{"id":1751,"date":"2019-08-08T14:47:55","date_gmt":"2019-08-08T18:47:55","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=back-matter&p=1751"},"modified":"2019-08-08T16:25:41","modified_gmt":"2019-08-08T20:25:41","slug":"answer-key-10-1","status":"publish","type":"back-matter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/back-matter\/answer-key-10-1\/","title":{"raw":"Answer Key 10.1","rendered":"Answer Key 10.1"},"content":{"raw":"[latexpage]\r\n
    \r\n \t
  1. \\(\\begin{array}{rrrrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n\\sqrt{2x+3}&-&3&=&0 \\\\\r\n&+&3&&+3 \\\\\r\n\\midrule\r\n(\\sqrt{2x+3})^2&&&=&\\phantom{+}(3)^2 \\\\ \\\\\r\n2x&+&3&=&9 \\\\\r\n&-&3&&-3 \\\\\r\n\\midrule\r\n&&\\dfrac{2x}{2}&=&\\dfrac{6}{2} \\\\ \\\\\r\n&&x&=&3\r\n\\end{array}\\)<\/li>\r\n \t
  2. \\(\\begin{array}{rrrrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n\\sqrt{5x+1}&-&4&=&0 \\\\\r\n&+&4&=&+4 \\\\\r\n\\midrule\r\n(\\sqrt{5x+1})^2&&&=&(4)^2 \\\\ \\\\\r\n5x&+&1&=&16 \\\\\r\n&-&1&&-1 \\\\\r\n\\midrule\r\n&&\\dfrac{5x}{5}&=&\\dfrac{15}{5} \\\\ \\\\\r\n&&x&=&3\r\n\\end{array}\\)<\/li>\r\n \t
  3. \\(\\begin{array}{rrrrrrrrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n\\sqrt{6x-5}&-&x&=&0&&&& \\\\\r\n&+&x&&+x&&&& \\\\\r\n\\midrule\r\n(\\sqrt{6x-5})^2&&&=&(x)^2&&&& \\\\ \\\\\r\n6x&-&5&=&x^2&&&& \\\\\r\n-6x&+&5&&&-&6x&+&5 \\\\\r\n\\midrule\r\n&&0&=&x^2&-&6x&+&5 \\\\\r\n&&0&=&(x&-&5)(x&-&1) \\\\ \\\\\r\n&&x&=&5,&1&&&\r\n\\end{array}\\)<\/li>\r\n \t
  4. \\((\\sqrt{7x+8})^2=(x)^2\\)\r\n\r\n\\(\\begin{array}{rrrrrrrrr}\r\n7x&+&8&=&x^2&&&& \\\\\r\n-7x&-&8&&&-&7x&-&8 \\\\\r\n\\midrule\r\n&&0&=&x^2&-&7x&-&8 \\\\\r\n&&0&=&(x&-&8)(x&+&1) \\\\ \\\\\r\n&&x&=&-1,&8&&&\r\n\\end{array}\\)<\/li>\r\n \t
  5. \\((\\sqrt{3+x})^2=(\\sqrt{6x+13})^2\\)\r\n\r\n\\(\\begin{array}{rrrrrrr}\r\n3&+&x&=&6x&+&13 \\\\\r\n-3&-&6x&&-6x&-&3 \\\\\r\n\\midrule\r\n&&\\dfrac{-5x}{-5}&=&\\dfrac{10}{-5}&& \\\\ \\\\\r\n&&x&=&-2&&\r\n\\end{array}\\)<\/li>\r\n \t
  6. \\((\\sqrt{x-1})^2=(\\sqrt{7-x})^2\\)\r\n\r\n\\(\\begin{array}{rrrrrrr}\r\nx&-&1&=&7&-&x \\\\\r\n+x&+&1&&+1&+&x \\\\\r\n\\midrule\r\n&&\\dfrac{2x}{2}&=&\\dfrac{8}{2}&& \\\\ \\\\\r\n&&x&=&4&&\r\n\\end{array}\\)<\/li>\r\n \t
  7. \\((\\sqrt[3]{3-3x})^3=(\\sqrt[3]{2x-5})^3\\)\r\n\r\n\\(\\begin{array}{rrrrrrr}\r\n3&-&3x&=&2x&-&5 \\\\\r\n-3&-&2x&&-2x&-&3 \\\\\r\n\\midrule\r\n&&\\dfrac{-5x}{-5}&=&\\dfrac{-8}{-5}&& \\\\ \\\\\r\n&&x&=&\\dfrac{8}{5}&&\r\n\\end{array}\\)<\/li>\r\n \t
  8. \\((\\sqrt[4]{3x-2})^4=(\\sqrt[4]{x+4})^4\\)\r\n\r\n\\(\\begin{array}{rrrrrrr}\r\n3x&-&2&=&x&+&4 \\\\\r\n-x&+&2&&-x&+&2 \\\\\r\n\\midrule\r\n&&\\dfrac{2x}{2}&=&\\dfrac{6}{2}&& \\\\ \\\\\r\n&&x&=&3&&\r\n\\end{array}\\)<\/li>\r\n \t
  9. \\((\\sqrt{x+7})^2\\ge (2)^2\\)\r\n\r\n\\(\\begin{array}{rrrrr}\r\nx&+&7&\\ge &4 \\\\\r\n&-&7&&-7 \\\\\r\n\\midrule\r\n&&x&\\ge &-3\r\n\\end{array}\\)<\/li>\r\n \t
  10. \\((\\sqrt{x-2})^2\\le (4)^2\\)\r\n\r\n\\(\\begin{array}{rrrrr}\r\nx&-&2&\\le &16 \\\\\r\n&+&2&&+2 \\\\\r\n\\midrule\r\n&&x&\\le &18\r\n\\end{array}\\)<\/li>\r\n \t
  11. \\((3)^2 < (\\sqrt{3x+6})^2 \\le (6)^2\\)\r\n\r\n\\(\\begin{array}{rrrcrrr}\r\n9&<&3x&+&6&\\le &36 \\\\\r\n-6&&&-&6&&-6 \\\\\r\n\\midrule\r\n\\dfrac{3}{3}&<&&\\dfrac{3x}{3}&&\\le &\\dfrac{30}{3} \\\\ \\\\\r\n1&<&&x&&\\le &10\r\n\\end{array}\\)<\/li>\r\n \t
  12. \\((0)^2 < (\\sqrt{x+5})^2 < (5)^2\\)\r\n\r\n\\(\\begin{array}{rrrcrrr}\r\n0&<&x&+&5&<&25 \\\\\r\n-5&&&-&5&&-5 \\\\\r\n\\midrule\r\n-5&<&&x&&<&20\r\n\\end{array}\\)<\/li>\r\n<\/ol>","rendered":"
      \n
    1. \"\begin{array}{rrrrr}<\/li>\n
    2. \"\begin{array}{rrrrr}<\/li>\n
    3. \"\begin{array}{rrrrrrrrr}<\/li>\n
    4. \"(\sqrt{7x+8})^2=(x)^2\"\n

      \"\begin{array}{rrrrrrrrr}<\/li>\n

    5. \"(\sqrt{3+x})^2=(\sqrt{6x+13})^2\"\n

      \"\begin{array}{rrrrrrr}<\/li>\n

    6. \"(\sqrt{x-1})^2=(\sqrt{7-x})^2\"\n

      \"\begin{array}{rrrrrrr}<\/li>\n

    7. \"(\sqrt[3]{3-3x})^3=(\sqrt[3]{2x-5})^3\"\n

      \"\begin{array}{rrrrrrr}<\/li>\n

    8. \"(\sqrt[4]{3x-2})^4=(\sqrt[4]{x+4})^4\"\n

      \"\begin{array}{rrrrrrr}<\/li>\n

    9. \"(\sqrt{x+7})^2\ge (2)^2\"\n

      \"\begin{array}{rrrrr}<\/li>\n

    10. \"(\sqrt{x-2})^2\le (4)^2\"\n

      \"\begin{array}{rrrrr}<\/li>\n

    11. \"(3)^2 < (\sqrt{3x+6})^2 \le (6)^2\"\n

      \"\begin{array}{rrrcrrr}<\/li>\n

    12. \"(0)^2 < (\sqrt{x+5})^2 < (5)^2\"\n

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