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Multivariable Expressions Practice Questions (page 2)

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Updated on Oct 3, 2011

Practice 2

  1. The terms 3r2 and 4t are unlike, so this expression cannot be simplified. Substitute 3 for r and –2 for t:  

    3(3)2 + 4(–2) = 3(9) –8 = 27 – 8 = 19

  2. This expression has two r terms and two t terms.  

    Combine the r terms: –r + 9r = 8r.  

    Combine the t terms: 5t – 2t = 3t.  

    The expression is now 8r + 3t.  

    Substitute 3 for r and –2 for t:  

    8(3) + 3(–2) = 24 –6 = 18

  3. This expression has two r terms, one t term, and one t2 term.  

    Combine the r terms: 6r –2r = 4r.  

    The expression is now 4r –3t2 + 4t.  

    Substitute 3 for r and –2 for t:  

    4(3) –3(–2)2 + 4(–2) = 4(3) –3(4) –8 = 12 –12 –8 = –8

  4. Use the distributive law to find –2r(8 + t). Multiply –2r by 8 and multiply –2r by t:   –2r(8 + t) = –16r + –2rt  

    This expression, 4rt –16r –2rt, has one r term and two rt terms.  

    Combine the rt terms: 4rt –2rt = 2rt.  

    The expression is now 2rt –16r.  

    Substitute 3 for r and –2 for t:  

    2(3)(–2) –16(3) = 6(–2) –48 = –12 –48 = –60

  5. This expression has two a terms, two b terms, and one c term.  

    Combine the a terms: 2a –5a = –3a.  

    Combine the b terms: 3b –4b = –b.  

    The expression is now –3ab –4c.  

    Substitute –3 for a, 6 for b, and 1 for c:  

    –3(–3) –(6) –4(1) = 9 –6 –4 = 3 –4 = –1

  6. Use the distributive law to find 4(ab + 2c).

    Multiply each term in parentheses by 4:  

    4(ab + 2c) = 4a –4b + 8c  

    Use the distributive law again to find c(a –3). Multiply both terms in parentheses by c:  

    c(a –3) = ac –3c  

    This expression, 4a –4b + 8c + ac –3c, has one a term, one b term, two c terms, and one ac term.  

    Combine the c terms: 8c –3c = 5c.  

    The expression is now 4a –4b + 5c + ac.  

    Substitute –3 for a, 6 for b, and 1 for c:  

    4(–3) –4(6) + 5(1) + (–3)(1) = –12 –24 + 5 –3 = –36 + 5 –3 = –31 –3 = –34

  7. Use the distributive law to find b(3a + c). Multiply each term in parentheses by b:  

    b(3a +c) = 3ab + bc  

    This expression, 3c2 –3ab + 3ab + bc, has one c2 term, two ab terms, and one bc term.  

    Combine the ab terms: –3ab + 3ab = 0.  

    The expression is now 3c2 + bc.  

    Substitute 6 for b and 1 for c:

    3(1)2 + (6)(1) = 3(1) + (6)(1) = 3 + 6 = 9
  8. Use the distributive law to find 2b(ca). Multiply each term in parentheses by 2b:

    2b(ca) = 2bc –2ab

    This expression, 5b + 2bc –2ab + 3ab, has one b term, one bc term, and two ab terms.

    Combine the ab terms: –2ab + 3ab = ab.

    The expression is now 5b + 2bc + ab.

    Substitute –3 for a, 6 for b, and 1 for c:

    5(6) + 2(6)(1) + (–3)(6) = 30 + 12(1) –18 = 30 + 12 –18 = 42 –18 = 24

  9. Use the distributive law to find 6(3w + x). Multiply both terms in parentheses by 6:

    6(3w + x) = 18w + 6x

    Use the distributive law again to find –3(wz). Multiply both terms in parentheses by –3:

    –3(wz) = –3w + 3z

    This expression, 18w + 6x + 2y –3w + 3z, has two w terms, one x term, one y term, and one z term.  

    Combine the w terms: 18w –3w = 15w.

    The expression is now 15w + 6x + 2y + 3z.

    Substitute 1 for w, –2 for x, 3 for y, and –4 for z:

    15(1) + 6(–2) + 2(3) + 3(–4) = 15 –12 + 6 –12 = 3 + 6 –12 = 9 –12 = –3

  10. Use the distributive law to find 5(z + 4). Multiply both terms in parentheses by 5:

    5(z + 4) = 5z + 20

     Use the distributive law again to find –10(12z + y). Multiply both terms in parentheses by –10:

    This expression, w2 –3x + 5z + 20 –5z –10y, has one w2 term, one x term, one y term, two z terms, and one constant.

     

    Combine the z terms: 5z –5z = 0.

     

    The expression is now w2 –3x + 20 –10y.

     

    Substitute 1 for w, –2 for x, and 3 for y:

    (1)2 –3(–2) + 20 –10(3) = 1 + 6 + 20 –30 = 7 + 20 –30 = 27 –30 = –3

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