{"id":213562,"date":"2024-02-15T11:07:29","date_gmt":"2024-02-15T17:07:29","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=213562"},"modified":"2025-06-30T15:33:25","modified_gmt":"2025-06-30T20:33:25","slug":"email-18-ged-qod-answer","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/email-18-ged-qod-answer\/","title":{"rendered":"GED Question of the Day"},"content":{"rendered":"
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<\/span>Answer<\/h2>\n
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  1. 103 and 104<\/span><\/li>\n<\/ol>\n
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    First, the variables have to be defined. Let \\(x\\) be the first integer; therefore, \\(x + 1\\) is the second integer. This is a two-step problem.<\/p>\n

    The sum of three times the first and two less than the second is translated into the following expression: \\(3x + (x + 1 – 2)\\). This expression is set equal to 411 to obtain \\(3x + (x + 1 – 2) = 411\\).<\/p>\n

    The left-hand side is simplified to obtain \\(4x – 1 = 411\\). To solve for \\(x\\), first add 1 to both sides and then divide both sides by 4 to obtain \\(x = 103\\). The next consecutive integer is 104. <\/p><\/div>\n<\/div>\n

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    \n Review the question<\/span>
    \n \uff0b<\/span>\n <\/div>\n
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    Two consecutive integers exist such that the sum of three times the first and two less than the second is equal to 411. What are those integers?<\/p>\n

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    1. 103 and 104<\/li>\n
    2. 104 and 105<\/li>\n
    3. 102 and 103<\/li>\n
    4. 100 and 101<\/li>\n<\/ol><\/div>\n<\/p><\/div>\n<\/div>\n<\/div>\n

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