By LearningExpress Editors
Updated on Oct 3, 2011
To review these concepts, go to Addition and Subtraction of Terms Study Guide.
Addition and Subtraction of Terms Practice Questions
Practice 1
For each problem, combine pairs of signs and solve.
 4 + (+10) =
 –3 – (+9) =
 7 – (–1) =
 0 + (–6) =
 9 + (–8) – (–2) =
 15 – (+2) + (–4) =
 –18 + (+20) – (–5) =
Practice 2
 6a^{5} + 2a^{5} =
 –2p + 2p =
 23q^{12} + (+11q^{12}) =
 3b^{2} +b^{2} + 10b^{2} =
 10a^{3} – (–5a^{3}) =
 –9t^{8} + 8t^{8} + (+13t^{8}) =
 15y^{4} + 12y^{4} – (–17y^{4}) =
Practice 3
 11g^{9} – 9g^{9} =
 4j^{6} – 5j^{6} =
 18n^{4} – (+13n^{4}) =
 h^{2} + (–7h^{2}) =
 –8z^{3} – z^{3} –z^{3} =
 16t^{15} – 9t^{15} – (+t^{15}) =
 3r^{–2} – 4r^{–2} + (–7r^{–2}) =
Practice 4
For each problem, decide if the terms can be combined or not.
 10k^{2} – 10k
 n + 19n
 8x^{–3} + (–8x^{3})
 –11y^{2} + (–15z^{2})
 a^{5}b^{4}c^{2} – (–2a^{5}b^{4}c^{2})
Solutions
Practice 1
 Combine the two plus signs into one plus sign and add: 4 + (+10) = 4 +10 = 14.
 Combine the minus sign and the plus sign into a minus sign and subtract: –3 – (+9) = –3 – 9 = –12.
 Combine the two minus signs into one plus sign and add: 7 – (–1) = 7 + 1 = 8.
 Combine the plus sign and the minus sign into a minus sign and subtract: 0 + (–6) = 0 – 6 = –6.
 Combine the first pair of signs into a minus sign, since the two signs are different: 9 + (–8) – (–2) = 9 – 8 – (–2). Combine the second pair of signs into a plus sign, because the two signs are the same: 9 – 8 – (–2) = 9 – 8 + 2 = 1 + 2 = 3.
 Combine the first pair of signs into a minus sign, since the two signs are different: 15 – (+2) + (–4) = 15 – 2 + (–4). Combine the second pair of signs into a minus sign, because the two signs are also different: 15 – 2 + (–4) = 15 – 2 – 4 = 13 – 4 = 9.
 Combine the first pair of signs into a plus sign, since the two signs are the same: –18 + (+20) – (–5) = –18 + 20 – (–5). Combine the second pair of signs into a plus sign, because the two signs are also the same: –18 + 20 + 5 = 2 + 5 = 7.
Practice 2
 Each term has a base of a and an exponent of 5, so the base and exponent of your answer is a^{5}. Add the coefficients: 6 + 2 = 8, so 6a^{5} + 2a^{5} = 8a^{5}.
 Each term has a base of p and an exponent of 1. Remember, if a base does not appear to have an exponent, then it has an exponent of 1. The base of your answer is p. Add the coefficients: –2 + 2 = 0, so –2p + 2p = 0p, or simply 0.
 Each term has a base of q and an exponent of 12, so the base and exponent of your answer is q^{12}. Combine the two plus signs into one plus sign: 23q^{12} + (+11q^{12}) = 23q^{12} + 11q^{12}. Add the coefficients of each term: 23 + 11 = 34, so 23q^{12} + 11q^{12} = 34q^{12}.
 Each term has a base of b and an exponent of 2, so the base and exponent of your answer is b^{2}. Add the coefficients of each term. Remember, if there is no number before the base of a term, the coefficient of that term is 1: 3 + 1 + 10 = 14, so 3b^{2} + b^{2} + 10b^{2} = 14b^{2}.
 Each term has a base of a and an exponent of 3, so the base and exponent of your answer is a^{3}. Combine the two minus signs into one plus sign: 10a^{3} – (–5a^{3}) = 10a^{3} + 5a^{3}. Add the coefficients of each term: 10 + 5 = 15, so 10a^{3} + 5a^{3} = 15a^{3}.
 Each term has a base of t and an exponent of 8, so the base and exponent of your answer is t^{8}. Combine the two plus signs into one plus sign: –9t^{8} + 8t^{8} + (+13t^{8}) = –9t^{8} + 8t^{8} + 13t^{8}. Add the coefficients of each term: –9 + 8 + 13 = 12, so 9t^{8} + 8t^{8} + 13t^{8} = 12t^{8}.
 Each term has a base of y and an exponent of 4, so the base and exponent of your answer is y^{4}. Combine the two minus signs into one plus sign: 15y^{4} + 12y^{4} – (–17y^{4}) = 15y^{4} + 12y^{4} + 17y^{4}. Add the coefficients of each term: 15 + 12 + 17 = 44, so 15y^{4} + 12y^{4} + 17y^{4} = 44y^{4}.

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From Algebra in 15 Minutues A Day. Copyright © 2009 by LearningExpress, LLC. All Rights Reserved.
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