Math I · A-REI.1

Explaining Equation-Solving Steps as Logical Consequences of Equality

Justifying equation steps turns algebra from guessing into an argument, which is the foundation of proof, debugging, and trustworthy quantitative work.

Concept Algebra
Domain Reasoning with Equations and Inequalities
Read time 9 minutes

What this learning objective is really asking you to learn

This objective asks students to understand equation solving as reasoning, not as a collection of memorized moves. It is one thing to solve \(3x + 5 = 20\) and get \(x = 5\). It is a deeper thing to explain why each step is allowed and why the final answer must be connected to the original equation.

An equation is a statement that two expressions have the same value. Solving an equation means finding all values of the variable that make the statement true. Every step in solving should follow logically from the previous statement. If you add the same number to both sides, equality is preserved. If you subtract the same expression from both sides, equality is preserved. If you multiply or divide both sides by the same nonzero number, equality is preserved. If you simplify equivalent expressions, equality is preserved.

This objective is about making that logic visible. Students should be able to say things like:

  • “I subtracted 5 from both sides because equal quantities remain equal when the same quantity is subtracted from each.”
  • “I divided both sides by 3 because the variable was multiplied by 3 and 3 is nonzero.”
  • “I used the distributive property to rewrite \(3(x - 2)\) as \(3x - 6\).”
  • “I checked the solution in the original equation to confirm it works.”

This may sound slow at first, but it builds mathematical power. A student who can justify steps is less likely to make illegal moves, more able to catch mistakes, and better prepared for proof, functions, systems, and advanced algebra.

Why students should learn this math

Many students can imitate equation-solving procedures for a while. The trouble comes when equations become unfamiliar. If students only memorize moves, they often break down when the structure changes. If they understand the logic, they can adapt.

For example, a memorized rule might tell students to “move the 5 to the other side and change the sign.” But that phrase hides the reason. What is actually happening is subtracting 5 from both sides. The “change the sign” shortcut may work in a simple case, but it can produce errors when expressions are more complicated. Understanding equality is safer than memorizing motion.

This objective also matters outside math class. Logical step-by-step transformation is the foundation of debugging computer code, checking financial calculations, interpreting scientific formulas, and evaluating arguments. In any technical field, it is not enough to have a result. You need a chain of reasoning that others can inspect.

When students ask, “Why do I have to show my work?” the honest answer is not “because the teacher said so.” The reason is that mathematics is a language of justified claims. Showing work reveals whether the answer came from valid reasoning. In advanced math, science, engineering, and programming, the method is often as important as the answer because the method must be trusted, reused, or corrected.

This objective also supports intellectual independence. If you know why a method works, you are not dependent on remembering exactly which procedure fits which problem. You can rebuild the method from principles. That is a major difference between fragile learning and durable learning.

The historical machinery: equality, balance, and proof

Equation solving has often been explained through the metaphor of balance. If two sides are equal, doing the same thing to both sides keeps them equal. This idea is ancient in spirit, even though the modern equals sign is relatively recent. Algebraic traditions from different cultures developed methods for restoring and balancing equations. The Arabic term associated with algebra, al-jabr, is often connected with completion or restoration, while al-muqabala is connected with balancing or comparison. The school phrase “do the same thing to both sides” is a modern classroom version of an old structural idea.

The equals sign itself was introduced by Robert Recorde in the sixteenth century. Before such symbols became standard, equality was often written in words. A compact symbol for equality made algebra easier to write and manipulate. But the symbol can also mislead students if they see it only as a command to calculate. In arithmetic, students often experience \(=\) as “write the answer next.” In algebra, \(=\) means “these two expressions have the same value.” That distinction is crucial.

Mathematics also inherited a culture of proof from Greek geometry, especially Euclid’s logical organization of definitions, assumptions, and propositions. Equation solving in school is not usually a formal proof in the Euclidean style, but it shares the same expectation: each claim should follow from accepted properties and previous claims.

This objective is where algebra and proof begin to meet. The student is not merely performing operations. The student is constructing a valid argument that the solution set has been preserved and that the final answer satisfies the original condition.

Where this fits in the big map of mathematics

A-REI.1 belongs to the domain “Reasoning with Equations and Inequalities.” That word reasoning matters. Algebra is not just symbol manipulation. It is the study of relationships and transformations that preserve truth.

In the larger map of mathematics, this objective supports proof, functions, systems, modeling, and advanced equation solving. When students later solve systems by elimination, they will need to understand why replacing one equation with a combination of equations can preserve solutions. When they solve quadratic equations, they will need to understand which transformations are reversible and which may introduce extra solutions. When they solve rational or radical equations, they will need to check for extraneous solutions. When they prove identities, they will need to transform expressions while preserving equivalence.

This objective also connects to computer science. An algorithm is a sequence of justified steps. If one step is invalid, the output cannot be trusted. Equation solving is one of the first algorithms students learn in symbolic form. Explaining the steps develops algorithmic thinking.

It connects to logic as well. Some transformations create equivalent equations: they have exactly the same solutions. Other transformations produce equations that are only consequences of the original and may have extra solutions. For example, squaring both sides can introduce extraneous solutions. Dividing by an expression containing a variable can lose solutions if that expression could be zero. A student who understands reasoning can manage these dangers.

How to execute the skill technically

The technical core is the idea of equivalent equations. Two equations are equivalent if they have the same solution set. Most basic equation-solving steps are designed to produce an equivalent equation that is simpler.

Here are common legal moves and the reasoning behind them.

Simplifying expressions: You may combine like terms or use the distributive property because the expression keeps the same value for every allowed value of the variable. For example, \(3(x - 2)\) is equivalent to \(3x - 6\).

Adding or subtracting the same expression on both sides: If \(A = B\), then \(A + C = B + C\) and \(A - C = B - C\). Equality is preserved because both sides change by the same amount.

Multiplying or dividing both sides by the same nonzero number: If \(A = B\), then \(kA = kB\) for any number \(k\). If \(k \ne 0\), then \(A/k = B/k\). Division by zero is not allowed because it is undefined and destroys the logic.

Substitution: If two expressions are equal, one can replace the other. This is used constantly in algebra and later in functions, geometry, and proof.

Factoring and expanding: These are expression rewrites based on the distributive property. They do not change the value of the expression; they change its form.

Checking: Substituting the final answer into the original equation verifies that the answer actually works. This is especially important when a step might not be reversible.

The important technical distinction is between transformations that preserve equivalence and transformations that may change the solution set. Adding the same expression to both sides preserves equivalence. Multiplying by zero does not, because it turns both sides into zero and makes every value look like a solution. Squaring both sides can introduce extra solutions because different numbers can have the same square. For example, \(x = 3\) implies \(x^2 = 9\), but \(x^2 = 9\) allows both \(x = 3\) and \(x = -3\).

A strong algebra student learns not only what to do but what each step does to the solution set.

A worked example with justifications

Solve and justify each step:

\[3(x - 2) + 5 = 2x + 11\].

Start with the original equation:

\[3(x - 2) + 5 = 2x + 11\].

Use the distributive property:

\[3x - 6 + 5 = 2x + 11\].

This is justified because \(3(x - 2)\) and \(3x - 6\) are equivalent expressions.

Combine like terms on the left:

\[3x - 1 = 2x + 11\].

This is justified because \(-6 + 5 = -1\).

Subtract 2x from both sides:

\[x - 1 = 11\].

This is justified because subtracting the same expression from equal quantities preserves equality.

Add 1 to both sides:

\[x = 12\].

This is justified because adding the same quantity to equal quantities preserves equality.

Check in the original equation:

Left side: \(3(12 - 2) + 5 = 3(10) + 5 = 35\). Right side: \(2(12) + 11 = 24 + 11 = 35\).

Both sides are equal, so \(x = 12\) is a solution.

Notice that the solution is not trusted because it appears at the bottom of the work. It is trusted because every step followed logically and the result checks in the original equation.

A second example: why illegal steps are dangerous

Consider the equation

\[x(x - 2) = 3(x - 2)\].

A tempting move is to divide both sides by \((x - 2)\), giving \(x = 3\). And \(x = 3\) is indeed a solution. But this division may have lost a solution because \((x - 2)\) could be zero. If \(x = 2\), then both sides of the original equation are zero:

Left side: \(2(2 - 2) = 2(0) = 0\). Right side: \(3(2 - 2) = 3(0) = 0\).

So \(x = 2\) is also a solution.

The safer method is to bring all terms to one side:

\[x(x - 2) - 3(x - 2) = 0\].

Factor:

\[(x - 2)(x - 3) = 0\].

Use the zero product property:

\(x - 2 = 0\) or \(x - 3 = 0\).

So

\(x = 2\) or \(x = 3\).

This example teaches a powerful lesson: dividing by an expression containing a variable can eliminate cases where that expression equals zero. Justification is not decoration. It protects the solution set.

Equations as solution-set machines

A helpful way to think about equation solving is to imagine each equation as describing a set of allowed values. The original equation has a solution set. Each algebraic step should ideally create a simpler equation with the same solution set. When you finally reach \(x = 12\), the solution set is obvious.

This view prevents a lot of shallow mistakes. For example, from \(2x = x\), a student might divide by \(x\) and get \(2 = 1\), which seems impossible. The mistake is dividing by a variable that might be zero. The original equation \(2x = x\) has the solution \(x = 0\). Dividing by \(x\) eliminated the only solution. A solution-set view makes the danger visible.

This also explains why checking matters. If a transformation may have enlarged the solution set, checking removes extraneous solutions. If a transformation may have shrunk the solution set, more careful reasoning is needed to recover lost cases.

Common mistakes and how to avoid them

One common mistake is treating the equals sign like a signal to “do something” rather than as a statement of equality. In algebra, both sides of the equation are expressions, and the equation claims they are equal for certain values.

Another mistake is performing an operation on only one side. If you subtract 5 from the left side, you must subtract 5 from the right side. The balance metaphor is useful here, but the deeper idea is equality preservation.

Students also make sign errors when distributing or subtracting expressions. For example, subtracting \((2x + 3)\) means subtracting both terms: \(-2x - 3\). Justification through properties helps slow the mind enough to avoid these errors.

Another mistake is using shortcuts without understanding. “Move it across and change the sign” can work as a shorthand for adding or subtracting both sides, but it can become dangerous if students forget what is really happening. Shortcuts should be earned by understanding, not used as substitutes for understanding.

Finally, students sometimes believe checking is only for weak students. In reality, checking is what strong mathematicians, scientists, engineers, and programmers do. It is quality control.

The big takeaway

This objective teaches that solving equations is a logical argument. Each step should follow from the previous one by a property of equality or an equivalent expression rewrite. Students learn not only to get an answer but to justify why the answer is trustworthy. This is the beginning of proof-minded algebra, and it is one of the most important upgrades in mathematical maturity. The real-world value is clear: in any field where decisions depend on calculations, a result without reasoning is fragile. A result with a valid chain of reasoning can be checked, trusted, and improved.

Problem Library

Problems in the App From This Objective

147 problems across 12 archetypes in the app.

name addition, subtraction, multiplication, division property.
12 problems Warmup Practice Mixed Review Assessment
Problem 1

In the step from x + 7 = 12 to x = 5, identify the property of equality used.

Problem 2

In the step from 3x = 18 to x = 6, identify the property of equality used.

Problem 3

In the step from x/4 = 9 to x = 36, identify the property of equality used.

Problem 4

In the step from x - 5 = 10 to x = 15, identify the property of equality used.

Problem 5

In the step from -4 + y = 7 to y = 11, identify the property of equality used.

Problem 6

In the step from 15 = x + 3 to 12 = x, identify the property of equality used.

Problem 7

In the step from (1/2)x = 8 to x = 16, identify the property of equality used.

Problem 8

In the step from -5y = 20 to y = -4, identify the property of equality used.

Open in simulator
Problem 9

In the step from x - 3 = 2x - 10 to x = 2x - 7, identify the property of equality used.

Problem 10

In the step from 5x + 2 = 17 to 5x = 15, identify the property of equality used.

Problem 11

In the step from x / (-3) = 7 to x = -21, identify the property of equality used.

Problem 12

In the step from 0.5z = 10 to z = 20, identify the property of equality used.

connect algebraic transformation to valid reasoning.
12 problems Warmup Practice Mixed Review Assessment
Problem 13

Fill in the missing justification for this solution step: 2(x + 3) = 14 -> 2x + 6 = 14.

Open in simulator
Problem 14

Fill in the missing justification for this solution step: 2x + 6 = 14 -> 2x = 8.

Problem 15

Fill in the missing justification for this solution step: 2x = 8 -> x = 4.

Problem 16

Fill in the missing justification for this solution step: x - 5 = 12 -> x = 17.

Problem 17

Fill in the missing justification for this solution step: x / 3 = 7 -> x = 21.

Problem 18

Fill in the missing justification for this solution step: 3x + 4x - 2 = 19 -> 7x - 2 = 19.

Problem 19

Fill in the missing justification for this solution step: -2(y - 4) = 10 -> -2y + 8 = 10.

Problem 20

Fill in the missing justification for this solution step: 5 + x - 3 = 10 -> x + 5 - 3 = 10.

Problem 21

Fill in the missing justification for this solution step: x + (2 + 3) = 10 -> x + 5 = 10.

Problem 22

Fill in the missing justification for this solution step: x + 7 - 7 = 15 - 7 -> x + 0 = 8.

Problem 23

Fill in the missing justification for this solution step: -4x = 20 -> x = -5.

Problem 24

Fill in the missing justification for this solution step: 18 = 3x -> 3x = 18.

audit equality-preserving moves.
12 problems Warmup Practice Mixed Review Assessment
Problem 25

Find the first invalid step in this proposed solution: 3(x + 2) = 15, 3x + 2 = 15, 3x = 13, x = 13/3.

Problem 26

Find the first invalid step in this proposed solution: 2x - 5 = 9, 2x = 4, x = 2.

Problem 27

Find the first invalid step in this proposed solution: 4x = 0, x = 4.

Problem 28

Find the first invalid step in this proposed solution: 5x + 3 = 18, 5x = 18 + 3, 5x = 21, x = 21/5.

Problem 29

Find the first invalid step in this proposed solution: -3x = 12, x = 12 / 3, x = 4.

Open in simulator
Problem 30

Find the first invalid step in this proposed solution: 2x + 3x + 5 = 20, 5x + 5 = 20, 10x = 20, x = 2.

Problem 31

Find the first invalid step in this proposed solution: -(x - 4) = 7, -x - 4 = 7, -x = 11, x = -11.

Problem 32

Find the first invalid step in this proposed solution: 2(x + 1) = 8, 2x + 1 = 8, 2x = 7, x = 7/2.

Problem 33

Find the first invalid step in this proposed solution: 2x + 6 = 10, 8x = 10, x = 10/8, x = 5/4.

Problem 34

Find the first invalid step in this proposed solution: x/2 + 1 = 5, x + 1 = 10, x = 9.

Problem 35

Find the first invalid step in this proposed solution: 3x = 2x + 5, 3x + 2x = 5, 5x = 5, x = 1.

Problem 36

Find the first invalid step in this proposed solution: 6x = 18, x = 18/6, x = 2.

reason about equivalence.
12 problems Warmup Practice Mixed Review Assessment
Problem 37

Explain why applying adding 6 to both sides of an equation preserves the solution set.

Problem 38

Explain why applying subtracting 4 to both sides of an equation preserves the solution set.

Problem 39

Explain why applying multiplying by 3 to both sides of an equation preserves the solution set.

Problem 40

Explain why applying adding 10 to both sides of an equation preserves the solution set.

Problem 41

Explain why applying subtracting 1 to both sides of an equation preserves the solution set.

Problem 42

Explain why applying multiplying by -5 to both sides of an equation preserves the solution set.

Problem 43

Explain why applying dividing by 2 to both sides of an equation preserves the solution set.

Problem 44

Explain why applying adding -3 to both sides of an equation preserves the solution set.

Problem 45

Explain why applying subtracting -8 to both sides of an equation preserves the solution set.

Problem 46

Explain why applying multiplying by 0.5 to both sides of an equation preserves the solution set.

Problem 47

Explain why applying dividing by -4 to both sides of an equation preserves the solution set.

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Problem 48

Explain why applying adding 7.5 to both sides of an equation preserves the solution set.

recognize equivalent strategies and efficiency.
12 problems Warmup Practice Mixed Review Assessment
Problem 49

Compare solution path A: distribute 4(x + 2) = 20, then solve with solution path B: divide both sides by 4 first, then solve x + 2 = 5. Are both correct, and which is more efficient?

Problem 50

Compare solution path A: clear fractions first in x/3 + 2 = 5 with solution path B: subtract 2 first, then multiply by 3. Are both correct, and which is more efficient?

Problem 51

Compare solution path A: distribute 5(x - 3) = 40, then solve with solution path B: divide both sides by 5 first, then solve x - 3 = 8. Are both correct, and which is more efficient?

Problem 52

Compare solution path A: distribute 3(x + 5) = 21, then solve with solution path B: divide both sides by 3 first, then solve x + 5 = 7. Are both correct, and which is more efficient?

Problem 53

Compare solution path A: subtract 1/2 from both sides of x + 1/2 = 3/4, then solve with solution path B: multiply the entire equation by 4 to clear fractions, then solve. Are both correct, and which is more efficient?

Problem 54

Compare solution path A: subtract 7 from both sides of 6x - 2x + 7 = 19, then combine like terms with solution path B: combine like terms (6x - 2x) first, then solve 4x + 7 = 19. Are both correct, and which is more efficient?

Problem 55

Compare solution path A: add 5 to both sides of -2x - 5 = 11, then divide by -2 with solution path B: add 2x to both sides and subtract 11 from both sides to get -16 = 2x, then solve. Are both correct, and which is more efficient?

Problem 56

Compare solution path A: distribute 2(x - 4) + 6 = 10, then solve with solution path B: subtract 6 from both sides first, then divide by 2, then solve x - 4 = 2. Are both correct, and which is more efficient?

Problem 57

Compare solution path A: subtract 0.5 from both sides of 0.2x + 0.5 = 1.5, then divide by 0.2 with solution path B: multiply the entire equation by 10 to clear decimals, then solve 2x + 5 = 15. Are both correct, and which is more efficient?

Problem 58

Compare solution path A: subtract 3x from both sides of 5x + 2 = 3x + 10, then solve with solution path B: subtract 5x from both sides of 5x + 2 = 3x + 10, then solve. Are both correct, and which is more efficient?

Problem 59

Compare solution path A: distribute 2(x + 3) - 4 = 8, then solve with solution path B: add 4 to both sides first, then divide by 2, then solve x + 3 = 6. Are both correct, and which is more efficient?

Problem 60

Compare solution path A: add x/5 and 2x/5 to get 3x/5 = 9, then solve with solution path B: multiply the entire equation by 5 to clear fractions, then solve x + 2x = 45. Are both correct, and which is more efficient?

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verify final value in original equation.
12 problems Warmup Practice Mixed Review Assessment
Problem 61

Verify that x = 6 solves 2x + 5 = 17 by substitution.

Problem 62

Verify that x = -2 solves 3x - 4 = -10 by substitution.

Problem 63

Verify that x = 8 solves x/2 + 7 = 11 by substitution.

Problem 64

Verify that x = 15 solves x + 10 = 25 by substitution.

Problem 65

Verify that x = 5 solves 4x = 20 by substitution.

Problem 66

Verify that x = 21 solves x / 3 = 7 by substitution.

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Problem 67

Verify that x = -3 solves 5x + 3 = -12 by substitution.

Problem 68

Verify that x = 3 solves 7x - 9 = 12 by substitution.

Problem 69

Verify that x = 20 solves x/4 - 2 = 3 by substitution.

Problem 70

Verify that x = 5 solves 0.5x + 1 = 3.5 by substitution.

Problem 71

Verify that x = -2 solves -6x + 1 = 13 by substitution.

Problem 72

Verify that x = 6 solves 2x + 1 = x + 7 by substitution.

interpret contradiction or identity after simplification.
12 problems Warmup Practice Mixed Review Assessment
Problem 73

Solve 2(x + 3) = 2x + 6 and explain whether it has one solution, no solution, or infinitely many solutions.

Problem 74

Solve 3x + 4 = 3x - 2 and explain whether it has one solution, no solution, or infinitely many solutions.

Problem 75

Solve 5x - 1 = 14 and explain whether it has one solution, no solution, or infinitely many solutions.

Problem 76

Solve 4x + 8 = 4(x + 2) and explain whether it has one solution, no solution, or infinitely many solutions.

Problem 77

Solve 2(3x - 1) + 5 = 6x + 3 and explain whether it has one solution, no solution, or infinitely many solutions.

Problem 78

Solve 7x - (2x + 5) = 5x - 5 and explain whether it has one solution, no solution, or infinitely many solutions.

Problem 79

Solve 5x + 10 = 5(x + 3) and explain whether it has one solution, no solution, or infinitely many solutions.

Open in simulator
Problem 80

Solve 2x - 7 = 2x + 1 and explain whether it has one solution, no solution, or infinitely many solutions.

Problem 81

Solve 3(x - 4) = 3x + 5 and explain whether it has one solution, no solution, or infinitely many solutions.

Problem 82

Solve 6x - 3 = 15 and explain whether it has one solution, no solution, or infinitely many solutions.

Problem 83

Solve 2(x + 4) = 18 and explain whether it has one solution, no solution, or infinitely many solutions.

Problem 84

Solve 4x + 7 = x + 13 and explain whether it has one solution, no solution, or infinitely many solutions.

communicate steps as a logical chain.
15 problems Warmup Practice Mixed Review Assessment
Problem 85

Write a proof-style explanation for solving 3x + 7 = 22.

Problem 86

Write a proof-style explanation for solving 2(x - 4) = 18.

Problem 87

Write a proof-style explanation for solving 5x - 4 = 31.

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Problem 88

Write a proof-style explanation for solving 4x + 10 = 30.

Problem 89

Write a proof-style explanation for solving 6x - 9 = 21.

Problem 90

Write a proof-style explanation for solving 2x + 5 = 17.

Problem 91

Write a proof-style explanation for solving 3(x + 2) = 15.

Problem 92

Write a proof-style explanation for solving 7x - 12 = 9.

Problem 93

Write a proof-style explanation for solving 5(x - 3) = 20.

Problem 94

Write a proof-style explanation for solving 8 + 2x = 18.

Problem 95

Write a proof-style explanation for solving 10 - 3x = 1.

Problem 96

Write a proof-style explanation for solving (x / 2) + 5 = 9.

Problem 97

Write a proof-style explanation for solving 4x + 2 = 22.

Problem 98

Write a proof-style explanation for solving 9 - x = 5.

Problem 99

Write a proof-style explanation for solving (x / 3) - 1 = 2.

distinguish reversible from non-reversible operations.
12 problems Warmup Practice Mixed Review Assessment
Problem 100

Does the transformation from x + 3 = 7 to x = 4 create an equivalent equation? Explain.

Problem 101

Does the transformation from x = 2 to x^2 = 4 create an equivalent equation? Explain.

Problem 102

Does the transformation from x + 1 = 5 to 0 = 0 create an equivalent equation? Explain.

Problem 103

Does the transformation from 2x = 10 to x = 5 create an equivalent equation? Explain.

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Problem 104

Does the transformation from x - 5 = 2x + 1 to -x - 5 = 1 create an equivalent equation? Explain.

Problem 105

Does the transformation from 3(x + 2) = 15 to 3x + 6 = 15 create an equivalent equation? Explain.

Problem 106

Does the transformation from x(x - 1) = 2(x - 1) to x = 2 create an equivalent equation? Explain.

Problem 107

Does the transformation from x = -2 to x^2 = 4 create an equivalent equation? Explain.

Problem 108

Does the transformation from 5x - 2x + 1 = 7 to 3x + 1 = 7 create an equivalent equation? Explain.

Problem 109

Does the transformation from x^2 = 9 to x = 3 create an equivalent equation? Explain.

Problem 110

Does the transformation from x - 4 = 6 to x = 10 create an equivalent equation? Explain.

Problem 111

Does the transformation from x = pi/2 to sin(x) = 1 create an equivalent equation? Explain.

explain multiplying both sides by a common denominator.
12 problems Warmup Practice Mixed Review Assessment
Problem 112

Explain and perform clearing denominators in x/3 + 2 = 5.

Problem 113

Explain and perform clearing denominators in x/4 - 1/2 = 3.

Problem 114

Explain and perform clearing denominators in 2x/5 + 1 = 7.

Problem 115

Explain and perform clearing denominators in x/2 + 3 = 7.

Problem 116

Explain and perform clearing denominators in x/3 - 1/6 = 2.

Problem 117

Explain and perform clearing denominators in 3x/4 - 2 = 4.

Problem 118

Explain and perform clearing denominators in x/5 + 1 = x/2.

Problem 119

Explain and perform clearing denominators in x/2 - x/4 + 1 = 3.

Problem 120

Explain and perform clearing denominators in (x+1)/3 = 4.

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Problem 121

Explain and perform clearing denominators in x/2 + x/3 = 5.

Problem 122

Explain and perform clearing denominators in x/4 - 3 = -1/2.

Problem 123

Explain and perform clearing denominators in (x-2)/5 = 1.

critique operation sequence and repair it.
12 problems Warmup Practice Mixed Review Assessment
Problem 124

A student solves 3x + 6 = 21 using the wrong operation order: divide by 3 first to get x + 6 = 7. Explain the error and repair the solution.

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Problem 125

A student solves 2(x - 4) = 18 using the wrong operation order: subtract 4 first from inside parentheses. Explain the error and repair the solution.

Problem 126

A student solves (x + 5)/2 = 9 using the wrong operation order: multiply only x by 2 to get x + 5 = 18 is skipped and x = 18 - 5 is claimed after changing structure. Explain the error and repair the solution.

Problem 127

A student solves 4x - 8 = 12 using the wrong operation order: divide by 4 first to get x - 8 = 3. Explain the error and repair the solution.

Problem 128

A student solves 3(x + 5) = 27 using the wrong operation order: add 5 first from inside parentheses. Explain the error and repair the solution.

Problem 129

A student solves (x - 7)/3 = 5 using the wrong operation order: subtract 7 from x first. Explain the error and repair the solution.

Problem 130

A student solves -2x + 10 = 4 using the wrong operation order: divide by -2 first to get x + 10 = -2. Explain the error and repair the solution.

Problem 131

A student solves (2x + 1)/3 = 7 using the wrong operation order: subtract 1 from 2x first. Explain the error and repair the solution.

Problem 132

A student solves 5 + 2x = 15 using the wrong operation order: divide by 2 first to get 5/2 + x = 15/2. Explain the error and repair the solution.

Problem 133

A student solves x/4 + 3 = 7 using the wrong operation order: multiply by 4 first to get x + 12 = 7. Explain the error and repair the solution.

Problem 134

A student solves - (x - 6) = 10 using the wrong operation order: subtract 6 from x first. Explain the error and repair the solution.

Problem 135

A student solves (10 - 2x)/2 = 3 using the wrong operation order: subtract 10 from 2x first. Explain the error and repair the solution.

show a sequence of equivalence-preserving transformations.
12 problems Warmup Practice Mixed Review Assessment
Problem 136

Show that 2(x + 3) = 14 and 2x + 6 = 14 have the same solution set by using equivalence-preserving transformations.

Problem 137

Show that 3x - 9 = 12 and x - 3 = 4 have the same solution set by using equivalence-preserving transformations.

Problem 138

Show that x/4 + 2 = 7 and x + 8 = 28 have the same solution set by using equivalence-preserving transformations.

Problem 139

Show that x - 5 = 10 and x = 15 have the same solution set by using equivalence-preserving transformations.

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Problem 140

Show that x + 7 = 12 and x = 5 have the same solution set by using equivalence-preserving transformations.

Problem 141

Show that -x = 8 and x = -8 have the same solution set by using equivalence-preserving transformations.

Problem 142

Show that 4x + 2x - 1 = 11 and 6x - 1 = 11 have the same solution set by using equivalence-preserving transformations.

Problem 143

Show that 5x - 10 = 20 and 5(x - 2) = 20 have the same solution set by using equivalence-preserving transformations.

Problem 144

Show that 3x = 15 - 2x and 5x = 15 have the same solution set by using equivalence-preserving transformations.

Problem 145

Show that x + (3 + 4) = 10 and x + 7 = 10 have the same solution set by using equivalence-preserving transformations.

Problem 146

Show that x + 1/2 = 3/2 and 2x + 1 = 3 have the same solution set by using equivalence-preserving transformations.

Problem 147

Show that 5 + x = 12 and x + 5 = 12 have the same solution set by using equivalence-preserving transformations.