{"id":2785,"date":"2025-08-13T18:39:57","date_gmt":"2025-08-13T18:39:57","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=2785"},"modified":"2025-10-22T17:05:01","modified_gmt":"2025-10-22T17:05:01","slug":"parametric-functions-and-vectors-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/parametric-functions-and-vectors-background-youll-need-1\/","title":{"raw":"Parametric Functions and Vectors: Background You'll Need 1","rendered":"Parametric Functions and Vectors: Background You&#8217;ll Need 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Substitute an algebraic expression into another function<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<p data-start=\"571\" data-end=\"785\">Sometimes we work with two related equations that share a variable. We can connect them by solving one equation for that variable and substituting the expression into the other. This is called substitution.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p data-start=\"754\" data-end=\"871\">Given [latex]x = 2t + 1[\/latex], and [latex]y = 3x - 4[\/latex], find [latex]y[\/latex] in terms of [latex]t[\/latex].<\/p>\r\n<p data-start=\"873\" data-end=\"1017\">We already know what [latex]x[\/latex] equals, so substitute [latex]2t + 1[\/latex] wherever we see [latex]x[\/latex] in [latex]y = 3x - 4[\/latex]:<\/p>\r\n<p data-start=\"1019\" data-end=\"1110\">[latex]\r\n\\begin{align}\r\ny &amp;= 3(2t + 1) - 4 \\\\\r\n&amp;= 6t + 3 - 4 \\\\\r\n&amp;= 6t - 1\r\n\\end{align}\r\n[\/latex]<\/p>\r\n<p data-start=\"1112\" data-end=\"1141\">So [latex]y = 6t - 1[\/latex].<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p data-start=\"801\" data-end=\"907\">Given [latex]x = 3t - 2[\/latex], solve for [latex]t[\/latex] and substitute into [latex]y = 2t + 1[\/latex].<\/p>\r\n<p data-start=\"909\" data-end=\"969\">First, solve [latex]x = 3t - 2[\/latex] for [latex]t[\/latex]:<\/p>\r\n<p data-start=\"971\" data-end=\"1065\">[latex]\r\n\\begin{align}\r\nx &amp;= 3t - 2 \\\\\r\nx + 2 &amp;= 3t \\\\\r\nt &amp;= \\dfrac{x + 2}{3}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n<p data-start=\"1067\" data-end=\"1150\">Now substitute this expression for [latex]t[\/latex] into [latex]y = 2t + 1[\/latex]:<\/p>\r\n<p data-start=\"1152\" data-end=\"1344\">[latex]\r\n\\begin{align}\r\ny &amp;= 2\\left(\\dfrac{x + 2}{3}\\right) + 1 \\\\\r\n&amp;= \\dfrac{2(x + 2)}{3} + 1 \\\\\r\n&amp;= \\dfrac{2x + 4}{3} + 1 \\\\\r\n&amp;= \\dfrac{2x + 4}{3} +\\dfrac{3}{3} \\\\\r\n&amp;= \\dfrac{2x + 7}{3}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n<p data-start=\"1346\" data-end=\"1397\">So [latex]y = \\dfrac{2}{3}x + \\dfrac{7}{3}[\/latex].<\/p>\r\n<p data-start=\"1399\" data-end=\"1478\">This process connects the two equations into one, eliminating [latex]t[\/latex].<\/p>\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]313898[\/ohm_question]<\/section>\r\n<ol data-start=\"1354\" data-end=\"1645\">\r\n \t<li style=\"list-style-type: none;\" data-start=\"1457\" data-end=\"1645\"><\/li>\r\n<\/ol>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Substitute an algebraic expression into another function<\/span><\/li>\n<\/ul>\n<\/section>\n<p data-start=\"571\" data-end=\"785\">Sometimes we work with two related equations that share a variable. We can connect them by solving one equation for that variable and substituting the expression into the other. This is called substitution.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p data-start=\"754\" data-end=\"871\">Given [latex]x = 2t + 1[\/latex], and [latex]y = 3x - 4[\/latex], find [latex]y[\/latex] in terms of [latex]t[\/latex].<\/p>\n<p data-start=\"873\" data-end=\"1017\">We already know what [latex]x[\/latex] equals, so substitute [latex]2t + 1[\/latex] wherever we see [latex]x[\/latex] in [latex]y = 3x - 4[\/latex]:<\/p>\n<p data-start=\"1019\" data-end=\"1110\">[latex]\\begin{align}  y &= 3(2t + 1) - 4 \\\\  &= 6t + 3 - 4 \\\\  &= 6t - 1  \\end{align}[\/latex]<\/p>\n<p data-start=\"1112\" data-end=\"1141\">So [latex]y = 6t - 1[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p data-start=\"801\" data-end=\"907\">Given [latex]x = 3t - 2[\/latex], solve for [latex]t[\/latex] and substitute into [latex]y = 2t + 1[\/latex].<\/p>\n<p data-start=\"909\" data-end=\"969\">First, solve [latex]x = 3t - 2[\/latex] for [latex]t[\/latex]:<\/p>\n<p data-start=\"971\" data-end=\"1065\">[latex]\\begin{align}  x &= 3t - 2 \\\\  x + 2 &= 3t \\\\  t &= \\dfrac{x + 2}{3}  \\end{align}[\/latex]<\/p>\n<p data-start=\"1067\" data-end=\"1150\">Now substitute this expression for [latex]t[\/latex] into [latex]y = 2t + 1[\/latex]:<\/p>\n<p data-start=\"1152\" data-end=\"1344\">[latex]\\begin{align}  y &= 2\\left(\\dfrac{x + 2}{3}\\right) + 1 \\\\  &= \\dfrac{2(x + 2)}{3} + 1 \\\\  &= \\dfrac{2x + 4}{3} + 1 \\\\  &= \\dfrac{2x + 4}{3} +\\dfrac{3}{3} \\\\  &= \\dfrac{2x + 7}{3}  \\end{align}[\/latex]<\/p>\n<p data-start=\"1346\" data-end=\"1397\">So [latex]y = \\dfrac{2}{3}x + \\dfrac{7}{3}[\/latex].<\/p>\n<p data-start=\"1399\" data-end=\"1478\">This process connects the two equations into one, eliminating [latex]t[\/latex].<\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm313898\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=313898&theme=lumen&iframe_resize_id=ohm313898&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<ol data-start=\"1354\" data-end=\"1645\">\n<li style=\"list-style-type: none;\" data-start=\"1457\" data-end=\"1645\"><\/li>\n<\/ol>\n","protected":false},"author":67,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":520,"module-header":"background_you_need","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2785"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/67"}],"version-history":[{"count":7,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2785\/revisions"}],"predecessor-version":[{"id":4819,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2785\/revisions\/4819"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/520"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2785\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=2785"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=2785"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=2785"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=2785"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}