Learning goals: Use equivalent expressions carefully; solve equations and inequalities; describe solution sets with interval notation.
Structure before calculation
An expression names a quantity; an equation states that two expressions are equal. Simplifying \(3(x-2)+2x\) gives \(5x-6\), but it does not determine \(x\). Solving \(5x-6=14\) asks which values make the statement true. Here \(5x=20\), so \(x=4\). Substitute into the original expression to check: \(3(4-2)+2(4)=14\).
The distributive property connects expansion and factoring. For example, \(6x+9=3(2x+3)\), and \(x^2-9=(x-3)(x+3)\). Factoring is not decoration: a product equal to zero lets us identify zero factors. Conversely, a sum equal to zero does not require each summand to be zero.
What makes a transformation legitimate?
Adding the same quantity to both sides of an equation preserves its solutions. Multiplying by a nonzero constant does too. Dividing by an expression requires it to be nonzero. In \(\frac{x^2-1}{x-1}=x+1\), cancellation is valid only for \(x\ne1\). The simplified formula does not restore an input excluded by the original expression. Similarly, \(\sqrt{x^2}=|x|\), because the principal square root is nonnegative.
Ordered numbers and intervals
The number line turns inequalities into sets. The notation \([-2,5)\) includes \(-2\) but excludes \(5\); it means \(-2\le x\lt5\). Parentheses always accompany infinity because infinity is not a real-number endpoint. Intersection means “and,” so both conditions must hold. Union means “or,” so either condition suffices.
Solve \(-3x+7\ge16\): subtract seven to get \(-3x\ge9\), then divide by \(-3\) and reverse the inequality to obtain \(x\le-3\). Multiplication by a negative reverses order on the number line. Check a value such as \(-4\), which gives \(19\ge16\), and a rejected value such as zero, which gives the false statement \(7\ge16\).
See “and” and “or” on a number line
Pause and explain: Does \(x=5\) belong to the intersection? Does it belong to the union? Explain using both original intervals.
Check your reasoning
Five is not in A because A ends with a parenthesis, but it is in B. Therefore it fails the “both” requirement for \(A\cap B\), yet satisfies the “at least one” requirement for \(A\cup B\). This is why the intersection has an open endpoint at five while the union includes it.
Absolute value is distance
The expression \(|x-a|\) is the distance from \(x\) to \(a\). Thus \(|x-2|\le3\) describes all points at most three units from two: \[-3\le x-2\le3\quad\Longrightarrow\quad -1\le x\le5.\] In contrast, \(|x-2|\gt3\) describes the two exterior regions \(x\lt-1\) or \(x\gt5\). Do not join these exterior conditions with “and”: no number lies in both regions.
Absolute value measures distance
Explore: select a case. With the controls focused, use the arrow keys to switch cases.
Selected case. \(|x-2|\le3\) gives \([-1,5]\). Both endpoints have distance exactly three and are included.
Selected case. \(|x-2|\gt3\) gives \((-\infty,-1)\cup(5,\infty)\). Arrows indicate continuation beyond the picture; the endpoints are excluded.
Pause and explain: Where does \(x=-2\) belong? Check its distance instead of relying only on the shading.
Check your reasoning
\(|-2-2|=|-4|=4\). Since four is greater than three, negative two belongs to the outer left ray. It does not satisfy the “at most three” condition.
Efficient checking
Use exact fractions until the last step, especially when rounding could move an endpoint. Label exclusions and check boundary points separately. A decimal approximation can support a calculation, but it does not replace an exact solution when the question asks for one. Explain both the algebraic transformation and its effect on the solution set.
Practice: attempt before reading the solutions
Show your reasoning, state restrictions, and check your result. For explanation questions, use complete mathematical sentences.
- Simplify \(4(2x-3)-3(x+2)\).
- Solve \(\frac{x}{3}+2=7\).
- Solve \(5-2x\lt11\) and give interval notation.
- Solve \(-1\le2x+3\lt9\).
- Solve \(|x+4|\le2\).
- Solve \(|2x-1|\gt5\).
- Simplify \(\frac{x^2-4}{x-2}\) and state the restriction.
- Evaluate \(\sqrt{x^2}\) when \(x=-7\).
Worked solutions and reasoning
Compare these steps with your own work. The solution numbers match the practice questions above.
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Distribute first: \(8x-12-3x-6=5x-18\). The minus sign applies to both terms inside the second parentheses.
The second product is \(-3(x+2)=-3x-6\), not \(-3x+6\). Combine like terms only after distributing: the x-coefficients give \(8-3=5\), and the constants give \(-12-6=-18\).
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Subtract two, giving \(x/3=5\), then multiply by three: \(x=15\). Substitution gives \(5+2=7\).
Use inverse operations on both sides to preserve equality. Subtracting two undoes the addition, and multiplying by three undoes division by three. The final substitution is a check in the original equation, not merely in a rearranged version.
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\(-2x\lt6\), so \(x\gt-3\) after dividing by a negative. The interval is \((-3,\infty)\).
Dividing by \(-2\) reverses the inequality because multiplying numbers by a negative reverses their order. The endpoint \(-3\) gives equality \(11=11\), so it is excluded. As a check, \(x=0\) gives \(5<11\).
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Subtract three from all parts: \(-4\le2x\lt6\). Divide by two: \(-2\le x\lt3\), or \([-2,3)\).
The same operation must be applied to all three parts. Division by the positive number two preserves both inequality directions. At \(x=-2\), the middle expression equals \(-1\), which is allowed; at \(x=3\), it equals nine, which is excluded.
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The distance from \(-4\) is at most two. Equivalently \(-2\le x+4\le2\), so the solution is \([-6,-2]\).
An absolute value at most two means the enclosed quantity lies between \(-2\) and \(2\), inclusive. Subtract four from all three parts. Both endpoints give absolute value two, while the center \(x=-4\) gives zero.
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Use \(2x-1\gt5\) or \(2x-1\lt-5\). Thus \(x\gt3\) or \(x\lt-2\), giving \((-\infty,-2)\cup(3,\infty)\).
An absolute value greater than five means the enclosed quantity is either above five or below negative five. Solve the two branches separately, then join them with “or.” The endpoints produce exactly five and are excluded by the strict inequality.
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Factor the numerator: \((x-2)(x+2)\). Cancellation gives \(x+2\) only when \(x\ne2\); the original denominator excludes two.
Cancellation divides numerator and denominator by the same nonzero factor; it does not erase the original restriction. At \(x=2\), the original expression is undefined even though \(x+2\) equals four. The two formulas agree only on the original domain.
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\(\sqrt{49}=7\), consistent with \(\sqrt{x^2}=|x|\), not with the incorrect rule \(\sqrt{x^2}=x\) for all real numbers.
The radical symbol denotes the nonnegative square root. Although both \(7\) and \(-7\) square to 49, \(\sqrt{49}\) denotes seven alone. This distinguishes evaluating a radical from solving the equation \(u^2=49\).
Reflection
Choose one problem you initially found difficult. Explain the key idea, identify a possible mistake, and write a new problem that uses the same idea. Solve your new problem and check it independently.