Math I · A-REI.3

Solving Linear Equations and Inequalities in One Variable, Including Literal Linear Equations

Linear equations and inequalities are the basic control language for unknown quantities, budgets, rates, thresholds, and formulas with parameters.

Concept Algebra
Domain Reasoning with Equations and Inequalities
Read time 10 minutes

What this learning objective is really asking you to learn

This objective is about mastering one of the central moves of algebra: isolating a variable by using legal, logical steps. A linear equation in one variable is an equation where the variable appears to the first power and is not multiplied by itself or placed inside more complicated operations. Examples include \(3x + 7 = 22\), \(5 - 2x = 13\), and \(4(x - 3) = 2x + 10\). A linear inequality is similar, but it uses a comparison symbol such as \(<\), \(>\), \(\le\), or \(\ge\). Examples include \(3x + 7 \le 22\) and \(5 - 2x > 13\).

A literal equation is an equation with several letters where you solve for one letter in terms of the others. For example, the distance formula \(d = rt\) can be solved for \(r\) as \(r = d/t\), or for \(t\) as \(t = d/r\). The formula for perimeter of a rectangle, \(P = 2L + 2W\), can be solved for \(L\) as \(L = (P - 2W)/2\), or more simply \(L = P/2 - W\). Literal equations are sometimes called formula rearrangements, but the important idea is that letters can represent known quantities, unknown quantities, parameters, or variables depending on the situation.

This objective is not just “get x by itself.” It is about understanding why each step preserves truth. If \(3x + 7 = 22\), subtracting 7 from both sides gives \(3x = 15\) because equal quantities remain equal when the same amount is subtracted from each. Dividing both sides by 3 gives \(x = 5\) because equal quantities remain equal when both are divided by the same nonzero number. Solving is a process of transforming an equation into an equivalent, simpler equation.

For inequalities, the same idea mostly holds, but with an important twist: multiplying or dividing by a negative number reverses the inequality sign. If \(-2x < 10\), dividing by -2 gives \(x > -5\), not \(x < -5\). The reversal is not a random rule. It comes from the order of numbers on the number line. Since \(2 < 5\), multiplying by -1 gives \(-2 > -5\). Negative multiplication flips order.

The goal of this objective is fluency with linear reasoning. Students should be able to solve efficiently, check their answers, graph inequality solutions on a number line, and interpret the answer in context. They should also be able to handle letters as coefficients, not panic when an equation contains more than one symbol, and understand that formulas are not magic. Formulas are equations that can be rearranged to answer different questions.

Why students should learn this math

Linear equations are the workhorse of practical mathematics. They appear wherever there is a constant rate, fixed fee, steady change, uniform speed, hourly wage, unit price, or proportional adjustment with an added starting value. If you earn $18 per hour plus a $40 bonus, your pay is linear. If a taxi charges a base fee plus a per-mile rate, the cost is linear. If water drains at a constant rate, the amount remaining is linear. If a temperature conversion follows \(F = 9/5 C + 32\), that formula is linear. If a recipe is scaled by a constant factor, the relationship is often linear.

Students should learn this math because it gives them control over unknowns in everyday decisions. If a job pays $16 per hour and you need $240, the equation \(16h = 240\) tells you the needed hours. If a phone plan costs $35 plus $0.10 per text after a limit, an inequality can tell you how many extra texts stay under a budget. If a car travels at 60 miles per hour, \(d = rt\) lets you solve for distance, rate, or time depending on what you need.

Literal equations are especially important because real life does not always ask for the variable that is already isolated. A formula is like a machine with multiple access points. \(A = lw\) might be introduced as a way to find area, but if you know the area and width, you can solve for length. \(I = Prt\) can find interest, but it can also find principal, rate, or time. \(V = IR\), Ohm's law, can find voltage, current, or resistance. A student who can rearrange formulas is not trapped by the way a formula is printed.

This is a major step toward mathematical independence. Students often memorize formulas as if they are one-way instructions. But equations are relationships. Solving for a different variable means looking at the relationship from another angle. In science, engineering, finance, and data work, formulas are constantly rearranged to isolate the quantity of interest. A physicist may solve for acceleration, time, force, or mass depending on the question. An engineer may solve for resistance, voltage, or current. A nurse may solve for dosage or volume. A contractor may solve for length, area, cost, or number of materials.

Inequalities matter because decisions usually involve limits. You rarely need exactly $100 of groceries; you need to spend no more than $100. A machine does not need to be exactly 80 degrees; it must stay below a danger threshold. A scholarship may require at least a certain score. A sports team may need more than a certain number of wins. Inequalities express these real limits.

Learning this objective also strengthens reasoning. Each algebraic step is a claim: this new statement has the same solution as the old one. When students learn to solve carefully, they learn a form of logical discipline. They learn not to move symbols randomly. They learn to preserve meaning.

The historical machinery: the rise of symbolic equation solving

The desire to solve linear equations is ancient. People have always needed to solve problems like “A number plus 7 is 20,” “Three equal shares and 5 more make 29,” or “A worker earns a fixed amount per day; how many days are needed?” Ancient Egyptian, Babylonian, Chinese, Indian, Greek, and Islamic mathematicians all dealt with problems that we would now describe as linear or algebraic.

What changed over time was the language. Early algebra was often rhetorical, written in words. Later it became syncopated, using abbreviations. Eventually it became symbolic, using letters and operation signs. Symbolic algebra made general methods easier to see. Instead of solving each word problem as a separate story, mathematicians could study forms like \(ax + b = c\) and understand a method that works whenever \(a\) is not zero.

Al-Khwarizmi's work in the ninth century helped organize algebra as a systematic discipline. His methods were not written in modern notation, but they emphasized transformations that restored and balanced equations. Over later centuries, European mathematicians developed the symbolic notation students now use. The equals sign appeared in the sixteenth century through Robert Recorde. The use of letters for known and unknown quantities became more standardized through the work of mathematicians such as François Viète and René Descartes.

Literal equations are tied to this symbolic revolution. Once letters could stand for general quantities, formulas became flexible. A formula such as \(d = rt\) is not one problem; it is a whole family of problems. The letters allow a relationship to be stored and reused. This is one of the great powers of algebra: it compresses many cases into one structure.

Inequalities also became more formal as mathematics grew to include bounds, approximations, and optimization. The symbols \(<\) and \(>\) are relatively recent compared with the ancient idea of comparison, but the concept is fundamental. Modern science depends on inequalities for error bounds, tolerances, estimates, constraints, and safety margins.

When students solve linear equations and inequalities today, they are learning the most accessible part of a long historical development: the transformation of unknown-number problems into a symbolic system that can be applied across contexts.

Where this fits in the big map of mathematics

Linear equations are one of the foundation stones of algebra. They come before systems of linear equations, linear functions, slope, intercepts, modeling, coordinate geometry, and linear programming. If a student is shaky with one-variable linear equations, many later topics feel unstable.

This objective also connects directly to functions. The equation \(3x + 7 = 22\) can be seen as asking when the function \(f(x) = 3x + 7\) reaches the value 22. Inequalities such as \(3x + 7 \le 22\) ask for all inputs that produce outputs at or below 22. This is an early version of solving function conditions.

Literal equations connect to modeling and science. Formulas are the skeleton of applied mathematics. Rearranging them prepares students for physics, chemistry, finance, engineering, geometry, and statistics. In later math, solving for one variable in terms of others appears in inverse functions, parameter analysis, regression formulas, trigonometric identities, and calculus.

Inequality solving connects to number lines, intervals, domains, constraints, and optimization. When students solve \(2x - 5 \ge 11\), they get \(x \ge 8\), which is not one number but a ray on the number line. That prepares them for interval notation and for understanding domains of functions. It also prepares them for two-variable inequalities, where solution sets become regions instead of rays.

In the larger map, this objective is part of the move from arithmetic to structure. Arithmetic asks, “What is the result of these operations?” Algebra asks, “What values make this relationship true?” Advanced mathematics asks even broader questions: “What structures preserve truth? What transformations are legal? What can be generalized?” Solving linear equations is the first serious training ground for those questions.

How to execute the skill technically

The basic strategy for solving a linear equation is to undo operations in a logical order while keeping the equation balanced. Start by simplifying each side: distribute, combine like terms, and clear parentheses. Then move variable terms to one side and constant terms to the other. Finally, divide by the coefficient of the variable.

Consider \(4(x - 3) = 2x + 10\). Distribute first: \(4x - 12 = 2x + 10\). Subtract 2x from both sides: \(2x - 12 = 10\). Add 12 to both sides: \(2x = 22\). Divide by 2: \(x = 11\). Check by substitution: left side is \(4(11 - 3) = 32\); right side is \(2(11) + 10 = 32\). The solution works.

For equations with fractions, clearing denominators can make the work cleaner. If \(x/3 + 5 = 11\), subtracting 5 gives \(x/3 = 6\), then multiplying by 3 gives \(x = 18\). If there are multiple fractions, multiply every term by the least common denominator. The key phrase is every term. Multiplying only one part of the equation breaks equivalence.

For inequalities, solve similarly, but reverse the inequality sign when multiplying or dividing by a negative. For \(-3x + 4 \le 16\), subtract 4: \(-3x \le 12\). Divide by -3 and reverse the sign: \(x \ge -4\). A check helps: choose \(x = 0\), which should work because \(0 \ge -4\). Substitute into the original inequality: \(-3(0) + 4 \le 16\), so \(4 \le 16\), true. Choose \(x = -5\), which should not work. Substitute: \(15 + 4 \le 16\), so \(19 \le 16\), false.

Graphing one-variable inequalities on a number line is part of the skill. Use an open circle for \(<\) or \(>\), and a closed circle for \(\le\) or \(\ge\). Shade in the direction of the allowed values. For \(x \ge -4\), place a closed circle at -4 and shade to the right.

Literal equations require the same logic, but the symbols may look more intimidating. To solve \(P = 2L + 2W\) for \(L\), subtract 2W from both sides: \(P - 2W = 2L\). Divide by 2: \((P - 2W)/2 = L\). This can be written as \(L = (P - 2W)/2\) or \(L = P/2 - W\). The letters \(P\) and \(W\) behave like quantities. You do not need their numerical values to isolate \(L\).

When solving literal equations, students should pay attention to restrictions. If solving \(d = rt\) for \(t\), the result is \(t = d/r\), but this assumes \(r \ne 0\). Division by zero is not allowed. In context, a zero rate might mean no movement, making time impossible to determine from distance.

A worked example: solving a linear inequality in context

A student has $75 to spend on a club event. The room costs $30 to reserve, and snacks cost $4 per person. How many people can attend?

Let \(p\) be the number of people. The cost is \(30 + 4p\), and the budget condition is \(30 + 4p \le 75\). Subtract 30: \(4p \le 45\). Divide by 4: \(p \le 11.25\). Since people must be whole numbers, at most 11 people can attend. The algebraic solution is a range, but the context turns it into a greatest whole-number choice.

This example shows why inequalities matter. The student is not trying to spend exactly $75. The student is trying not to exceed $75. The solution is a set of allowable choices.

A worked example: rearranging a formula

The formula for simple interest is \(I = Prt\), where \(I\) is interest, \(P\) is principal, \(r\) is annual interest rate, and \(t\) is time in years. Suppose you want to solve for the rate \(r\).

Start with \(I = Prt\). Divide both sides by \(Pt\): \(I/(Pt) = r\). So \(r = I/(Pt)\), assuming \(P\) and \(t\) are not zero. This rearranged formula answers a different question: if you know the interest earned, the starting amount, and the time, what rate produced it?

The same equation has become a different tool because we isolated a different quantity.

Common mistakes and how to avoid them

One common mistake is changing only one side of an equation. If you add, subtract, multiply, or divide, you must preserve equality by doing the same legal operation to both sides. Another mistake is distributing incorrectly, especially with negative signs: \(-2(x - 5)\) is \(-2x + 10\), not \(-2x - 10\).

In inequalities, the most common mistake is forgetting to reverse the sign when multiplying or dividing by a negative. Students can catch this by checking one value from their proposed solution.

In literal equations, students often panic because there are many letters. The solution is to focus on the target variable. Treat all other symbols as quantities that move according to the same algebraic rules. Another mistake is dividing by an expression without noting that it cannot be zero. This becomes more important in advanced algebra.

The deepest mistake is solving mechanically without understanding equivalence. Each step should create a statement with the same solution set, until the solution is obvious.

What students should be able to say

A student who has mastered this objective should be able to say: “I can solve linear equations and inequalities by using inverse operations and preserving equivalence. I know that inequality signs reverse when I multiply or divide by a negative number. I can graph inequality solutions on a number line and interpret them in context. I can also rearrange formulas to solve for a chosen variable, even when other letters remain in the answer.”

That is algebraic control. It gives students the ability to move inside formulas instead of being trapped by them.

Problem Library

Problems in the App From This Objective

180 problems across 15 archetypes in the app.

inverse operations.
12 problems Warmup Practice Mixed Review Assessment
Problem 1

Solve the linear equation 3x + 4 = 19.

Problem 2

Solve the linear equation x/2 - 7 = 3.

Problem 3

Solve the linear equation -4x = 28.

Problem 4

Solve the linear equation 5x = 30.

Problem 5

Solve the linear equation x/3 = 7.

Problem 6

Solve the linear equation x + 8 = 15.

Problem 7

Solve the linear equation x - 10 = 2.

Problem 8

Solve the linear equation -2x + 5 = 11.

Problem 9

Solve the linear equation x/4 - 1 = 5.

Problem 10

Solve the linear equation 7x - 3 = 18.

Open in simulator
Problem 11

Solve the linear equation x/5 + 9 = 4.

Problem 12

Solve the linear equation -6x + 10 = -8.

distribute before isolating variable.
12 problems Warmup Practice Mixed Review Assessment
Problem 13

Solve the equation requiring distribution: 2(x + 5) = 24.

Problem 14

Solve the equation requiring distribution: -3(x - 4) = 18.

Problem 15

Solve the equation requiring distribution: 4(x + 1) - 3 = 17.

Problem 16

Solve the equation requiring distribution: 5(x - 2) = 15.

Problem 17

Solve the equation requiring distribution: -2(x + 7) = 6.

Problem 18

Solve the equation requiring distribution: 3(x - 1) = 2x + 5.

Problem 19

Solve the equation requiring distribution: 10 - 2(x + 3) = 4.

Problem 20

Solve the equation requiring distribution: (x + 8) / 2 = 5.

Problem 21

Solve the equation requiring distribution: (2x - 6) / 3 = 4.

Problem 22

Solve the equation requiring distribution: 0.5(x - 4) = 3.

Problem 23

Solve the equation requiring distribution: -4(x - 5) = 8.

Problem 24

Solve the equation requiring distribution: 7 + 3(2x - 1) = 40.

Open in simulator
collect variable terms and constants.
12 problems Warmup Practice Mixed Review Assessment
Problem 25

Solve 5x + 2 = 2x + 17, which has variables on both sides, and classify the result.

Problem 26

Solve 3x + 4 = 3x - 8, which has variables on both sides, and classify the result.

Problem 27

Solve 2(x + 1) = 2x + 2, which has variables on both sides, and classify the result.

Problem 28

Solve 4x - 7 = x + 5, which has variables on both sides, and classify the result.

Problem 29

Solve 7x - 3 = 7x + 10, which has variables on both sides, and classify the result.

Problem 30

Solve 3(x - 2) = 3x - 6, which has variables on both sides, and classify the result.

Problem 31

Solve 6y + 1 = 2y - 11, which has variables on both sides, and classify the result.

Problem 32

Solve 4(z + 2) = 4z - 5, which has variables on both sides, and classify the result.

Problem 33

Solve 5m + 10 = 5(m + 2), which has variables on both sides, and classify the result.

Open in simulator
Problem 34

Solve 2(p - 3) = 5p + 9, which has variables on both sides, and classify the result.

Problem 35

Solve 1/2 a + 3 = 1/4 a + 5, which has variables on both sides, and classify the result.

Problem 36

Solve 9k + 1 - 2k = 7k + 5, which has variables on both sides, and classify the result.

reverse inequality when multiplying/dividing by negative.
12 problems Warmup Practice Mixed Review Assessment
Problem 37

Solve and graph the inequality 2x + 3 < 11.

Problem 38

Solve and graph the inequality -3x >= 12.

Problem 39

Solve and graph the inequality x/2 - 5 <= 1.

Problem 40

Solve and graph the inequality x - 7 > -2.

Problem 41

Solve and graph the inequality 4x < 20.

Problem 42

Solve and graph the inequality -5x <= 25.

Problem 43

Solve and graph the inequality 3x + 5 >= 14.

Open in simulator
Problem 44

Solve and graph the inequality -2x + 1 < 7.

Problem 45

Solve and graph the inequality x / -3 > 2.

Problem 46

Solve and graph the inequality (x + 4) / 2 > 3.

Problem 47

Solve and graph the inequality 10 - 4x <= 2.

Problem 48

Solve and graph the inequality 15 < 3x.

handle `and` versus `or` solution sets.
12 problems Warmup Practice Mixed Review Assessment
Problem 49

Solve the compound inequality 3 < x + 2 <= 8.

Problem 50

Solve the compound inequality x < -2 or x >= 5.

Problem 51

Solve the compound inequality -4 <= 2x < 10.

Problem 52

Solve the compound inequality 1 <= 3x - 2 < 10.

Problem 53

Solve the compound inequality -5 < -x + 1 <= 3.

Problem 54

Solve the compound inequality 7 < 2x + 3 < 15.

Problem 55

Solve the compound inequality 10 >= -2x + 4 > 0.

Problem 56

Solve the compound inequality -1 < x - 5 <= 2.

Problem 57

Solve the compound inequality x <= 0 or x > 7.

Problem 58

Solve the compound inequality x + 3 < 1 or x - 2 >= 4.

Open in simulator
Problem 59

Solve the compound inequality 2x - 1 < 5 or -x + 3 <= -2.

Problem 60

Solve the compound inequality -3x > 9 or x/2 + 1 >= 3.

rearrange symbols rather than numbers.
12 problems Warmup Practice Mixed Review Assessment
Problem 61

Solve the literal linear equation y = mx + b for x.

Problem 62

Solve the literal linear equation P = 2l + 2w for w.

Problem 63

Solve the literal linear equation C = a + bt for t.

Problem 64

Solve the literal linear equation A = lw for l.

Problem 65

Solve the literal linear equation v = d/t for d.

Open in simulator
Problem 66

Solve the literal linear equation V = IR for I.

Problem 67

Solve the literal linear equation A = (h/2)(b1 + b2) for h.

Problem 68

Solve the literal linear equation ax + by = c for x.

Problem 69

Solve the literal linear equation d = rt for r.

Problem 70

Solve the literal linear equation I = Prt for P.

Problem 71

Solve the literal linear equation ax + b = cx + d for x.

Problem 72

Solve the literal linear equation A = P(1 + rt) for t.

model, solve, interpret boundary.
12 problems Warmup Practice Mixed Review Assessment
Problem 73

Solve and interpret the context inequality: A budget is 50 dollars and each ticket costs 8 dollars. How many tickets can be bought?.

Open in simulator
Problem 74

Solve and interpret the context inequality: A student needs at least 90 points and earns 12 points per task. How many tasks are needed?.

Problem 75

Solve and interpret the context inequality: An elevator holds at most 1200 pounds and each box weighs 75 pounds. How many boxes?.

Problem 76

Solve and interpret the context inequality: A baker has 350 grams of flour and each cake requires 40 grams. How many cakes can be made?.

Problem 77

Solve and interpret the context inequality: A charity needs to raise at least $500, and each donor contributes $45. How many donors are needed?.

Problem 78

Solve and interpret the context inequality: A truck can carry a maximum of 2000 kg. Each crate weighs 180 kg. How many crates can it carry?.

Problem 79

Solve and interpret the context inequality: A project requires at least 150 hours of work, and each worker can contribute 35 hours per week. How many workers are needed for a week?.

Problem 80

Solve and interpret the context inequality: A shelf is 120 cm long, and each book is 3.5 cm wide. How many books can fit on the shelf?.

Problem 81

Solve and interpret the context inequality: A factory has 800 meters of fabric. Each dress requires 2.2 meters of fabric. How many dresses can be made?.

Problem 82

Solve and interpret the context inequality: A student needs at least 70 points to pass. Each correct answer is worth 3 points. How many correct answers are needed?.

Problem 83

Solve and interpret the context inequality: A car has 10 liters of fuel. It consumes fuel at a rate of 0.75 liters per 10 km. How many full 10 km segments can it travel?.

Problem 84

Solve and interpret the context inequality: 100 people need to be transported. Each van can hold 12 people. How many vans are needed?.

catch sign, distribution, inequality-direction errors.
12 problems Warmup Practice Mixed Review Assessment
Problem 85

Diagnose the error in the worked solution -2x < 8, so x < -4 and correct it.

Problem 86

Diagnose the error in the worked solution 3(x + 2)=15, so 3x + 2 = 15 and correct it.

Problem 87

Diagnose the error in the worked solution 2x + 5 = 17, so 2x = 22 and correct it.

Problem 88

Diagnose the error in the worked solution (x + 3)^2 = x^2 + 9 and correct it.

Problem 89

Diagnose the error in the worked solution 3x + 4y = 7xy and correct it.

Problem 90

Diagnose the error in the worked solution 1/2 + 1/3 = 2/5 and correct it.

Problem 91

Diagnose the error in the worked solution (x + 5) / 5 = x + 1 and correct it.

Problem 92

Diagnose the error in the worked solution 5 - (x - 2) = 5 - x - 2 = 3 - x and correct it.

Problem 93

Diagnose the error in the worked solution x^2 * x^3 = x^6 and correct it.

Open in simulator
Problem 94

Diagnose the error in the worked solution -3^2 = 9 and correct it.

Problem 95

Diagnose the error in the worked solution 4x = 12, so x = 12 - 4 = 8 and correct it.

Problem 96

Diagnose the error in the worked solution sqrt(9 + 16) = sqrt(9) + sqrt(16) = 3 + 4 = 7 and correct it.

multiply by common denominator and solve.
12 problems Warmup Practice Mixed Review Assessment
Problem 97

Solve by clearing denominators: x/3 + x/6 = 5.

Problem 98

Solve by clearing denominators: x/4 - 1/2 = 3.

Problem 99

Solve by clearing denominators: 2x/5 + x/10 = 3.

Problem 100

Solve by clearing denominators: x/2 + x/3 = 10.

Problem 101

Solve by clearing denominators: x/5 - x/10 = 1.

Problem 102

Solve by clearing denominators: x/2 + x/4 = 9.

Problem 103

Solve by clearing denominators: x/3 - 1/6 = 2.

Problem 104

Solve by clearing denominators: x/7 + x/14 = 3.

Problem 105

Solve by clearing denominators: x/8 - 1/4 = 1/2.

Open in simulator
Problem 106

Solve by clearing denominators: x/9 + x/3 = 8.

Problem 107

Solve by clearing denominators: x/6 + x/12 = 3/4.

Problem 108

Solve by clearing denominators: x/10 - 1/5 = 1/2.

preserve inequality direction and exact/decimal reasoning.
12 problems Warmup Practice Mixed Review Assessment
Problem 109

Solve the inequality with fractions or decimals: 0.5x + 3 <= 8.

Problem 110

Solve the inequality with fractions or decimals: x/4 - 2 > 1.

Problem 111

Solve the inequality with fractions or decimals: -0.2x >= 4.

Problem 112

Solve the inequality with fractions or decimals: 0.7x - 1 < 6.

Problem 113

Solve the inequality with fractions or decimals: x/3 + 5 >= 7.

Problem 114

Solve the inequality with fractions or decimals: -0.5x + 2 > 7.

Problem 115

Solve the inequality with fractions or decimals: -x/2 - 1 <= 3.

Problem 116

Solve the inequality with fractions or decimals: 0.25x + 1.5 <= 2.5.

Open in simulator
Problem 117

Solve the inequality with fractions or decimals: 2x/5 - 3 < 1.

Problem 118

Solve the inequality with fractions or decimals: -0.1x + 0.5 >= 1.5.

Problem 119

Solve the inequality with fractions or decimals: -3x/4 + 2 > 5.

Problem 120

Solve the inequality with fractions or decimals: x/5 + 0.8 < 1.2.

simplify each side before isolating variable.
12 problems Warmup Practice Mixed Review Assessment
Problem 121

Solve after combining like terms on both sides: 3x + 2x + 4 = 24.

Open in simulator
Problem 122

Solve after combining like terms on both sides: 5x - 2 + x = 2x + 10.

Problem 123

Solve after combining like terms on both sides: 7 + 2x - 3 = x + 9.

Problem 124

Solve after combining like terms on both sides: 4x + 6x - 10 = 30.

Problem 125

Solve after combining like terms on both sides: 2x + 5 + 3x = x + 15 - 2.

Problem 126

Solve after combining like terms on both sides: 8x - 3x + 7 = 22.

Problem 127

Solve after combining like terms on both sides: 6x - 4 + 2x = 3x + 16 - 5.

Problem 128

Solve after combining like terms on both sides: 10 - 3x + 5x = 4x + 2 - x.

Problem 129

Solve after combining like terms on both sides: 12 = 7x - 2x + 2.

Problem 130

Solve after combining like terms on both sides: 9x - 2 - 4x + 5 = 18.

Problem 131

Solve after combining like terms on both sides: 7x + 3 - 2x - 1 = 22.

Problem 132

Solve after combining like terms on both sides: 15x - 7 + 5x = 10x + 23 - 10.

interpret final equivalent statement.
12 problems Warmup Practice Mixed Review Assessment
Problem 133

Classify 4x + 2 = 4x + 9 as one solution, no solution, or infinitely many solutions.

Problem 134

Classify 3(x + 2) = 3x + 6 as one solution, no solution, or infinitely many solutions.

Problem 135

Classify 2x - 1 = 9 as one solution, no solution, or infinitely many solutions.

Problem 136

Classify 5x - 3 = 5x + 7 as one solution, no solution, or infinitely many solutions.

Problem 137

Classify 2(x + 4) = 2x - 1 as one solution, no solution, or infinitely many solutions.

Problem 138

Classify x + 10 = x - 2 as one solution, no solution, or infinitely many solutions.

Open in simulator
Problem 139

Classify 7x + 5 = 7x + 5 as one solution, no solution, or infinitely many solutions.

Problem 140

Classify 4(x - 1) = 4x - 4 as one solution, no solution, or infinitely many solutions.

Problem 141

Classify x + 3 = x + 3 as one solution, no solution, or infinitely many solutions.

Problem 142

Classify 3x + 5 = 11 as one solution, no solution, or infinitely many solutions.

Problem 143

Classify 5x - 10 = 0 as one solution, no solution, or infinitely many solutions.

Problem 144

Classify x / 2 + 3 = 7 as one solution, no solution, or infinitely many solutions.

translate among inequality, graph, and interval notation.
12 problems Warmup Practice Mixed Review Assessment
Problem 145

Write the solution x < 4 in interval notation and describe the number-line graph.

Problem 146

Write the solution x >= -2 in interval notation and describe the number-line graph.

Open in simulator
Problem 147

Write the solution -1 < x <= 5 in interval notation and describe the number-line graph.

Problem 148

Write the solution x > 7 in interval notation and describe the number-line graph.

Problem 149

Write the solution x <= 0 in interval notation and describe the number-line graph.

Problem 150

Write the solution 3 < x < 9 in interval notation and describe the number-line graph.

Problem 151

Write the solution -5 <= x <= -1 in interval notation and describe the number-line graph.

Problem 152

Write the solution 0 <= x < 10 in interval notation and describe the number-line graph.

Problem 153

Write the solution x > -3 in interval notation and describe the number-line graph.

Problem 154

Write the solution x <= 15 in interval notation and describe the number-line graph.

Problem 155

Write the solution -8 < x <= 2 in interval notation and describe the number-line graph.

Problem 156

Write the solution 1 <= x < 6 in interval notation and describe the number-line graph.

reverse inequality direction.
12 problems Warmup Practice Mixed Review Assessment
Problem 157

Solve the inequality -4x < 20, where multiplying or dividing by a negative is required.

Problem 158

Solve the inequality -x/3 >= 6, where multiplying or dividing by a negative is required.

Problem 159

Solve the inequality 7 - 2x <= 15, where multiplying or dividing by a negative is required.

Problem 160

Solve the inequality -x > 5, where multiplying or dividing by a negative is required.

Problem 161

Solve the inequality -5x >= 25, where multiplying or dividing by a negative is required.

Problem 162

Solve the inequality 10 - 3x < 19, where multiplying or dividing by a negative is required.

Problem 163

Solve the inequality 8 - x/2 >= 12, where multiplying or dividing by a negative is required.

Problem 164

Solve the inequality 18 < -6x, where multiplying or dividing by a negative is required.

Problem 165

Solve the inequality 20 >= 5 - 5x, where multiplying or dividing by a negative is required.

Problem 166

Solve the inequality -2x/5 < 10, where multiplying or dividing by a negative is required.

Open in simulator
Problem 167

Solve the inequality 4 - 3x > 16 + x, where multiplying or dividing by a negative is required.

Problem 168

Solve the inequality -2(x + 3) <= 10, where multiplying or dividing by a negative is required.

decide whether to distribute, clear fractions, combine terms, or move variables first.
12 problems Warmup Practice Mixed Review Assessment
Problem 169

Choose the best first method for solving 3(x + 4) = 27 and explain why.

Problem 170

Choose the best first method for solving x/3 + x/6 = 5 and explain why.

Problem 171

Choose the best first method for solving 2x + 5x - 4 = 31 and explain why.

Problem 172

Choose the best first method for solving 4x + 7 = 2x + 19 and explain why.

Problem 173

Choose the best first method for solving 2(x + 1) = 4x and explain why.

Problem 174

Choose the best first method for solving y/2 - 1 = y/4 and explain why.

Open in simulator
Problem 175

Choose the best first method for solving 7w - 3w + 10 = 2 and explain why.

Problem 176

Choose the best first method for solving 3z - 8 = z + 12 and explain why.

Problem 177

Choose the best first method for solving (x + 7)/4 = 3 and explain why.

Problem 178

Choose the best first method for solving 5 + 3x - 2 = 10 and explain why.

Problem 179

Choose the best first method for solving 1/2 + m/3 = 5/6 and explain why.

Problem 180

Choose the best first method for solving 6p - 1 = 9p + 11 and explain why.