Explanation Given: The system of equations is given below. { x − y + 2 z = − 1 2 x + 2 y + 3 z = − 3 x + 3 y + z = − 2 Method used: Left-to-Right Elimination Method (or Back substitution): 1. If necessary, interchange two equations or use multiplication to make the coefficient of x in the first equation a 1. 2. Add a multiple of the first equation to each of the following equations so that the coefficients of x in second and third equations become 0. 3. Multiply (or divide) both sides of the second equation by a number that makes the coefficient of y in the second equation equal to 1. 4. Add a multiple of the (new) second equation to the (new) third equation so that the coefficient of y in the newest third equation becomes 0. 5. Multiply (or divide) both sides of the third equation by a number that makes the coefficient of z in the third equation equal to 1. This gives the solution for z in the system of equations. 6. Use the solution for z to solve y in the second equation. Then substitute values for y and z to solve for x in the first equation. (This is called back substitution) Calculation: Use the method stated above and solved the system of equations for x , y , and z as follows. The given system of equations is as follows. { x − y + 2 z = − 1 2 x + 2 y + 3 z = − 3 x + 3 y + z = − 2 Perform the operations E q 2 → E q 2 − 2 E q 1 and E q 3 → E q 3 − E q 1 , and obtain the new equivalent system of equations as follows. { x − y + 2 z = − 1 4 y − z = − 1 4 y − z = − 1 Perform the row operations E q 3 → E q 3 − E q 2 , then the new equivalent system is, { x − y + 2 z = − 1 4 y − z = − 1 0 = 0 From the above system observe that the third equation is reduced to 0 = 0 , and the variable z is not a leading variable in any of the two equations. Hence, the variables x and y can be solve in terms of z as follows. Use back substitution in the second equation. y = 1 4 ( z − 1 ) Substitute y = 1 4 ( z − 1 ) in the first equation and solve for x in terms of z

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Chapter 7.1, Problem 21E

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Chapter 7.1, Problem 20E

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An important application of systems of equations arises in connection with supply and demand. As the price of a product increases, the demand for that product decreases. However, at higher prices, suppliers are willing to produce greater quantities of the product. Exercises 97–98 involve supply and demand. 97. A chain of electronics stores sells hand-held color televisions. The weekly demand and supply models are given as follows: Number sold Demand model per week N = -5p + 750 Price of television Number supplied to the chain per week N = 2.5p. 1apow hjddns a. How many hand-held color televisions can be sold and supplied at $120 per television? b. Find the price at which supply and demand are equal. At this price, how many televisions can be supplied and sold each week?

Chapter 7 Solutions

EBK COLLEGE ALGEBRA IN CONTEXT

Ch. 7.1 - In Exercises 14, solve the systems of equations....Ch. 7.1 - Prob. 2E Ch. 7.1 - Prob. 3E Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - Prob. 8E Ch. 7.1 - Prob. 9E Ch. 7.1 - Prob. 10E

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The system of equations if possible.

Solution Summary: The author explains the Left-to-Right Elimination Method (or Back substitution) used to solve the system of equations.

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Chapter 7 Solutions