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Hawkes Learning Systems: College Algebra

Hawkes Learning Systems: College Algebra. Section 2.1b: Applications of Linear Equations in One Variable. Objectives. Solving linear equations for one variable. Interlude: distance and interest problems. Review. In Section 2.1a , we learned that:

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Hawkes Learning Systems: College Algebra

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  1. Hawkes Learning Systems:College Algebra Section 2.1b: Applications of Linear Equations in One Variable

  2. Objectives • Solving linear equations for one variable. • Interlude: distance and interest problems.

  3. Review In Section 2.1a, we learned that: • An equation is a statement that two expressions are equal • There are three types of equations: • An Identity is an equation that is true for all real numbers. • A Contradiction is an equation that is never true. • A Conditional equation is an equation that is true for some values of the variable(s) and false for others. • The solution set is the set of values by which the variable(s) can be replaced to make the equation true. • Two equations that have the same solution set are called equivalent equations.

  4. Solving Linear Equations for One Variable • One common task in applied mathematics is to solve a given equation in two or more variables for one of the variables. • Solving for a variable means to transform the equation into an equivalent one in which the specified variable is isolated on one side. • This is accomplished by the same methods we used to solve equations in Section 2.1a.

  5. Example 1: Solving Linear Equations for One Variable Solve the following equations for the specified variable. . Solve for . Step 1: add to both sides of the equation. Step 2: divide by on both sides of the equation.

  6. Example 2: Solving Linear Equations for One Variable . Solve for .

  7. Example 3: Solving Linear Equations for One Variable . Solve for .

  8. Interlude: Distance and Interest Problems Good examples of linear equations arise from certain distance and simple interest problems. The basic distance formula is where is distance traveled at rate for time . The simple interest formula is where is the interest earned on principal invested at rate for time .

  9. Example 4: Distance Problems A riverboat travels downstream at an average speed of 20 miles per hour. How long will it take for the boat to travel 110 miles? Application of the distance equation. miles Given. miles per hour hours, or 5 hours and 30 minutes

  10. Example 5: Distance Problems Two trucks leave a warehouse at the same time. One travels due west at an average speed of 61 miles per hour, and the other travels due east at an average speed of 53 miles per hour. After how many hours will the two trucks be 456 miles apart? Given. Plug values in. Combine like terms, and solve. hours

  11. Example 6: Interest Problems Sarah invested $10,000 in a global technology mutual fund on January 1st. On July 1st, her stock is worth $11,400. What effective annual rate of return has she earned so far? Application of the interest equation. Given. and , or

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