Answer Key 2.6

  1. (3) + 1 + (4) − (1)
    3 + 1 + 4 − 1
    7
  2. (5)2 + (5) − (1)
    25 + 5 − 1
    29
  3. (6) − [(6)(5) ÷ 6]
    6 − [30 ÷ 6]
    6 − 5
    1
  4. [6 + (4) − (1)] ÷ 3
    [6 + 4 − 1] ÷ 3
    9 ÷ 3
    3
  5. (5)2 − ((3) − 1)
    25 − (3 − 1)
    25 − 2
    23
  6. (6) + 6(4) − 4(4)
    6 + 24 − 16
    14
  7. 5(4) + (2)(5) ÷ 2
    20 + 10 ÷ 2
    20 + 5
    25
  8. 5((6) + (2)) + 1 + (5)
    5(6 + 2) + 1 + 5
    5(8) + 6
    40 + 6
    46
  9. [4 − ((6) − (4))] ÷ 2 + (6)
    [4 − (6 − 4)] ÷ 2 + 6
    [4 − 2] ÷ 2 + 6
    2 ÷ 2 + 6
    1 + 6
    7
  10. (4) + (5) − 1
    4 + 5 − 1
    8
  11. \begin{array}{rrl} \\ \\ \\ \dfrac{ab}{a}&=&\dfrac{c}{a} \\ \\ b&=&\dfrac{c}{a} \end{array}
  12. \begin{array}{l} \\ \\ \\ \left(g=\dfrac{h}{i}\right)(i) \\ \\ h=gi \end{array}
  13. \begin{array}{l} \\ \\ \\ \\ \left(\left(\dfrac{f}{g}\right)x=b\right)\dfrac{g}{f} \\ \\ x=\dfrac{bg}{f} \end{array}
  14. \begin{array}{l} \\ \\ \\ \left(p=\dfrac{3y}{q}\right)\left(\dfrac{q}{3}\right) \\ \\ y=\dfrac{pq}{3} \end{array}
  15. \begin{array}{l} \\ \\ \\ \left(3x=\dfrac{a}{b}\right)\left(\dfrac{1}{3}\right) \\ \\ x=\dfrac{a}{3b} \end{array}
  16. \begin{array}{l} \\ \\ \\ \left(\dfrac{ym}{b}=\dfrac{c}{d}\right)\left(\dfrac{b}{m}\right) \\ \\ y=\dfrac{bc}{dm} \end{array}
  17. \begin{array}{l} \\ \\ \\ \left(V=\dfrac{4}{3}\pi r^3\right)\left(\dfrac{3}{4r^3}\right) \\ \\ \pi=\dfrac{3V}{4r^3} \end{array}
  18. \begin{array}{l} \\ \\ \left(E=mv^2\right)\left(\dfrac{1}{v^2}\right) \\ \\ m=\dfrac{E}{v^2} \end{array}
  19. \begin{array}{l} \\ \\ \\ \left(c=\dfrac{4y}{m+n}\right)\left(\dfrac{m+n}{4}\right) \\ \\ y=\dfrac{c(m+n)}{4} \end{array}
  20. \begin{array}{l} \\ \\ \\ \left(\dfrac{rs}{a-3}=k\right)\left(\dfrac{a-3}{s}\right) \\ \\ r=\dfrac{k(a-3)}{s} \end{array}
  21. \begin{array}{l} \\ \\ \\ \left(V=\dfrac{\pi Dn}{12}\right)\left(\dfrac{12}{\pi n}\right) \\ \\ D=\dfrac{12V}{\pi n} \end{array}
  22. \begin{array}{rrl} \\ \\ \\ \\ \\ \\ F\phantom{+kL}&=&kR-kL \\ +kL&&\phantom{kR}+kL \\ \midrule \dfrac{F+kL}{k}&=&\dfrac{kR}{k} \\ \\ R&=&\dfrac{F+kL}{k} \end{array}
  23. \begin{array}{rrl} \\ \\ \\ \\ \\ \\ P\phantom{-np}&=&\phantom{-}np-nc \\ -np&&-np \\ \midrule \dfrac{P-np}{-n}&=&\dfrac{-nc}{-n} \\ \\ c&=&\dfrac{P-np}{-n} \end{array}
  24. \begin{array}{rrrrr} \\ \\ \\ S\phantom{-2B}&=&L&+&2B \\ -2B&&&-&2B \\ \midrule L&=&S&-&2B \end{array}
  25. \phantom{1}
    \left(T=\dfrac{D-d}{L}\right)(L) \\
    \begin{array}{rrrrr} TL\phantom{+d}&=&D&-&d \\ +d&&&+&d \\ \midrule D&=&TL&+&d \end{array}
  26. \phantom{1}
    \left(I=\dfrac{E_a-E_q}{R}\right)(R) \\
    \begin{array}{rrrrr} IR&=&E_a&-&E_q \\ +E_q&&&+&E_q \\ \midrule E_a&=&IR&+&E_q \end{array}
  27. \begin{array}{rrl} \\ \\ \\ \\ \dfrac{L}{1+at}&=&\dfrac{L_o(1+at)}{1+at} \\ \\ L_o&=&\dfrac{L}{1+at} \end{array}
  28. \begin{array}{rrl} \\ \\ \\ \\ \\ \\ 2m+p&=&\phantom{-}4m+q \\ -2m-q&&-2m-q \\ \midrule \dfrac{p-q}{2}&=&\dfrac{2m}{2} \\ \\ m&=&\dfrac{p-q}{2} \end{array}
  29. \phantom{1}
    \left(\dfrac{k-m}{r}=q\right)(r) \\
    \begin{array}{rrl} k-m&=&qr \\ +m&&\phantom{qr}+m \\ \midrule k&=&qr+m \end{array}
  30. \begin{array}{rrrrr} \\ \\ \\ \\ \\ \\ R\phantom{-b}&=&aT&+&b \\ -b&&&-&b \\ \midrule \dfrac{R-b}{a}&=&\dfrac{aT}{a}&& \\ \\ T&=&\dfrac{R-b}{a}&& \end{array}
  31. \begin{array}{rrl} \\ \\ \\ \\ \\ \\ Q_1\phantom{+PQ_1}&=&PQ_2-PQ_1 \\ +PQ_1&&\phantom{PQ_2}+PQ_1 \\ \midrule \dfrac{Q_1+PQ_1}{P}&=&\dfrac{PQ_2}{P} \\ \\ Q_2&=&\dfrac{Q_1+PQ_1}{P} \end{array}
  32. \begin{array}{rrl} \\ \\ \\ \\ \\ \\ L\phantom{-\pi r_2-2d}&=&\pi r_1+\pi r_2+2d \\ -\pi r_2-2d&&\phantom{\pi r_1}-\pi r_2-2d \\ \midrule \dfrac{L-\pi r_2-2d}{\pi}&=&\dfrac{\pi r_1}{\pi} \\ \\ r_1&=&\dfrac{L-\pi r_2-2d}{\pi} \end{array}
  33. \phantom{1}
    \left(R=\dfrac{kA(T+T_1)}{d}\right)\left(\dfrac{d}{kA}\right) \\
    \begin{array}{rrl} \dfrac{Rd}{kA}\phantom{-T}&=&\phantom{-}T+T_1 \\ -T&&-T \\ \midrule T_1&=&\dfrac{Rd}{kA}-T \end{array}
  34. \phantom{1}
    \left(P=\dfrac{V_1(V_2-V_1)}{g}\right)\left(\dfrac{g}{V_1}\right) \\
    \begin{array}{rrl} \dfrac{Pg}{V_1}\phantom{+V_1}&=&V_2-V_1 \\ +V_1&&\phantom{V_2}+V_1 \\ \midrule V_2&=&\dfrac{Pg}{V_1}+V_1 \end{array}

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Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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