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Extra info for Algebra in 15 Minutes a Day
The exponent of h in your answer is 14. (10g3h5)(–2g5h9) = –20g8h14. 5. Multiply the coefﬁcients: (9)(9) = 81. The ﬁrst term has bases of a, b, and c (the second term also has bases of a and c), so your answer has bases of a, b, and c. Next, add the exponents of a from each term: 6 + 2 = 8. The exponent of a in your answer is 8. Because the second term does not contain b, the exponent of b in your answer will be the same as the exponent of b in the ﬁrst term, 11. Finally, add the exponents of c from each term: 4 + 2 = 6.
Combine the two different signs into one minus sign: 3r–2 – 4r–2 + (–7r–2) = 3r–2 – 4r–2 – 7r–2. Now, work with the coefﬁcients. Subtract 4 and 7 from 3: 3 – 4 – 7 = –8, so 3r–2 – 4r–2 – 7r–2 = –8r–2. Practice 4 1. 10k2 and 10k have the same base, but different exponents, so 10k cannot be subtracted from 10k2. 2. n and 19n have the same base (n) and the same exponent (1), so n can be added to 19n. 3. 8x–3 and –8x3 have the same base (x), but different exponents; 3 and –3 are not the same exponent, so 8x–3 and –8x3 cannot be combined.
7. Each term has a base of y and an exponent of 4, so the base and exponent of your answer is y4. Combine the two minus signs into one plus sign: 15y4 + 12y4 – (–17y4) = 15y4 + 12y4 + 17y4. Add the coefﬁcients of each term: 15 + 12 + 17 = 44, so 15y4 + 12y4 + 17y4 = 44y4. Practice 3 1. Each term has a base of g and an exponent of 9, so the base and exponent of your answer is g9. Subtract the coefﬁcient of the second term from the coefﬁcient of the ﬁrst term: 11 – 9 = 2, so 11g9 – 9g9 = 2g9. 2. Each term has a base of j and an exponent of 6, so the base and exponent of your answer is j 6.