{"id":35988,"date":"2025-06-25T00:59:24","date_gmt":"2025-06-25T00:59:24","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=35988"},"modified":"2025-06-25T00:59:25","modified_gmt":"2025-06-25T00:59:25","slug":"solve-the-equation","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/solve-the-equation\/","title":{"rendered":"Solve the equation"},"content":{"rendered":"\n

Solve the equation.\\ \\frac{3n}{n+2} – \\frac{5}{n-4} = \\frac{2n^2 – 14n}{n^2 – 2n – 8<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

We are given the equation:3nn+2\u22125n\u22124=2n2\u221214nn2\u22122n\u22128\\frac{3n}{n+2} – \\frac{5}{n-4} = \\frac{2n^2 – 14n}{n^2 – 2n – 8}n+23n\u200b\u2212n\u221245\u200b=n2\u22122n\u221282n2\u221214n\u200b<\/p>\n\n\n\n

\n\n\n\n

Step 1: Factor the denominator on the right-hand side<\/h3>\n\n\n\n

n2\u22122n\u22128=(n\u22124)(n+2)n^2 – 2n – 8 = (n – 4)(n + 2)n2\u22122n\u22128=(n\u22124)(n+2)<\/p>\n\n\n\n

So the equation becomes:3nn+2\u22125n\u22124=2n2\u221214n(n\u22124)(n+2)\\frac{3n}{n+2} – \\frac{5}{n-4} = \\frac{2n^2 – 14n}{(n-4)(n+2)}n+23n\u200b\u2212n\u221245\u200b=(n\u22124)(n+2)2n2\u221214n\u200b<\/p>\n\n\n\n

\n\n\n\n

Step 2: Find the least common denominator (LCD)<\/h3>\n\n\n\n

The LCD is (n+2)(n\u22124)(n + 2)(n – 4)(n+2)(n\u22124). We will write all terms with this denominator.<\/p>\n\n\n\n

Left-hand side:<\/strong>3nn+2=3n(n\u22124)(n+2)(n\u22124),5n\u22124=5(n+2)(n\u22124)(n+2)\\frac{3n}{n+2} = \\frac{3n(n – 4)}{(n + 2)(n – 4)}, \\quad \\frac{5}{n – 4} = \\frac{5(n + 2)}{(n – 4)(n + 2)}n+23n\u200b=(n+2)(n\u22124)3n(n\u22124)\u200b,n\u221245\u200b=(n\u22124)(n+2)5(n+2)\u200b<\/p>\n\n\n\n

Now substitute:3n(n\u22124)\u22125(n+2)(n+2)(n\u22124)=2n2\u221214n(n+2)(n\u22124)\\frac{3n(n – 4) – 5(n + 2)}{(n + 2)(n – 4)} = \\frac{2n^2 – 14n}{(n + 2)(n – 4)}(n+2)(n\u22124)3n(n\u22124)\u22125(n+2)\u200b=(n+2)(n\u22124)2n2\u221214n\u200b<\/p>\n\n\n\n

\n\n\n\n

Step 3: Simplify the numerator on the left-hand side<\/h3>\n\n\n\n

3n(n\u22124)=3n2\u221212n,5(n+2)=5n+103n(n – 4) = 3n^2 – 12n, \\quad 5(n + 2) = 5n + 103n(n\u22124)=3n2\u221212n,5(n+2)=5n+103n2\u221212n\u22125n\u221210=3n2\u221217n\u2212103n^2 – 12n – 5n – 10 = 3n^2 – 17n – 103n2\u221212n\u22125n\u221210=3n2\u221217n\u221210<\/p>\n\n\n\n

So the equation becomes:3n2\u221217n\u221210(n+2)(n\u22124)=2n2\u221214n(n+2)(n\u22124)\\frac{3n^2 – 17n – 10}{(n + 2)(n – 4)} = \\frac{2n^2 – 14n}{(n + 2)(n – 4)}(n+2)(n\u22124)3n2\u221217n\u221210\u200b=(n+2)(n\u22124)2n2\u221214n\u200b<\/p>\n\n\n\n

Since the denominators are the same, we equate the numerators:3n2\u221217n\u221210=2n2\u221214n3n^2 – 17n – 10 = 2n^2 – 14n3n2\u221217n\u221210=2n2\u221214n<\/p>\n\n\n\n

\n\n\n\n

Step 4: Bring all terms to one side<\/h3>\n\n\n\n

3n2\u221217n\u221210\u22122n2+14n=03n^2 – 17n – 10 – 2n^2 + 14n = 03n2\u221217n\u221210\u22122n2+14n=0n2\u22123n\u221210=0n^2 – 3n – 10 = 0n2\u22123n\u221210=0<\/p>\n\n\n\n

\n\n\n\n

Step 5: Solve the quadratic equation<\/h3>\n\n\n\n

Use the quadratic formula:n=\u2212(\u22123)\u00b1(\u22123)2\u22124(1)(\u221210)2(1)=3\u00b19+402=3\u00b1492n = \\frac{-(-3) \\pm \\sqrt{(-3)^2 – 4(1)(-10)}}{2(1)} = \\frac{3 \\pm \\sqrt{9 + 40}}{2} = \\frac{3 \\pm \\sqrt{49}}{2}n=2(1)\u2212(\u22123)\u00b1(\u22123)2\u22124(1)(\u221210)\u200b\u200b=23\u00b19+40\u200b\u200b=23\u00b149\u200b\u200bn=3\u00b172\u21d2n=102=5orn=\u221242=\u22122n = \\frac{3 \\pm 7}{2} \\Rightarrow n = \\frac{10}{2} = 5 \\quad \\text{or} \\quad n = \\frac{-4}{2} = -2n=23\u00b17\u200b\u21d2n=210\u200b=5orn=2\u22124\u200b=\u22122<\/p>\n\n\n\n

\n\n\n\n

Step 6: Check for extraneous solutions<\/h3>\n\n\n\n

The original equation contains denominators n+2n + 2n+2 and n\u22124n – 4n\u22124. We must exclude values that make any denominator zero.<\/p>\n\n\n\n

    \n
  • n=\u22122n = -2n=\u22122 makes n+2=0n + 2 = 0n+2=0 \u2192 Not allowed<\/li>\n\n\n\n
  • n=4n = 4n=4 makes n\u22124=0n – 4 = 0n\u22124=0 \u2192 Not allowed<\/li>\n<\/ul>\n\n\n\n

    Since n=\u22122n = -2n=\u22122 is not valid, we reject it.<\/p>\n\n\n\n

    \n\n\n\n

    \u2705 Final Answer:<\/h3>\n\n\n\n

    n=5\\boxed{n = 5}n=5\u200b<\/p>\n\n\n\n

    \n\n\n\n

    Explanation <\/h3>\n\n\n\n

    To solve the equation 3nn+2\u22125n\u22124=2n2\u221214nn2\u22122n\u22128\\frac{3n}{n+2} – \\frac{5}{n-4} = \\frac{2n^2 – 14n}{n^2 – 2n – 8}n+23n\u200b\u2212n\u221245\u200b=n2\u22122n\u221282n2\u221214n\u200b, we start by factoring the denominator on the right. The quadratic n2\u22122n\u22128n^2 – 2n – 8n2\u22122n\u22128 factors as (n\u22124)(n+2)(n – 4)(n + 2)(n\u22124)(n+2). Recognizing this is helpful because it matches the denominators on the left. We then rewrite each term so that they all have a common denominator: (n+2)(n\u22124)(n + 2)(n – 4)(n+2)(n\u22124). This allows us to combine and compare the numerators directly.<\/p>\n\n\n\n

    We expand the numerators: 3n(n\u22124)3n(n – 4)3n(n\u22124) gives 3n2\u221212n3n^2 – 12n3n2\u221212n, and 5(n+2)5(n + 2)5(n+2) gives 5n+105n + 105n+10. Subtracting the second expression from the first gives 3n2\u221217n\u2212103n^2 – 17n – 103n2\u221217n\u221210. The right-hand side numerator remains 2n2\u221214n2n^2 – 14n2n2\u221214n. Now that both sides of the equation have the same denominator, we equate the numerators: 3n2\u221217n\u221210=2n2\u221214n3n^2 – 17n – 10 = 2n^2 – 14n3n2\u221217n\u221210=2n2\u221214n.<\/p>\n\n\n\n

    Bringing all terms to one side yields the quadratic n2\u22123n\u221210=0n^2 – 3n – 10 = 0n2\u22123n\u221210=0. Solving with the quadratic formula, we find n=5n = 5n=5 and n=\u22122n = -2n=\u22122. However, we must check for values that make any denominator in the original equation zero. Here, n=\u22122n = -2n=\u22122 makes the term n+2=0n + 2 = 0n+2=0, which is undefined. So we reject that solution.<\/p>\n\n\n\n

    The only valid solution is n=5n = 5n=5, which satisfies all parts of the original equation without making any denominator zero.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    Solve the equation.\\ \\frac{3n}{n+2} – \\frac{5}{n-4} = \\frac{2n^2 – 14n}{n^2 – 2n – 8 The Correct Answer and Explanation is: We are given the equation:3nn+2\u22125n\u22124=2n2\u221214nn2\u22122n\u22128\\frac{3n}{n+2} – \\frac{5}{n-4} = \\frac{2n^2 – 14n}{n^2 – 2n – 8}n+23n\u200b\u2212n\u221245\u200b=n2\u22122n\u221282n2\u221214n\u200b Step 1: Factor the denominator on the right-hand side n2\u22122n\u22128=(n\u22124)(n+2)n^2 – 2n – 8 = (n – 4)(n + 2)n2\u22122n\u22128=(n\u22124)(n+2) […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-35988","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35988","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=35988"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35988\/revisions"}],"predecessor-version":[{"id":35991,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35988\/revisions\/35991"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=35988"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=35988"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=35988"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}