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 \(7\ge16\).

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.

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.

1. Simplify \(4(2x-3)-3(x+2)\).
2. Solve \(\frac{x}{3}+2=7\).
3. Solve \(5-2x\lt11\) and give interval notation.
4. Solve \(-1\le2x+3\lt9\).
5. Solve \(|x+4|\le2\).
6. Solve \(|2x-1|\gt5\).
7. Simplify \(\frac{x^2-4}{x-2}\) and state the restriction.
8. Evaluate \(\sqrt{x^2}\) when \(x=-7\).

Worked solutions and reasoning

1. Distribute first: \(8x-12-3x-6=5x-18\). The minus sign applies to both terms inside the second parentheses.
2. Subtract two, giving \(x/3=5\), then multiply by three: \(x=15\). Substitution gives \(5+2=7\).
3. \(-2x\lt6\), so \(x\gt-3\) after dividing by a negative. The interval is \((-3,\infty)\).
4. Subtract three from all parts: \(-4\le2x\lt6\). Divide by two: \(-2\le x\lt3\), or \([-2,3)\).
5. The distance from \(-4\) is at most two. Equivalently \(-2\le x+4\le2\), so the solution is \([-6,-2]\).
6. Use \(2x-1\gt5\) or \(2x-1\lt-5\). Thus \(x\gt3\) or \(x\lt-2\), giving \((-\infty,-2)\cup(3,\infty)\).
7. Factor the numerator: \((x-2)(x+2)\). Cancellation gives \(x+2\) only when \(x\ne2\); the original denominator excludes two.
8. \(\sqrt{49}=7\), consistent with \(\sqrt{x^2}=|x|\), not with the incorrect rule \(\sqrt{x^2}=x\) for all real numbers.

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.