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book College Algebra in Context with Applications for the Managerial, Life, and Social Sciences 3rd Edition by Ronald J Harshbarger, Lisa Yocco cover

College Algebra in Context with Applications for the Managerial, Life, and Social Sciences 3rd Edition by Ronald J Harshbarger, Lisa Yocco

النسخة 3الرقم المعياري الدولي: 032157060X
book College Algebra in Context with Applications for the Managerial, Life, and Social Sciences 3rd Edition by Ronald J Harshbarger, Lisa Yocco cover

College Algebra in Context with Applications for the Managerial, Life, and Social Sciences 3rd Edition by Ronald J Harshbarger, Lisa Yocco

النسخة 3الرقم المعياري الدولي: 032157060X
تمرين 6
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الخطوة 1 من2

Consider the equation    <div class=answer> Consider the equation   . Let us solve this equation for <i>x</i> algebraically and graphically. First solve this equation algebraically.   The least common denominator of 2 and 3 is 6. So, multiply both sides by 6.     Remove parentheses using the distributive property.     Add   to both sides.     Subtract 15 from both sides.     Divide both sides by 28.   Thus, the algebraic solution of the equation   is   . .

Let us solve this equation for x algebraically and graphically.

First solve this equation algebraically.

    <div class=answer> Consider the equation   . Let us solve this equation for <i>x</i> algebraically and graphically. First solve this equation algebraically.   The least common denominator of 2 and 3 is 6. So, multiply both sides by 6.     Remove parentheses using the distributive property.     Add   to both sides.     Subtract 15 from both sides.     Divide both sides by 28.   Thus, the algebraic solution of the equation   is   .

The least common denominator of 2 and 3 is 6.

So, multiply both sides by 6.

    <div class=answer> Consider the equation   . Let us solve this equation for <i>x</i> algebraically and graphically. First solve this equation algebraically.   The least common denominator of 2 and 3 is 6. So, multiply both sides by 6.     Remove parentheses using the distributive property.     Add   to both sides.     Subtract 15 from both sides.     Divide both sides by 28.   Thus, the algebraic solution of the equation   is   .

    <div class=answer> Consider the equation   . Let us solve this equation for <i>x</i> algebraically and graphically. First solve this equation algebraically.   The least common denominator of 2 and 3 is 6. So, multiply both sides by 6.     Remove parentheses using the distributive property.     Add   to both sides.     Subtract 15 from both sides.     Divide both sides by 28.   Thus, the algebraic solution of the equation   is   . Remove parentheses using the distributive property.

    <div class=answer> Consider the equation   . Let us solve this equation for <i>x</i> algebraically and graphically. First solve this equation algebraically.   The least common denominator of 2 and 3 is 6. So, multiply both sides by 6.     Remove parentheses using the distributive property.     Add   to both sides.     Subtract 15 from both sides.     Divide both sides by 28.   Thus, the algebraic solution of the equation   is   .

    <div class=answer> Consider the equation   . Let us solve this equation for <i>x</i> algebraically and graphically. First solve this equation algebraically.   The least common denominator of 2 and 3 is 6. So, multiply both sides by 6.     Remove parentheses using the distributive property.     Add   to both sides.     Subtract 15 from both sides.     Divide both sides by 28.   Thus, the algebraic solution of the equation   is   . Add     <div class=answer> Consider the equation   . Let us solve this equation for <i>x</i> algebraically and graphically. First solve this equation algebraically.   The least common denominator of 2 and 3 is 6. So, multiply both sides by 6.     Remove parentheses using the distributive property.     Add   to both sides.     Subtract 15 from both sides.     Divide both sides by 28.   Thus, the algebraic solution of the equation   is   . to both sides.

    <div class=answer> Consider the equation   . Let us solve this equation for <i>x</i> algebraically and graphically. First solve this equation algebraically.   The least common denominator of 2 and 3 is 6. So, multiply both sides by 6.     Remove parentheses using the distributive property.     Add   to both sides.     Subtract 15 from both sides.     Divide both sides by 28.   Thus, the algebraic solution of the equation   is   .

    <div class=answer> Consider the equation   . Let us solve this equation for <i>x</i> algebraically and graphically. First solve this equation algebraically.   The least common denominator of 2 and 3 is 6. So, multiply both sides by 6.     Remove parentheses using the distributive property.     Add   to both sides.     Subtract 15 from both sides.     Divide both sides by 28.   Thus, the algebraic solution of the equation   is   . Subtract 15 from both sides.

    <div class=answer> Consider the equation   . Let us solve this equation for <i>x</i> algebraically and graphically. First solve this equation algebraically.   The least common denominator of 2 and 3 is 6. So, multiply both sides by 6.     Remove parentheses using the distributive property.     Add   to both sides.     Subtract 15 from both sides.     Divide both sides by 28.   Thus, the algebraic solution of the equation   is   .

    <div class=answer> Consider the equation   . Let us solve this equation for <i>x</i> algebraically and graphically. First solve this equation algebraically.   The least common denominator of 2 and 3 is 6. So, multiply both sides by 6.     Remove parentheses using the distributive property.     Add   to both sides.     Subtract 15 from both sides.     Divide both sides by 28.   Thus, the algebraic solution of the equation   is   . Divide both sides by 28.

    <div class=answer> Consider the equation   . Let us solve this equation for <i>x</i> algebraically and graphically. First solve this equation algebraically.   The least common denominator of 2 and 3 is 6. So, multiply both sides by 6.     Remove parentheses using the distributive property.     Add   to both sides.     Subtract 15 from both sides.     Divide both sides by 28.   Thus, the algebraic solution of the equation   is   .

Thus, the algebraic solution of the equation    <div class=answer> Consider the equation   . Let us solve this equation for <i>x</i> algebraically and graphically. First solve this equation algebraically.   The least common denominator of 2 and 3 is 6. So, multiply both sides by 6.     Remove parentheses using the distributive property.     Add   to both sides.     Subtract 15 from both sides.     Divide both sides by 28.   Thus, the algebraic solution of the equation   is   . is    <div class=answer> Consider the equation   . Let us solve this equation for <i>x</i> algebraically and graphically. First solve this equation algebraically.   The least common denominator of 2 and 3 is 6. So, multiply both sides by 6.     Remove parentheses using the distributive property.     Add   to both sides.     Subtract 15 from both sides.     Divide both sides by 28.   Thus, the algebraic solution of the equation   is   . .


الخطوة 2 من 2

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College Algebra in Context with Applications for the Managerial, Life, and Social Sciences 3rd Edition by Ronald J Harshbarger, Lisa Yocco
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