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Worksheet Grade 8 Math

Mortlake Alum Feed: Equations With Variables on Both Sides

A grade 8 worksheet from a fictional plant's chemical log, where students expand parentheses, clear fractions and decimals, and solve for one variable.

Objectives

  1. Students will solve linear equations with the variable on both sides by collecting like terms.
  2. Students will expand parentheses with the distributive property before solving an equation.
  3. Students will clear a fraction or a decimal coefficient by multiplying both sides by a common factor.

Vocabulary

Alum
Alum is the most common chemical that water plants add to make tiny dirt particles clump together. A pinch of alum can help a whole jar of muddy water turn clear.

Also: distributive property, like terms, coefficient, day tank

Teacher overview

Alum is a coagulant that pulls fine particles together so they settle out, and operators at the fictional Mortlake water plant set its feed rate by hand after each jar test. Their chemical log records the settings as equations, some with parentheses from a doubled or tripled recipe and some with fractional or decimal coefficients from a pump dial. Every equation on this worksheet has the variable on both sides, which is what the standard calls for. The error to watch for is distributing to only the first term inside a set of parentheses, so 5 times the quantity s minus 2 becomes 5s minus 2 instead of 5s minus 10.

Mortlake Alum Feed: Equations With Variables on Both Sides — Grade 8 Math
Name: ____________________________ Date: ______________

Student Worksheet

In our scenario, every equation below comes from the chemical log at the fictional Mortlake water plant. Solve each one and show your steps. Where an item names a unit, include it in your answer.

  1. Solve 9x + 14 = 5x + 42.
  2. Solve 6(x + 3) = 4x + 30.
  3. Solve 5(x - 4) = 3(x + 2).
  4. Solve 0.4x + 18 = 0.1x + 33.
  5. Solve (1/2)x + 9 = (1/5)x + 18.
  6. Solve 2(3x + 5) = 4(x + 8).
  7. Day tank A holds 16 gallons of alum solution and gains 7 gallons each hour. Day tank B holds 46 gallons and gains 4 gallons each hour. Write an equation and find the hour when the two tanks hold the same amount.
  8. How many gallons does each day tank hold at the hour you found in item 7?
  9. Solve 3(b + 4) = 2b + 26 for b.
  10. The lime feed setting s must satisfy 5(s - 2) = 3s + 16. Solve for s.
  11. Check your answer to item 10 by putting it back into 5(s - 2) = 3s + 16. What single number do both sides equal?
  12. Explain why multiplying both sides of the equation in item 5 by 10 is a helpful first step.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: 7. Subtract 5x from both sides to get 4x + 14 = 42. Subtract 14 from both sides to get 4x = 28. Divide both sides by 4 to get x = 7.
  2. Answer: 6. Distribute on the left to get 6x + 18 = 4x + 30. Subtract 4x from both sides to get 2x + 18 = 30. Subtract 18 to get 2x = 12, then divide by 2 to get x = 6.
  3. Answer: 13. Distribute on both sides to get 5x - 20 = 3x + 6. Subtract 3x to get 2x - 20 = 6.
    Add 20 to get 2x = 26, then divide by 2 to get x = 13.
  4. Answer: 50. Subtract 0.1x from both sides to get 0.3x + 18 = 33. Subtract 18 to get 0.3x = 15.
    Divide both sides by 0.3 to get x = 50.
  5. Answer: 30. Multiply every term by 10 to clear the fractions, which gives 5x + 90 = 2x + 180.
    Subtract 2x to get 3x + 90 = 180. Subtract 90 to get 3x = 90, then divide by 3 to get x = 30.
  6. Answer: 11. Distribute on both sides to get 6x + 10 = 4x + 32. Subtract 4x to get 2x + 10 = 32. Subtract 10 to get 2x = 22, then divide by 2 to get x = 11.
  7. Answer: 10 hours. The equation is 7h + 16 = 4h + 46. Subtract 4h to get 3h + 16 = 46. Subtract 16 to get 3h = 30, then divide by 3 to get h = 10 hours.
  8. Answer: 86 gallons. Substitute h = 10 into either side. Day tank A gives 7(10) + 16 = 86 gallons, and day tank B gives 4(10) + 46 = 86 gallons, which confirms the answer to item 7.
  9. Answer: 14. Distribute on the left to get 3b + 12 = 2b + 26. Subtract 2b to get b + 12 = 26, then subtract 12 to get b = 14.
  10. Answer: 13. Distribute on the left to get 5s - 10 = 3s + 16. Subtract 3s to get 2s - 10 = 16. Add 10 to get 2s = 26, then divide by 2 to get s = 13.
  11. Answer: 55. The left side is 5(13 - 2), which is 5 times 11, or 55. The right side is 3(13) + 16, which is 39 plus 16, or 55. Both sides equal 55, so s = 13 is correct.
  12. Answer: Multiplying by 10 clears both fractions at once, because 10 is a common multiple of 2 and 5. The equation becomes 5x + 90 = 2x + 180, which has whole-number coefficients and is easier to solve without rounding errors.

Differentiation

Support

Give these students a two-column recording sheet with the left column labeled "what I did" and the right column labeled "result," and pre-fill the first step for items 2, 3, and 6 from the fictional Mortlake log. Assign items 1, 2, 3, 7, and 8, and provide an arrow diagram showing the distributive property reaching both terms inside the parentheses.

Extension

Ask students to write an equation with parentheses on both sides whose solution is exactly 8, then trade with a partner and solve. They should also write a second equation, again with parentheses on both sides, whose solution is a negative number, and explain what a negative hour would mean in the day tank scenario.

ELL supports

Pre-teach alum, day tank, and setting with a labeled sketch of a small tank feeding a pipe, and post the phrase "the same amount" next to an equals sign. Provide the sentence frame for item 12: "Multiplying by 10 helps because ___, so the new equation has ___." Let students read the fictional Mortlake equations aloud in pairs before solving.

Extensions

Have students rewrite items 2, 3, and 6 as word problems set at the fictional Mortlake plant, with a starting amount and an hourly rate for each side of the equation, then trade with a partner and solve from the words alone. Ask them to state which version was easier to solve and why, and to check that their word problem produces the same solution as the original equation.