Chapter Practice Test (Interactive)
Hint 1 — Relevant concept
Solve the equation for \(y\) to get slope-intercept form \(y = mx + b\), then pick a few \(x\)-values (multiples of the slope's denominator keep \(y\) an integer), compute \(y\), and plot the points (sec 2.1).
Hint 2 — Worked example
See Example 2 in sec 2.1 (Graphing an Equation in Two Variables by Plotting Points).
Hint 1 — Relevant concept
Set \(y = 0\) and solve for \(x\) to find the \(x\)-intercept; set \(x = 0\) and solve for \(y\) to find the \(y\)-intercept. Write each as an ordered pair (sec 2.1).
Hint 2 — Worked example
See Example 4 in sec 2.1 (Finding the Intercepts of the Given Equation).
Hint 1 — Relevant concept
Distance: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\), left as an exact radical. Midpoint: \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\). Watch the double negatives when subtracting (sec 2.1).
Hint 2 — Worked example
See Example 5 in sec 2.1 (Finding the Distance between Two Points) and Example 7 (Finding the Midpoint of the Line Segment).
Hint 1 — Relevant concept
A bracket means the endpoint is included (\(\le\) or \(\ge\)); a parenthesis means it is not (\(\lt\) or \(\gt\)). Infinity always gets a parenthesis (sec 2.7).
Hint 2 — Worked example
See Example 1 in sec 2.7 (Using Interval Notation to Express All Real Numbers Greater Than or Equal to a).
Hint 1 — Relevant concept
To solve a linear equation with the variable on both sides, collect all variable terms on one side and all constant terms on the other (sec 2.2).
Hint 2 — Worked example
See Example 1 in sec 2.2 (Solving an Equation in One Variable), which solves a similar equation by isolating \(x\).
Hint 1 — Relevant concept
First distribute each factor, then combine like terms on each side before isolating \(x\) (sec 2.2).
Hint 2 — Worked example
See Example 2 in sec 2.2 (Solving an Equation Algebraically When the Variable Appears on Both Sides), which distributes and collects terms to solve a linear equation.
Hint 1 — Relevant concept
Note the excluded value \(x \ne 0\), multiply every term by the LCD to clear the denominators, then solve the resulting quadratic by factoring and check that no solution is an excluded value (sec 2.2, 2.5).
Hint 2 — Worked example
See Example 4 in sec 2.2 (Solving a Rational Equation without Factoring).
Hint 1 — Relevant concept
The two denominators have no common factor, so the LCD is their product. State the excluded values, multiply the whole equation by the LCD, distribute, and solve the resulting quadratic. Reject any solution that equals an excluded value (sec 2.2).
Hint 2 — Worked example
See Example 6 in sec 2.2 (Solving Rational Equations with a Binomial in the Denominator) and Example 7 (Solving a Rational Equation with Factored Denominators and Stating Excluded Values).
Hint 1 — Relevant concept
Let \(s\) be the shortest side, write the other two sides in terms of \(s\), set the sum of the three sides equal to the perimeter, solve for \(s\), and back-substitute (sec 2.3).
Hint 2 — Worked example
See Example 1 in sec 2.3 (Modeling a Linear Equation to Solve an Unknown Number Problem).
Hint 1 — Relevant concept
Multiply through by the LCD of the fractions to get integer coefficients, write the quadratic in standard form, then use the quadratic formula. Simplify the radical and reduce the fraction (sec 2.5).
Hint 2 — Worked example
See Example 9 in sec 2.5 (Solve the Quadratic Equation Using the Quadratic Formula).
Hint 1 — Relevant concept
Solve an inequality like an equation: move the constant, then divide by the coefficient (the coefficient here is positive, so the inequality does not reverse). Write the answer as an inequality or in interval notation (sec 2.7).
Hint 2 — Worked example
See Example 5 in sec 2.7 (Solving an Inequality Algebraically).
Hint 1 — Relevant concept
\(|X| \lt k\) means \(-k \lt X \lt k\). Rewrite as a compound inequality and operate on all three parts at once (sec 2.7).
Hint 2 — Worked example
See Example 10 in sec 2.7 (Solving an Absolute Value Inequality).
Hint 1 — Relevant concept
\(|X| \ge k\) means \(X \le -k\) or \(X \ge k\). Solve the two inequalities separately and join them with "or" — a union of two intervals (sec 2.7).
Hint 2 — Worked example
See Example 11 in sec 2.7 and the Absolute Value Inequalities rule box just before Example 10.
Hint 1 — Relevant concept
Compute the slope \(m = \frac{y_2 - y_1}{x_2 - x_1}\), then use the point-slope form \(y - y_1 = m(x - x_1)\) with either point and simplify to \(y = mx + b\) (sec 2.2).
Hint 2 — Worked example
See Example 11 in sec 2.2 (Finding the Equation of a Line Passing Through Two Given Points).
Hint 1 — Relevant concept
An undefined slope means a vertical line: every point on it has the same \(x\)-coordinate, so the equation is \(x = c\) (sec 2.2).
Hint 2 — Worked example
See Example 13 in sec 2.2 (Finding the Equation of a Line Passing Through the Given Points).
Hint 1 — Relevant concept
The slope of a perpendicular line is the negative reciprocal of the given slope. Then use point-slope form with the given point and simplify (sec 2.2).
Hint 2 — Worked example
See Example 16 in sec 2.2 (Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point).
Hint 1 — Relevant concept
Add the real parts together and the imaginary parts together, then write the result as \(a + bi\) (sec 2.4).
Hint 2 — Worked example
See Example 3 in sec 2.4 (Adding and Subtracting Complex Numbers).
Hint 1 — Relevant concept
Recall that \(\sqrt{-1} = i\), so \(\sqrt{-a} = i\sqrt{a}\) for any positive real number \(a\). Simplify each radical separately, then combine like terms (sec 2.4).
Hint 2 — Worked example
See Example 1 in sec 2.4 (Expressing an Imaginary Number in Standard Form), which rewrites square roots of negative numbers using \(i\).
Hint 1 — Relevant concept
Distribute, replace \(i^2\) with \(-1\), then group the real and imaginary terms into \(a + bi\) form (sec 2.4).
Hint 2 — Worked example
See Example 4 in sec 2.4 (Multiplying a Complex Number by a Real Number) and Example 5 (Multiplying a Complex Number by a Complex Number).
Hint 1 — Relevant concept
To divide complex numbers, multiply numerator and denominator by the complex conjugate of the denominator. The conjugate of \(a + bi\) is \(a - bi\), and \((a + bi)(a - bi) = a^2 + b^2\) (sec 2.4).
Hint 2 — Worked example
See Example 7 in sec 2.4 (Dividing Complex Numbers), which divides two complex numbers by multiplying by the conjugate.
Hint 1 — Relevant concept
Use the quadratic formula. The discriminant is negative, so write \(\sqrt{-k}\) as \(i\sqrt{k}\), then split the fraction into real and imaginary parts to get \(a \pm bi\) (sec 2.5, 2.4).
Hint 2 — Worked example
See Example 10 in sec 2.5 (Solving a Quadratic Equation with the Quadratic Formula).
Hint 1 — Relevant concept
Isolate the squared expression, then use the square root property: if \(X^2 = k\) then \(X = \pm\sqrt{k}\). Solve the two resulting linear equations (sec 2.5).
Hint 2 — Worked example
See Example 7 in sec 2.5 (Solving a Quadratic Equation Using the Square Root Property).
Hint 1 — Relevant concept
Complete the square: take half the coefficient of \(x\), square it, add it to both sides, factor the left side as a perfect square, then apply the square root property (sec 2.5).
Hint 2 — Worked example
See Example 8 in sec 2.5 (Solving a Quadratic by Completing the Square).
Hint 1 — Relevant concept
Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a\), \(b\), \(c\) are the coefficients of \(ax^2 + bx + c = 0\). Simplify the radical and reduce (sec 2.5).
Hint 2 — Worked example
See Example 9 in sec 2.5 (Solve the Quadratic Equation Using the Quadratic Formula) or Example 10 (Solving a Quadratic Equation with the Quadratic Formula).
Hint 1 — Relevant concept
The radical is already isolated: square both sides, solve the resulting quadratic by factoring, then check each candidate in the original equation and discard any extraneous solution (sec 2.6).
Hint 2 — Worked example
See Example 6 in sec 2.6 (Solving an Equation with One Radical).
Hint 1 — Relevant concept
Isolate the radical first, then square both sides, solve the quadratic by factoring, and check both candidates in the original equation — one of them is extraneous (sec 2.6).
Hint 2 — Worked example
See Example 6 in sec 2.6 (Solving an Equation with One Radical) and Example 7 for the isolation step.
Hint 1 — Relevant concept
Raise both sides to the reciprocal power \(\tfrac{3}{2}\). Because the original exponent has an even numerator, both the positive and the negative cube root give a solution — check both (sec 2.6).
Hint 2 — Worked example
See Example 2 in sec 2.6 (Solve the Equation Including a Variable Raised to a Rational Exponent).
Hint 1 — Relevant concept
Factor by grouping: factor the first two terms and the last two terms, pull out the common binomial, then factor the remaining difference of squares. Set each factor to zero (sec 2.6).
Hint 2 — Worked example
See Example 5 in sec 2.6 (Solve a Polynomial by Grouping).
Hint 1 — Relevant concept
The equation is quadratic in form: substitute \(u\) for the repeated binomial, factor the quadratic in \(u\), then replace \(u\) and solve for \(x\). Do not expand the binomial (sec 2.6).
Hint 2 — Worked example
See Example 10 in sec 2.6 (Solving an Equation in Quadratic Form Containing a Binomial).