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Solving Systems Using Substitution 589. Lesson 10-2. When equations for lines in a system are in y = mx + b form, a method of solving called ...
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Solving Systems Using Substitution 589
Lesson 10-
When equations for lines in a system are in y = mx + b form, a method of solving called substitution can be very efficient. Example 1 illustrates this method.
Solve the system
y = 7 x + 25 y = −5x - 11
using substitution.
Solution Because 7x + 25 and −5x - 11 both equal y, they must equal each other. Substitute one of them for y in the other equation. 7x + 25 = –5x - 11 Substitution 12x + 25 = –11 Add 5x to both sides. 12x = –36 Subtract 25 from both sides. x = –3 Divide both sides by 12. Now you know x = −3. However, you must still solve for y. You can substitute −3 for x into either of the original equations. We choose the first equation. y = 7x + 25 y = 7(–3) + 25 y = –21 + 25 y = 4 The solution is x = –3 and y = 4, or just (–3, 4).
QY
Check A graph shows that the lines with equations y = 7 x + 25 and y = −5x - 11 intersect at (−3, 4).
Suppose two quantities are increasing or decreasing at different constant rates. Then each quantity can be described by an equation of the form y = mx + b. To find out when the quantities are equal, you can solve a system using substitution. Example 2 illustrates this idea.
BIG IDEA Substituting an expression that equals a single variable is an effective first step for solving some systems.
QY Check that (−3, 4) is a solution to y = −5x - 11.
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ļ 5 ļ 4 ļ 3 ļ 2 ļ 1 1 2 3 4 5
y
x
y = 5 x - 11
y =^7 x +^25
Find the greatest common factor. a. 15; 200 b. 1,500; 20, c. 14; 26; 53 d. 1,400; 2,600; 5,
Mental Math
590 Linear Systems
The Rapid Taxi Company charges $2.15 for a taxi ride plus 20¢ for each __^1 10 mile traveled. A competitor, Carl’s Cabs, charges $1.50 for a taxi ride plus 25¢ for each __ 101 mile traveled. For what distance do the rides cost the same? Solution Let d = the distance of a cab ride in tenths of a mile. Let C = the cost of a cab ride of distance d. Rapid Taxi: C = 2.15 + 0.20d Carl’s Cabs: C = 1.50 + 0.25d The rides cost the same when the values of C and d for Rapid Taxi equal the values for Carl’s Cabs, so we need to solve the system formed by these two equations. Substitute 2.15 + 0.20d for C in the second equation. 2.15 + 0.20d = 1.50 + 0.25d Now solve. 0.65 = 0.05d Add −1.50 and −0.20d to both sides. d = 13 Divide both sides by 0.05. The two companies charge the same amount for a ride that is 13 tenths of a mile long, or 1.3 miles long. Check Check to see if the cost will be the same for a ride of 13 tenths of a mile. The cost for Rapid Taxi is 2.15 + 0.20 · 13 = 2.15 + 2.60 = 4.75. The cost for Carl’s Cabs is 1.50 + 0.25 · 13 = 1.50 + 3.25 = 4.75. The cost is $4.75 from each company, so the answer checks.
In Example 2, Carl’s Cabs is cheaper at first, but as the number of miles increases, the prices become closer. Eventually the price for Carl’s Cabs catches up with Rapid Taxi’s price, and then Carl’s is more expensive than Rapid. The next example also involves “catching up.”
Bart was so confident that he could run faster than his little sister that he bragged, “I can beat you in a 50-meter race. I’m so sure that I’ll give you a 10-meter head start!” Bart could run at a speed of 4 meters per second, while his sister could run 3 meters per second. Could Bart catch up to his sister before the end of the race? Solution Let d be the distance that Bart and his sister have traveled after t seconds. Recall that distance = rate · time. For Bart, d = 4t.
Chapter 10
The average taxi fare in New York in 2006 was $9.65. Source: MSNBC
592 Linear Systems
Chapter 10
Approximately 124,900 acres of tomatoes were harvested in the United States in 2002. Source: U.S. Department of Agriculture
Solving Systems Using Substitution 593
In 14 and 15, a system that involves a quadratic equation is given. Each system has two solutions. a. Solve the system by substitution. b. Check your answers.
y = 1 __ 9 x^2 y = 4 x
y = 2 x^2 + 5 x - 3 y = x^2 - 2 x + 5
REVIEW
y = 20 x + 8 24 x - y = −
. Verify that (^) ( __^12 , 18 (^) ) is a
solution to the system, but that (1, 20) is not. (Lesson 10-1)
In 17 and 18, solve the system of equations by graphing. (Lesson 10-1)
Lesson 10-
QY ANSWER Does 4 = −5(−3) - 11? 4 = 15 - 11 4 = 4 Yes, it checks.