Algebra 1 Structure of Algebraic Expressions Worksheet Solutions (PDF)

This algebra 1 expressions answers for Algebra 1 covers structure of algebraic expressions (Common Core A-SSE.A.1a, A-SSE.A.1b), with a free PDF to print.

Algebra 1 · Common Core State Standards: A-SSE.A.1a, A-SSE.A.1b
Algebra 1 expressions answers: exercises 1 to 10
Exercise 1 – Coefficient of x
The term with \(x\) is \(6x\), so the number multiplying \(x\) is 6.
6
Exercise 2 – Count the terms
Terms are separated by + and – signs: \(4a^{2}\), \(-3a\), and \(8\).
3 terms
Exercise 3 – Name the constant
The constant term has no variable. In \(2x^{2} – 9x + 14\) it is the last term.
14
Exercise 4 – Factors in a product
A product is made of factors that are multiplied. Here \(5\) is multiplied by the group \((x – 2)\).
\(5\) and \((x – 2)\)
Exercise 5 – Evaluate an expression
Substitute \(-2\) for \(x\): \(3(-2)^{2} – 2(-2) + 5\).
Square first: \(3(4) + 4 + 5 = 12 + 4 + 5\).
\(21\)
Exercise 6 – Pick the like terms
Like terms have the same variable part. \(5x\) and \(-2x\) both have \(x\) to the first power.
\(4x^{2}\) has a different exponent, and \(3y\) has a different variable.
\(5x\) and \(-2x\)
Exercise 7 – Combine like terms
Group like terms: \((8x – 5x) + (3 + 9)\).
Combine: \(3x + 12\).
\(3x + 12\)
Exercise 8 – Distribute a number
Multiply 4 by each term inside: \(4 \cdot 3x = 12x\) and \(4 \cdot (-5) = -20\).
\(12x – 20\)
Exercise 9 – Distribute a negative
Multiply each term by \(-2\): \((-2)(x) = -2x\) and \((-2)(-6) = +12\).
\(-2x + 12\)
Exercise 10 – Distribute and combine
Distribute: \(6x + 3 + 4x – 7\).
Combine like terms: \((6x + 4x) + (3 – 7)\).
\(10x – 4\)
Exercises 11 and up
Exercise 11 – Subtract an expression
Subtracting changes the sign of every term in the second group: \(7x + 5 – 3x + 2\).
Combine: \((7x – 3x) + (5 + 2)\).
\(4x + 7\)
Exercise 12 – One quantity
Both terms contain the factor \((x – 3)\). Add the coefficients: \(2 + 5 = 7\).
\(7(x – 3)\)
Exercise 13 – Taxi fare meaning
\(2.25\) is the coefficient of \(m\): the cost for each mile. The table shows it, since the fare rises $2.25 per mile (\(\$5.75 \to \$8.00\)).
\(3.50\) is the constant: a fixed charge paid before any miles.
$2.25 per mile; $3.50 starting charge
Exercise 14 – Write an expression
3 times a number is \(3n\). Taking away 5 gives \(3n – 5\).
\(3n – 5\)
Exercise 15 – Gym membership
Rate \(\$18\) per month times \(m\) months, plus the one-time fee: \(18m + 25\).
Substitute \(m = 6\): \(18(6) + 25 = 108 + 25 = 133\).
\(18m + 25\); $133
Exercise 16 – Perimeter
Perimeter is \(2(3x + 2) + 2(x + 5)\).
Distribute: \(6x + 4 + 2x + 10\), which simplifies to \(8x + 14\).
For \(x = 4\): \(8(4) + 14 = 46\) feet.
\(8x + 14\); 46 feet
Exercise 17 – Bake sale
Money is \(1.50c + 2.25b\).
For \(c = 24\) and \(b = 18\): \(1.50(24) + 2.25(18) = 36 + 40.50 = 76.50\).
The term \(2.25b\) is the money earned from brownies alone.
\(1.50c + 2.25b\); $76.50
Exercise 18 – Sale price
\(0.85p\) is 0.85 times \(p\). Paying 85% of the price means a 15% discount, and the 4 is a shipping charge.
For \(p = 60\): \(0.85(60) + 4 = 51 + 4 = 55\).
15% discount; $55
Exercise 19 – Patio area
Garden area: \(4(x + 7) = 4x + 28\). Patio area: \(x \cdot x = x^{2}\).
Total: \(x^{2} + 4x + 28\).
For \(x = 5\): \(25 + 20 + 28 = 73\).
\(x^{2} + 4x + 28\); 73 square feet
Exercise 20 – Compare two plans
\((4 + 2.5m) – (1.5 + 3m) = 4 + 2.5m – 1.5 – 3m\).
Combine like terms: \(2.5 – 0.5m\).
For \(m = 3\): \(2.5 – 0.5(3) = 2.5 – 1.5 = 1\), so A costs $1 more.
\(2.5 – 0.5m\); $1
Exercise 21 – Fundraiser tickets
Distribute: \(3(2s + 3a) = 6s + 9a\), so the expressions are equivalent.
Substitute: \(6(40) + 9(25) = 240 + 225 = 465\).
Equivalent; $465
Key facts behind this algebra 1 expressions answers
- Terms are separated by + and – signs, and each term keeps its own sign.
- A coefficient multiplies a variable; a constant has no variable.
- Like terms have the same variable part and can be combined by adding coefficients.
- A grouped expression such as (x + 3) can be treated as one factor.
- In a model, a coefficient is often a rate and a constant is a starting amount.
How to use this algebra 1 expressions answers
Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.
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Download the algebra 1 expressions answers as a PDF with the button below and keep it with your notes.
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