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← Prealgebra levels

Prealgebra / LEVEL 5 · DIFFICULTY 2/5

Expressions, Equations & Inequalities

Represent unknown quantities and justify each transformation.

3 stages · 24 practice problems · two 6-question assessment forms

Choose an island to read its lesson.

  1. 1MINI QUEST 5.1Expressions & SubstitutionRead the lesson
  2. 2MINI QUEST 5.2Solving & Checking Linear EquationsRead the lesson
  3. 3MINI QUEST 5.3Inequalities & Feasible ValuesRead the lesson
  4. LEVEL CHECKCastle challenge

    6 questions across this level.

    Sign in for the level check →

STAGE 5.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Expressions & Substitution

Useful preparation: Mixed Numbers & Fraction Powers

The idea to remember

Substitute each variable's value using parentheses, then follow the order of operations.

Remember with 2 flashcards →

Goal: Build and evaluate expressions while combining only like terms.

Before you begin: Before starting, explain one example from each prerequisite. If an idea is unfamiliar, follow the prerequisite lesson links.

A variable holds a place for a quantity

An expression describes a calculation. It has a value once each variable is assigned a number; it is not an equation unless an equality is asserted.

3 x+4\displaystyle 3\, x + 43x+4

Substitute with parentheses

Replace every occurrence of the variable. Parentheses preserve the signs and grouping when the replacement is negative or fractional.

x2\displaystyle x^{2}x2 at x=−3\displaystyle x = -3x=−3 is (−3)2\displaystyle \left(-3\right)^{2}(−3)2

Like terms share a variable part

Combine coefficients only when the variable factors and exponents match. Distribution reveals the terms hidden in parentheses.

4 x+7 x−2=11 x−2\displaystyle 4\, x + 7\, x - 2 = 11\, x - 24x+7x−2=11x−2
WORKED EXAMPLE 1

Evaluate 2 x2−3 x\displaystyle 2\, x^{2} - 3\, x2x2−3x at x=−2\displaystyle x = -2x=−2.

  1. Substitute: 2 (−2)2−3 (−2)\displaystyle 2\, \left(-2\right)^{2} - 3\, \left(-2\right)2(−2)2−3(−2).
  2. Calculate 8+6\displaystyle 8 + 68+6.

14

WORKED EXAMPLE 2

A ride costs 6 dollars plus 4 dollars per kilometer. What is the cost for 5 kilometers?

  1. Model the cost as 6+4 k\displaystyle 6 + 4\, k6+4k.
  2. Substitute k=5\displaystyle k = 5k=5.

26 dollars

Common pitfalls

Possible mix-up: 3 x+2\displaystyle 3\, x + 23x+2 is 5x.

A constant and an x-term are not like terms.

Possible mix-up: Substitute −2 into x2\displaystyle x^{2}x2 as −22\displaystyle -2^{2}−22.

Use (−2)2\displaystyle \left(-2\right)^{2}(−2)2 so the negative is part of the base.

Explain it to yourself

Give two different stories represented by 5+3 n\displaystyle 5 + 3\, n5+3n.

Discuss together

Ask what the variable and each fixed number mean in the situation.

Preview the eight practice prompts
  1. Problem 1

    Evaluate 5 x−4\displaystyle 5\, x - 45x−4 at x=3\displaystyle x = 3x=3.

    Practise this exact problem →
  2. Problem 2

    Evaluate x2+2 x\displaystyle x^{2} + 2\, xx2+2x at x=−3\displaystyle x = -3x=−3.

    Practise this exact problem →
  3. Problem 3

    Combine 8 a−3 a+6 a\displaystyle 8\, a - 3\, a + 6\, a8a−3a+6a. What is the coefficient of a?

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  4. Problem 4

    Expand 4 (2 x−5)\displaystyle 4\, \left(2\, x - 5\right)4(2x−5). What is the constant term?

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  5. Problem 5

    Evaluate a+b2\displaystyle \frac{a + b}{2}2a+b​ when a=73\displaystyle a = \frac{7}{3}a=37​ and b=53\displaystyle b = \frac{5}{3}b=35​.

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  6. Problem 6

    Evaluate 3 (t−2)+2 (t+4)\displaystyle 3\, \left(t - 2\right) + 2\, \left(t + 4\right)3(t−2)+2(t+4) at t=6\displaystyle t = 6t=6.

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  7. Problem 7New context

    A bike rental charges $8 to start and $3 for each hour. What is the total cost in dollars for 7 hours?

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  8. Problem 8New context

    A row of n square frames uses 4n sticks before shared edges are removed. There are n−1\displaystyle n - 1n−1 shared edges, each counted twice. How many sticks remain for n=9\displaystyle n = 9n=9?

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Open stage 5.1 in the student workspace →

STAGE 5.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Solving & Checking Linear Equations

Useful preparation: Expressions & Substitution

The idea to remember

Perform the same valid operation on both sides to preserve equality. Check the solution in the original equation.

Remember with 2 flashcards →

Goal: Maintain equality while isolating an unknown quantity.

Before you begin: Before starting, explain one example from each prerequisite. If an idea is unfamiliar, follow the prerequisite lesson links.

START WITH SOMETHING YOU CAN MOVE

What is inside the mystery box?

Every star weighs one unit. Each mystery box has the same weight. Drag stars into the tray, or tap a star and then a “Move selected” button.

One small move: The two sides start with equal weight. Now move ONE star from the mystery side into the tray.
Mystery side
?
1 box + 3 stars
Star side
7 stars
▲
Star tray
Both sides start level. Nothing has been moved yet.
Next: connect equations to graphs

PARENT EQUATION

Linear parent: y=x\displaystyle y = xy=x

A straight line through (0, 0): one step right gives one step up.

How does the question’s equation change this graph?

In y=ax+b\displaystyle y = ax + by=ax+b, a scales the vertical change (and reflects across the x-axis when negative); b moves the line up or down.

Check: If a=0\displaystyle a = 0a=0, the result is a horizontal constant, not a tilted line.

First identify this family. Then read the coefficients and shifts in your actual equation; this reference does not replace the question’s given values.

AN EQUATION IS A RULE FOR MAKING POINTS

Change the rule. Watch the graph.

Pick an x value. The rule tells you its y value. A point (x, y) shows both numbers together. Drag across the graph or use the point slider. This is an adjustable learning example; use the coefficients in your own practice question when solving it.

Parent equation: y=x\displaystyle y = xy=x

Start with y=x\displaystyle y = xy=x. Multiply each y value by 2, then shift up 1. A negative multiplier reflects across the x-axis.

Trace a point along the graph
0%

Blue: y=2 x+1\displaystyle y = 2\, x + 1y=2x+1

y = 2x + 1; horizontal x-axis and vertical y-axis-10-10-8-8-6-6-4-4-2-2224466881010xy0

Moving point: (-4, -7)

Dashed gray: parent equation. Blue: current equation. View: −10 to 10 on both axes. The graph continues beyond this window. Purple marks your selected point.

Slope a: 2
Vertical intercept b: 1
Choose the point’s x: 1
Point (1, 3)
2×(1)+(1)=3\displaystyle 2 \times \left(1\right) + \left(1\right) = 32×(1)+(1)=3
What does moving one step right do?

Each step of 1 to the right changes y by 2. The line rises. Setting x=0\displaystyle x = 0x=0 gives y=1\displaystyle y = 1y=1, where the line meets the vertical axis.

Three points made by the blue rule
xSubstitute into the ruley
-12×(−1)+(1)\displaystyle 2 \times \left(-1\right) + \left(1\right)2×(−1)+(1)-1
02×(0)+(1)\displaystyle 2 \times \left(0\right) + \left(1\right)2×(0)+(1)1
12×(1)+(1)\displaystyle 2 \times \left(1\right) + \left(1\right)2×(1)+(1)3

An equation states a balance

Adding or subtracting the same quantity on both sides preserves equality. Multiplication or division also preserves equality when dividing only by nonzero numbers.

x+5=12⇒x=7\displaystyle x + 5 = 12 \Rightarrow x = 7x+5=12⇒x=7

Undo operations in reverse order

For a chain such as 3 x+4\displaystyle 3\, x + 43x+4, undo the final addition before the multiplication. With variables on both sides, collect like terms first.

3 x+4=19⇒3 x=15⇒x=5\displaystyle 3\, x + 4 = 19 \Rightarrow 3\, x = 15 \Rightarrow x = 53x+4=19⇒3x=15⇒x=5

Check in the original statement

A candidate answer must satisfy the equation before simplification and fit the story. Some equations are true for every x; others have no solution.

2 x+1=2 x+3\displaystyle 2\, x + 1 = 2\, x + 32x+1=2x+3 has no solution
WORKED EXAMPLE 1

Solve 4 (x−2)=3 x+9\displaystyle 4\, \left(x - 2\right) = 3\, x + 94(x−2)=3x+9.

  1. Expand: 4 x−8=3 x+9\displaystyle 4\, x - 8 = 3\, x + 94x−8=3x+9.
  2. Subtract 3x and add 8.
  3. Check: 4 (17−2)=60\displaystyle 4\, \left(17 - 2\right) = 604(17−2)=60 and 3 (17)+9=60\displaystyle 3\, \left(17\right) + 9 = 603(17)+9=60.

17

WORKED EXAMPLE 2

Three equal notebooks and a $4 folder cost $25. What is a notebook price?

  1. Write 3 n+4=25\displaystyle 3\, n + 4 = 253n+4=25.
  2. Subtract 4, then divide by 3.

7 dollars

Common pitfalls

Possible mix-up: Move a term without changing anything else.

Describe the same operation on both sides; this avoids sign errors.

Possible mix-up: Every equation has one answer.

Some equations have no solutions or infinitely many.

Explain it to yourself

How does substitution help detect an error when solving an equation?

Discuss together

Ask the learner to justify one line with an operation applied to both sides, then verify the result.

Preview the eight practice prompts
  1. Problem 1

    Solve x+13=31\displaystyle x + 13 = 31x+13=31.

    Practise this exact problem →
  2. Problem 2

    Solve 5 x−7=28\displaystyle 5\, x - 7 = 285x−7=28.

    Practise this exact problem →
  3. Problem 3

    Solve x4+3=8\displaystyle \frac{x}{4} + 3 = 84x​+3=8.

    Practise this exact problem →
  4. Problem 4

    Solve 3 (x+2)=2 x+17\displaystyle 3\, \left(x + 2\right) = 2\, x + 173(x+2)=2x+17.

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  5. Problem 5

    Solve (23) x−12=56\displaystyle \left(\frac{2}{3}\right)\, x - \frac{1}{2} = \frac{5}{6}(32​)x−21​=65​.

    Practise this exact problem →
  6. Problem 6

    Does 4 x+2=4 x−5\displaystyle 4\, x + 2 = 4\, x - 54x+2=4x−5 have any solution? Enter yes or no.

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  7. Problem 7New context

    Five equal tickets and a $6 booking charge cost $71. What is the price of one ticket in dollars?

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  8. Problem 8New context

    Two plans cost 9+4 h\displaystyle 9 + 4\, h9+4h dollars and 21+2 h\displaystyle 21 + 2\, h21+2h dollars for h hours. At what value of h do their costs match?

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Open stage 5.2 in the student workspace →

STAGE 5.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Inequalities & Feasible Values

Useful preparation: Solving & Checking Linear Equations

The idea to remember

Multiplying or dividing both sides of an inequality by a negative number reverses the inequality sign.

Remember with 2 flashcards →

Goal: Track inequality direction and interpret boundaries in context.

Before you begin: Before starting, explain one example from each prerequisite. If an idea is unfamiliar, follow the prerequisite lesson links.

Inequalities describe sets

A strict inequality excludes its boundary; a non-strict inequality includes it. One test value does not describe the whole solution set.

x<4\displaystyle x < 4x<4 excludes 4; x≤4\displaystyle x \le 4x≤4 includes 4

Negative scaling reverses order

Adding the same number preserves order. Multiplying or dividing by a negative reverses the inequality because it reverses positions on the number line.

−2 x<8⇒x>−4\displaystyle -2\, x < 8 \Rightarrow x > -4−2x<8⇒x>−4

Context can restrict the answer

A continuous solution interval may contain only certain allowed whole numbers. Check endpoint inclusion and any requirement that quantities be nonnegative.

3 n≤14\displaystyle 3\, n \le 143n≤14, n a nonnegative integer ⇒ n≤4\displaystyle n \le 4n≤4
WORKED EXAMPLE 1

Solve −3 x+2≥14\displaystyle -3\, x + 2 \ge 14−3x+2≥14.

  1. Subtract 2:−3 x≥12\displaystyle 2 : -3\, x \ge 122:−3x≥12.
  2. Divide by −3 and reverse the comparison.

x≤−4\displaystyle x \le -4x≤−4

WORKED EXAMPLE 2

A $9 entry fee leaves money for $4 rides within a $30 budget. What is the maximum number of rides?

  1. Write 9+4 r≤30\displaystyle 9 + 4\, r \le 309+4r≤30.
  2. r≤214=5.25\displaystyle r \le \frac{21}{4} = 5.25r≤421​=5.25.
  3. Only whole rides are allowed.

5

Common pitfalls

Possible mix-up: Dividing by a negative leaves the inequality direction unchanged.

Negative scaling reverses order.

Possible mix-up: A maximum of 5.7 items means 6 can be bought.

A budget bound must be rounded down to a feasible whole count.

Explain it to yourself

Explain why a boundary can satisfy ≤ but fail <.

Discuss together

Ask the learner to test the proposed maximum and the next larger whole number.

Preview the eight practice prompts
  1. Problem 1

    What is the greatest integer x satisfying x+7≤12\displaystyle x + 7 \le 12x+7≤12?

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  2. Problem 2

    What is the least integer x satisfying 2 x>9\displaystyle 2\, x > 92x>9?

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  3. Problem 3

    What is the greatest integer x satisfying −3 x≥12\displaystyle -3\, x \ge 12−3x≥12?

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  4. Problem 4

    Does x=6\displaystyle x = 6x=6 satisfy 4 x−5<19\displaystyle 4\, x - 5 < 194x−5<19? Enter yes or no.

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  5. Problem 5

    How many integers satisfy −2<x≤4\displaystyle -2 < x \le 4−2<x≤4?

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  6. Problem 6

    What is the least integer x satisfying −2 x+3<−8\displaystyle -2\, x + 3 < -8−2x+3<−8?

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  7. Problem 7New context

    A museum charges $11 entry plus $6 per workshop. With $40, how many whole workshops can a visitor afford at most?

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  8. Problem 8New context

    A delivery van already carries 85 kg. Each crate adds 18 kg, and total load must be strictly below 230 kg. What is the maximum whole number of crates?

    Practise this exact problem →
Open stage 5.3 in the student workspace →
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∑ Numerive

Learn

HomeTry a questionMy questsLearning SearchHomeworkUnderstand & rememberCurriculumFlashcardsUniversity

Exam preparation

Mock examsSATPSAT/NMSQTAMC 8 / 10 / 12

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My characterAchievementsFriendsStudy groups

Family & help

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Expressions, Equations & Inequalities: lessons and practice | Numerive Math