.50 = 5\) baskets

  10. Total cost = $95 per hour × number of hours, so \(570 = 95h\).

    Equivalent equation: \(h = 570 \div 95 = 6\) hours

Exercise 6.5 For each equation, create an equivalent equation and solve for the variable.

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

  2. \(\dfrac{3}{5}y = 21\)

  3. \(24 = \dfrac{4}{7}m\)

  4. \(\dfrac{5}{8}n = 30\)

  5. \(16 = \dfrac{2}{9}p\)

  6. \(\dfrac{7}{10}w = 35\)

  7. \(\dfrac{3}{4}k = 27\)

  8. \(40 = \dfrac{5}{6}t\)

  9. \(\dfrac{4}{9}r = 36\)

  10. \(28 = \dfrac{7}{12}s\)

  1. \(x = 18 \div \dfrac{2}{3} \Rightarrow x = 18 \cdot \dfrac{3}{2} \Rightarrow x = 27\)

  2. \(y = 21 \div \dfrac{3}{5} \Rightarrow y = 21 \cdot \dfrac{5}{3} \Rightarrow y = 35\)

  3. \(m = 24 \div \dfrac{4}{7} \Rightarrow m = 24 \cdot \dfrac{7}{4} \Rightarrow m = 42\)

  4. \(n = 30 \div \dfrac{5}{8} \Rightarrow n = 30 \cdot \dfrac{8}{5} \Rightarrow n = 48\)

  5. \(p = 16 \div \dfrac{2}{9} \Rightarrow p = 16 \cdot \dfrac{9}{2} \Rightarrow p = 72\)

  6. \(w = 35 \div \dfrac{7}{10} \Rightarrow w = 35 \cdot \dfrac{10}{7} \Rightarrow w = 50\)

  7. \(k = 27 \div \dfrac{3}{4} \Rightarrow k = 27 \cdot \dfrac{4}{3} \Rightarrow k = 36\)

  8. \(t = 40 \div \dfrac{5}{6} \Rightarrow t = 40 \cdot \dfrac{6}{5} \Rightarrow t = 48\)

  9. \(r = 36 \div \dfrac{4}{9} \Rightarrow r = 36 \cdot \dfrac{9}{4} \Rightarrow r = 81\)

  10. \(s = 28 \div \dfrac{7}{12} \Rightarrow s = 28 \cdot \dfrac{12}{7} \Rightarrow s = 48\)