{"id":1174,"date":"2021-12-02T19:36:51","date_gmt":"2021-12-03T00:36:51","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/2-2-solving-linear-equations\/"},"modified":"2023-08-30T12:12:35","modified_gmt":"2023-08-30T16:12:35","slug":"solving-linear-equations","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/solving-linear-equations\/","title":{"raw":"2.2 Solving Linear Equations","rendered":"2.2 Solving Linear Equations"},"content":{"raw":"When working with questions that require two or more steps to solve, do the reverse of the order of operations to solve for the variable.\r\n
\r\n

Example 2.2.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve [latex]4x+16=-4[\/latex] for [latex]x.[\/latex]\r\n

[latex]\\begin{array}{rrrrrl}\r\n4x& +& 16 &=&-4& \\\\\r\n&&-16&& -16&\\text{Subtract 16 from each side}\u00a0\\\\\r\n\\hline\r\n&&\\dfrac{4x}{4}& =& \\dfrac{-20}{4}&\\text{Divide each side by 4}\\\\ \\\\\r\n&&x& =& -5 & \\text{Solution}\r\n\\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nIn solving the above equation, notice that the general pattern followed was to do the opposite of the equation. [latex]4x[\/latex] was added to 16, so 16 was then subtracted from both sides. The variable [latex]x[\/latex] was multiplied by 4, so both sides were divided by 4.\r\n

\r\n

Example 2.2.2<\/p>\r\n\r\n<\/header>\r\n

\r\n
    \r\n \t
  1. [latex]\\begin{array}[t]{rrrrr}\r\n5x &+& 7 &=& 7 \\\\\r\n&-&7&&-7 \\\\\r\n\\hline\r\n&&\\dfrac{5x}{5} &=& \\dfrac{0}{5}\\\\ \\\\\r\n&&x& =& 0\r\n\\end{array}[\/latex]<\/li>\r\n \t
  2. [latex]\\begin{array}[t]{rrrrr}\r\n4 &- &2x& =& 10 \\\\\r\n-4&&&&-4\\\\\r\n\\hline\r\n&&\\dfrac{-2x}{-2}& = &\\dfrac{6}{-2}\\\\ \\\\\r\n&&x& =& -3\r\n\\end{array}[\/latex]<\/li>\r\n \t
  3. [latex]\\begin{array}[t]{rrrrr}\r\n-3x& -& 7& =& 8 \\\\\r\n&+&7& = & + 7 \\\\\r\n\\hline\r\n&&\\dfrac{-3x}{-3}& =& \\dfrac{15}{-3} \\\\ \\\\\r\n&&x &= &-5\r\n\\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n \r\n\r\n<\/div>\r\n

    Questions<\/h1>\r\nFor questions 1 to 20, solve each linear equation.\r\n
      \r\n \t
    1. [latex]5 + \\dfrac{n}{4} = 4[\/latex]<\/li>\r\n \t
    2. [latex]-2 = -2m + 12[\/latex]<\/li>\r\n \t
    3. [latex]102 = -7r + 4[\/latex]<\/li>\r\n \t
    4. [latex]27 = 21 - 3x[\/latex]<\/li>\r\n \t
    5. [latex]-8n + 3 = -77[\/latex]<\/li>\r\n \t
    6. [latex]-4 - b = 8[\/latex]<\/li>\r\n \t
    7. [latex]0 = -6v[\/latex]<\/li>\r\n \t
    8. [latex]-2 + \\dfrac{x}{2} = 4[\/latex]<\/li>\r\n \t
    9. [latex]-8 = \\dfrac{x}{5} - 6[\/latex]<\/li>\r\n \t
    10. [latex]-5 = \\dfrac{a}{4} - 1[\/latex]<\/li>\r\n \t
    11. \u00a0[latex]0 = -7 + \\dfrac{k}{2}[\/latex]<\/li>\r\n \t
    12. [latex]-6 = 15 + 3p[\/latex]<\/li>\r\n \t
    13. [latex]-12 + 3x = 0[\/latex]<\/li>\r\n \t
    14. [latex]-5m + 2 = 27[\/latex]<\/li>\r\n \t
    15. [latex]\\dfrac{b}{3} + 7 = 10[\/latex]<\/li>\r\n \t
    16. [latex]\\dfrac{x}{1} - 8 = -8[\/latex]<\/li>\r\n \t
    17. [latex]152 = 8n + 64[\/latex]<\/li>\r\n \t
    18. [latex]-11 = -8 + \\dfrac{v}{2}[\/latex]<\/li>\r\n \t
    19. [latex]-16 = 8a + 64[\/latex]<\/li>\r\n \t
    20. [latex]-2x - 3 = -29[\/latex]<\/li>\r\n<\/ol>\r\nAnswer Key 2.2<\/a>","rendered":"

      When working with questions that require two or more steps to solve, do the reverse of the order of operations to solve for the variable.<\/p>\n

      \n
      \n

      Example 2.2.1<\/p>\n<\/header>\n

      \n

      Solve [latex]4x+16=-4[\/latex] for [latex]x.[\/latex]<\/p>\n

      [latex]\\begin{array}{rrrrrl} 4x& +& 16 &=&-4& \\\\ &&-16&& -16&\\text{Subtract 16 from each side}\u00a0\\\\ \\hline &&\\dfrac{4x}{4}& =& \\dfrac{-20}{4}&\\text{Divide each side by 4}\\\\ \\\\ &&x& =& -5 & \\text{Solution} \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n

      In solving the above equation, notice that the general pattern followed was to do the opposite of the equation. [latex]4x[\/latex] was added to 16, so 16 was then subtracted from both sides. The variable [latex]x[\/latex] was multiplied by 4, so both sides were divided by 4.<\/p>\n

      \n
      \n

      Example 2.2.2<\/p>\n<\/header>\n

      \n
        \n
      1. [latex]\\begin{array}[t]{rrrrr} 5x &+& 7 &=& 7 \\\\ &-&7&&-7 \\\\ \\hline &&\\dfrac{5x}{5} &=& \\dfrac{0}{5}\\\\ \\\\ &&x& =& 0 \\end{array}[\/latex]<\/li>\n
      2. [latex]\\begin{array}[t]{rrrrr} 4 &- &2x& =& 10 \\\\ -4&&&&-4\\\\ \\hline &&\\dfrac{-2x}{-2}& = &\\dfrac{6}{-2}\\\\ \\\\ &&x& =& -3 \\end{array}[\/latex]<\/li>\n
      3. [latex]\\begin{array}[t]{rrrrr} -3x& -& 7& =& 8 \\\\ &+&7& = & + 7 \\\\ \\hline &&\\dfrac{-3x}{-3}& =& \\dfrac{15}{-3} \\\\ \\\\ &&x &= &-5 \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/div>\n

         <\/p>\n<\/div>\n

        Questions<\/h1>\n

        For questions 1 to 20, solve each linear equation.<\/p>\n

          \n
        1. [latex]5 + \\dfrac{n}{4} = 4[\/latex]<\/li>\n
        2. [latex]-2 = -2m + 12[\/latex]<\/li>\n
        3. [latex]102 = -7r + 4[\/latex]<\/li>\n
        4. [latex]27 = 21 - 3x[\/latex]<\/li>\n
        5. [latex]-8n + 3 = -77[\/latex]<\/li>\n
        6. [latex]-4 - b = 8[\/latex]<\/li>\n
        7. [latex]0 = -6v[\/latex]<\/li>\n
        8. [latex]-2 + \\dfrac{x}{2} = 4[\/latex]<\/li>\n
        9. [latex]-8 = \\dfrac{x}{5} - 6[\/latex]<\/li>\n
        10. [latex]-5 = \\dfrac{a}{4} - 1[\/latex]<\/li>\n
        11. \u00a0[latex]0 = -7 + \\dfrac{k}{2}[\/latex]<\/li>\n
        12. [latex]-6 = 15 + 3p[\/latex]<\/li>\n
        13. [latex]-12 + 3x = 0[\/latex]<\/li>\n
        14. [latex]-5m + 2 = 27[\/latex]<\/li>\n
        15. [latex]\\dfrac{b}{3} + 7 = 10[\/latex]<\/li>\n
        16. [latex]\\dfrac{x}{1} - 8 = -8[\/latex]<\/li>\n
        17. [latex]152 = 8n + 64[\/latex]<\/li>\n
        18. [latex]-11 = -8 + \\dfrac{v}{2}[\/latex]<\/li>\n
        19. [latex]-16 = 8a + 64[\/latex]<\/li>\n
        20. [latex]-2x - 3 = -29[\/latex]<\/li>\n<\/ol>\n

          Answer Key 2.2<\/a><\/p>\n","protected":false},"author":90,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-1174","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":1170,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1174","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":2,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1174\/revisions"}],"predecessor-version":[{"id":2073,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1174\/revisions\/2073"}],"part":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/parts\/1170"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1174\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=1174"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapter-type?post=1174"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=1174"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=1174"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}