Algebra Practice

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Discriminant	Result
$\Delta > 0$	two distinct real roots
$\Delta = 0$	one repeated real root
$\Delta < 0$	two complex roots

Difficulty

Direct

1.1Combine Like Terms in a Polynomial

Exam I | Problem 1.1 |

Like Terms

Simplify the expression and report the coefficient of $x$:

$$ 4x - 2x + 9x + 6 $$

Solution

Combine the like terms:

$$ 4x - 2x + 9x = 11x $$

So the simplified expression is

$$ 11x + 6 $$

The coefficient of $x$ is $11$.

1.2Solve a Two-Step Linear Equation

Exam I | Problem 1.2 |

Solve for $x$:

$$ 5x - 8 = 27 $$

Solution

Add $8$ to both sides:

$$ 5x = 35 $$

Divide both sides by $5$:

$$ x = 7 $$

1.3Evaluate an Expression with Powers and Parentheses

Exam I | Problem 1.3 |

Compute the value of

$$ 2^3 + 3(5 - 2) $$

Solution

Follow the order of operations:

$$ 2^3 = 8 $$

and

$$ 5 - 2 = 3 $$

Then multiply:

$$ 3(3) = 9 $$

Finally add:

$$ 8 + 9 = 17 $$

1.4Distribute a Constant Across a Binomial

Exam I | Problem 1.4 |

Distributing and factoring

After distributing, what is the coefficient of $x$ in the expression below?

$$ 4(3x - 5) $$

Solution

Distribute the $4$ to each term:

$$ 4(3x - 5) = 12x - 20 $$

The coefficient of $x$ is $12$.

1.5Evaluate a Linear Function

Exam I | Problem 1.5 |

Evaluating functions

$$ f(x) = 7x - 4, $$

what is $f(2)$?

Solution

Substitute $2$ for $x$:

$$ f(2) = 7(2) - 4 $$

$$ f(2) = 14 - 4 = 10 $$

1.6Solve a Fraction Equation

Exam I | Problem 1.6 |

Solve for $x$:

\frac{x}{6} + 2 = 9

Solution

Subtract $2$ from both sides:

\frac{x}{6} = 7

Multiply both sides by $6$:

$$ x = 42 $$

1.7Combine Fractional Coefficients

Exam I | Problem 1.7 |

Simplify the expression and give the coefficient of $x$:

\frac{2}{3}x + \frac{1}{3}x + 4

Solution

The $x$-terms have a common denominator:

\frac{2}{3}x + \frac{1}{3}x = \left(\frac{2}{3} + \frac{1}{3}\right)x = 1x

So the coefficient of $x$ is $1$.

1.8Solve a Bracketed Equation

Exam I | Problem 1.8 |

Solve for $x$:

$$ 3(x + 4) = 27 $$

Solution

Divide both sides by $3$:

$$ x + 4 = 9 $$

Subtract $4$ from both sides:

$$ x = 5 $$

1.9Use Inverse Operations on a Simple Ratio

Exam I | Problem 1.9 |

Solve for $x$:

\frac{x}{9} = 7

Solution

Multiply both sides by $9$:

$$ x = 63 $$

1.10Evaluate a Quadratic-Looking Expression

Exam I | Problem 1.10 |

$$ g(t) = 2t^2 - t, $$

what is $g(3)$?

Solution

Substitute $3$ for $t$:

$$ g(3) = 2(3^2) - 3 $$

Evaluate the exponent first:

$$ g(3) = 2(9) - 3 = 18 - 3 = 15 $$

Difficulty

Integrated

2.1Distribution and Like-Term Combination

Exam II | Problem 2.1 |

Solve for $x$:

$$ 4(2x - 3) + 5 = 3x + 18 $$

Solution

Expand the left-hand side:

$$ 8x - 12 + 5 = 3x + 18 $$

Combine constants:

$$ 8x - 7 = 3x + 18 $$

Move variable terms to one side:

$$ 5x - 7 = 18 $$

Add $7$:

$$ 5x = 25 $$

Divide by $5$:

$$ x = 5 $$

2.2Fraction Equation with Two Terms

Exam II | Problem 2.2 |

Solve for $x$:

\frac{x - 1}{4} + \frac{x + 3}{2} = 8

Solution

Multiply the entire equation by $4$ to clear the denominators:

$$ x - 1 + 2(x + 3) = 32 $$

Distribute:

$$ x - 1 + 2x + 6 = 32 $$

Combine like terms:

$$ 3x + 5 = 32 $$

Subtract $5$ and divide by $3$:

$$ 3x = 27 $$

$$ x = 9 $$

2.3Solve by Substitution in a System

Exam II | Problem 2.3 |

In the system

\begin{aligned} y &= x + 2 \\ x + y &= 14 \end{aligned}

what is the value of $x$?

Solution

Substitute $y = x + 2$ into the second equation:

$$ x + (x + 2) = 14 $$

Combine like terms:

$$ 2x + 2 = 14 $$

Subtract $2$:

$$ 2x = 12 $$

Divide by $2$:

$$ x = 6 $$

2.4Factor a Quadratic and Select the Smaller Root

Exam II | Problem 2.4 |

The equation

$$ x^2 - 7x + 12 = 0 $$

has two solutions. What is the smaller solution?

Solution

Factor the quadratic:

$$ x^2 - 7x + 12 = (x - 3)(x - 4) $$

Use the zero-product property:

x - 3 = 0 \quad \text{or} \quad x - 4 = 0

So the solutions are $x = 3$ and $x = 4$.

The smaller solution is $3$.

2.5Compute a Slope from Two Points

Exam II | Problem 2.5 |

Two points lie on a line:

(2, 5) \quad \text{and} \quad (8, 17)

What is the slope of the line?

Solution

Use the slope formula:

m = \frac{17 - 5}{8 - 2}

m = \frac{12}{6} = 2

2.6Nested Function Evaluation

Exam II | Problem 2.6 |

Evaluating functions

Let

$$ f(x) = 2x + 1 $$

and

$$ g(x) = 3x - 4. $$

What is the value of $f(g(2))$?

Solution

First evaluate the inner function:

$$ g(2) = 3(2) - 4 = 2 $$

Now substitute that result into $f$:

$$ f(2) = 2(2) + 1 = 5 $$

2.7Simplify and Evaluate a Rational Expression

Exam II | Problem 2.7 |

For $x \ne 3$, evaluate

\frac{x^2 - 9}{x - 3}

at $x = 5$.

Solution

Factor the numerator:

$$ x^2 - 9 = (x - 3)(x + 3) $$

So, for $x \ne 3$,

\frac{x^2 - 9}{x - 3} = x + 3

Now substitute $x = 5$:

$$ 5 + 3 = 8 $$

2.8Distribution on Both Sides

Exam II | Problem 2.8 |

Solve for $x$:

$$ 4(2x - 3) + 2 = 3x + 15 $$

Solution

Expand the left-hand side:

$$ 8x - 12 + 2 = 3x + 15 $$

Combine constants:

$$ 8x - 10 = 3x + 15 $$

Subtract $3x$ from both sides:

$$ 5x - 10 = 15 $$

Add $10$:

$$ 5x = 25 $$

Divide by $5$:

$$ x = 5 $$

Difficulty

Applied

3.1Model a Membership Fee

Final | Problem 3.1 |

Linear Equations

A climbing gym charges an upfront fee of \$18 plus \$9 per visit.

If the total bill is \$72, how many visits did the student make?

Solution

Let $v$ be the number of visits. Set up the equation:

$$ 18 + 9v = 72 $$

Subtract $18$:

$$ 9v = 54 $$

Divide by $9$:

$$ v = 6 $$

3.2Rectangle Dimensions from a Perimeter Condition

Final | Problem 3.2 |

Linear Equations

A rectangle has perimeter $54$ cm.

Its length is $5$ cm more than its width.

What is the width of the rectangle?

Solution

Let $w$ be the width. Then the length is $w + 5$.

Use the perimeter formula:

$$ 2w + 2(w + 5) = 54 $$

Expand and combine like terms:

$$ 4w + 10 = 54 $$

Subtract $10$:

$$ 4w = 44 $$

Divide by $4$:

$$ w = 11 $$

3.3Compare Ticket and Snack Prices

Final | Problem 3.3 |

At a concert, 2 tickets and 3 drinks cost \$34.

Also, 1 ticket and 1 drink cost \$13.

What is the price of one ticket?

Solution

Let $t$ be the ticket price and $d$ be the drink price.

Set up the system:

\begin{aligned} 2t + 3d &= 34 \\ t + d &= 13 \end{aligned}

Multiply the second equation by $2$:

$$ 2t + 2d = 26 $$

Subtract this from the first equation:

$$ d = 8 $$

Substitute back into $t + d = 13$:

$$ t + 8 = 13 $$

$$ t = 5 $$

3.4Constant Rate Change

Final | Problem 3.4 |

Linear Equations

A water tank starts with $120$ liters.

It gains $25$ liters per hour.

After how many hours will it contain $245$ liters?

Solution

Let $h$ be the number of hours. Write the equation:

$$ 120 + 25h = 245 $$

Subtract $120$:

$$ 25h = 125 $$

Divide by $25$:

$$ h = 5 $$

3.5Proportional Reasoning in a Recipe

Final | Problem 3.5 |

Fractions

A recipe uses $3$ cups of flour for $8$ servings.

If the same recipe is scaled to $16$ servings, how many cups of flour are needed?

Solution

Since $16$ is twice $8$, the flour amount also doubles.

3 \times 2 = 6

So the recipe needs $6$ cups of flour.

Difficulty

Challenge

4.1Identify the Perfect-Square Trinomial

Final | Problem 4.1 |

Common factoring methods

What value of $k$ makes

$$ x^2 - kx + 36 $$

a perfect square trinomial?

Solution

Compare the expression to the square of a binomial:

$$ (x - 6)^2 = x^2 - 12x + 36 $$

Matching coefficients shows that

$$ k = 12 $$

4.2Perpendicular Slope from Standard Form

Final | Problem 4.2 |

Linear Equations

The line

$$ 2x + y = 11 $$

has slope $-2$.

What is the slope of a line perpendicular to it?

Solution

Rewrite the line in slope-intercept form:

$$ y = -2x + 11 $$

So the slope is $-2$.

A perpendicular line has slope equal to the negative reciprocal:

m_\perp = \frac{1}{2}

In decimal form, that is $0.5$.

4.3Recover a Quadratic Constant from Its Roots

Final | Problem 4.3 |

The quadratic

$$ x^2 + bx + c $$

has roots $4$ and $-3$.

What is the value of $c$?

Solution

Write the quadratic using its roots:

$$ (x - 4)(x + 3) $$

Expand:

$$ x^2 - x - 12 $$

The constant term is

$$ c = -12 $$

4.4Cancel a Factor and Solve Carefully

Final | Problem 4.4 |

Solve for $x$, making sure to respect any restrictions:

\frac{x^2 - 25}{x - 5} = 12

Solution

First note the restriction:

x \ne 5

Factor the numerator:

$$ x^2 - 25 = (x - 5)(x + 5) $$

So for $x \ne 5$,

\frac{(x - 5)(x + 5)}{x - 5} = x + 5

Now solve:

$$ x + 5 = 12 $$

$$ x = 7 $$

Algebra Practice

Selected problems

Quadratic formula

Direct

1.1Combine Like Terms in a Polynomial

1.2Solve a Two-Step Linear Equation

1.3Evaluate an Expression with Powers and Parentheses

1.4Distribute a Constant Across a Binomial

1.5Evaluate a Linear Function

1.6Solve a Fraction Equation

1.7Combine Fractional Coefficients

1.8Solve a Bracketed Equation

1.9Use Inverse Operations on a Simple Ratio

1.10Evaluate a Quadratic-Looking Expression

Integrated

2.1Distribution and Like-Term Combination

2.2Fraction Equation with Two Terms

2.3Solve by Substitution in a System

2.4Factor a Quadratic and Select the Smaller Root

2.5Compute a Slope from Two Points

2.6Nested Function Evaluation

2.7Simplify and Evaluate a Rational Expression

2.8Distribution on Both Sides

Applied

3.1Model a Membership Fee

3.2Rectangle Dimensions from a Perimeter Condition

3.3Compare Ticket and Snack Prices

3.4Constant Rate Change

3.5Proportional Reasoning in a Recipe

Challenge

4.1Identify the Perfect-Square Trinomial

4.2Perpendicular Slope from Standard Form

4.3Recover a Quadratic Constant from Its Roots

4.4Cancel a Factor and Solve Carefully