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Examples and Guided Practice come from the Algebra 1 PowerPoint Presentations available at www.classzone.com

Examples and Guided Practice come from the Algebra 1 PowerPoint Presentations available at www.classzone.com

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## Examples and Guided Practice come from the Algebra 1 PowerPoint Presentations available at www.classzone.com

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**Examples and Guided Practice come from the Algebra 1**PowerPoint Presentations available at www.classzone.com**7 = 7**? = 3+ 2(2) 7 Check the intersection point EXAMPLE 1 Use the graph to solve the system. Then check your solution algebraically. x + 2y = 7 Equation 1 Equation2 3x – 2y = 5 SOLUTION The lines appear to intersect at the point (3, 2). Substitute3forxand2foryin each equation. CHECK x+ 2y= 7**5**3(3) – 2(2) 5 = 5 ANSWER Because the ordered pair (3, 2) is a solution of each equation, it is a solution of the system. ? = Check the intersection point EXAMPLE 1 3x– 2y= 5**Use the graph-and-check method**EXAMPLE 2 Solve the linear system: –x + y = –7 Equation 1 Equation2 x + 4y = –8 SOLUTION STEP1 Graph both equations.**–8 = –8**–(4) + (–3) –7 ? ? 4+ 4(–3) –8 = = –7= –7 Use the graph-and-check method EXAMPLE 2 STEP2 Estimate the point of intersection. The two lines appear to intersect at (4, – 3). STEP3 Check whether (4, –3) is a solution by substituting 4 for xand –3 for yin each of the original equations. Equation1 Equation2 –x+y=–7 x+4y=–8**ANSWER**Because (4, –3) is a solution of each equation, it is a solution of the linear system. Use the graph-and-check method EXAMPLE 2**1.**–5x + y = 0 5x + y = 10 ANSWER (1, 5) Use the graph-and-check method EXAMPLE 2 for Examples 1 and 2 GUIDED PRACTICE Solve the linear system by graphing. Check your solution.**2.**–x + 2y = 3 2x + y = 4 ANSWER (1, 2) Use the graph-and-check method EXAMPLE 2 for Examples 1 and 2 GUIDED PRACTICE Solve the linear system by graphing. Check your solution.**x – y = 5**3. 3x + y = 3 ANSWER (2, 3) Use the graph-and-check method EXAMPLE 2 for Examples 1 and 2 GUIDED PRACTICE Solve the linear system by graphing. Check your solution.**EXAMPLE 3**Standardized Test Practice The parks and recreation department in your town offers a season pass for $90. As a season pass holder, you pay $4 per session to use the town’s tennis courts. • • Without the season pass, you pay $13 per session to use the tennis courts.**A**B y = 4x y = 4x y = 90 + 13x y = 13x D C y = 13x y = 90 + 4x y = 90 + 4x y = 90 + 13x EXAMPLE 3 Standardized Test Practice Which system of equations can be used to find the number xof sessions of tennis after which the total cost ywith a season pass, including the cost of the pass, is the same as the total cost without a season pass?**y=13x**EXAMPLE 3 Standardized Test Practice SOLUTION Write a system of equations where yis the total cost (in dollars) for xsessions. EQUATION1**y = 90+ 4x**ANSWER The correct answer is C. A D B C EXAMPLE 3 Standardized Test Practice EQUATION 2**ANSWER**10 sessions for Example 3 GUIDED PRACTICE 4. Solve the linear system in Example 3 to find the number of sessions after which the total cost with a season pass, including the cost of the pass, is the same as the total cost without a season pass.**ANSWER**15 sessions for Example 3 GUIDED PRACTICE 5.WHAT IF?In Example 3, suppose a season pass costs $135. After how many sessions is the total cost with a season pass, including the cost of the pass, the same as the total cost without a season pass?**Solve a multi-step problem**EXAMPLE 4 RENTAL BUSINESS A business rents in-line skates and bicycles. During one day, the business has a total of 25 rentals and collects $450 for the rentals. Find the number of pairs of skates rented and the number of bicycles rented.**Solve a multi-step problem**EXAMPLE 4 SOLUTION STEP 1 Write a linear system. Let xbe the number of pairs of skates rented, and let ybe the number of bicycles rented. x + y = 25 Equation for number of rentals 15x + 30y = 450 Equation for money collected from rentals STEP2 Graph both equations.**ANSWER**The business rented 20 pairs of skates and 5 bicycles. ? ? 15(20) + 30(5) 450 20+5 25 = = 450 = 450 25 = 25 Solve a multi-step problem EXAMPLE 4 STEP 3 Estimate the point of intersection. The two lines appear to intersect at(20, 5). STEP 4 Check whether (20, 5) is a solution.**ANSWER**In Example 4, suppose the business has a total of 20 rentals and collects $420. Find the number of bicycles rented. 6. 8 bicycles Solve a multi-step problem EXAMPLE 4 for Example 4 GUIDED PRACTICE**EXAMPLE 1**Use the substitution method Solve the linear system: y = 3x + 2 Equation 1 x + 2y = 11 Equation 2 SOLUTION STEP 1 Solve for y. Equation 1 is already solved for y.**EXAMPLE 1**Use the substitution method STEP 2 Substitute 3x + 2 for y in Equation 2 and solve forx. x + 2y= 11 Write Equation 2. Substitute3x + 2 for y. x +2(3x + 2) = 11 7x + 4 = 11 Simplify. 7x = 7 Subtract 4 from each side. x = 1 Divide each side by 7.**ANSWER**The solution is (1, 5). EXAMPLE 1 Use the substitution method STEP3 Substitute 1 forxin the original Equation 1 to find the value of y. y = 3x+ 2 = 3(1) + 2 = 3 + 2 = 5**5=3(1) + 2**? ? 1 + 2 (5) = 11 5= 5 11=11 EXAMPLE 1 Use the substitution method GUIDED PRACTICE CHECK Substitute 1 for xand 5 for y in each of the original equations. y = 3x +2 x+2y = 11**EXAMPLE 2**Use the substitution method Solve the linear system: x – 2y = –6 Equation 1 4x + 6y = 4 Equation 2 SOLUTION STEP 1 Solve Equation 1 for x. x – 2y = –6 Write original Equation 1. x = 2y – 6 Revised Equation 1**EXAMPLE 2**Use the substitution method STEP2 Substitute 2y – 6 for xin Equation 2 and solve for y. 4x+ 6y = 4 Write Equation 2. Substitute2y – 6 forx. 4(2y – 6) + 6y = 4 8y –24 + 6y = 4 Distributive property 14y – 24 = 4 Simplify. 14y = 28 Add 24 to each side. y = 2 Divide each side by 14.**The solution is (–2, 2).**ANSWER EXAMPLE 2 Use the substitution method STEP3 Substitute 2 foryin the revised Equation 1 to find the value of x. x = 2y– 6 Revised Equation 1 x = 2(2) – 6 Substitute 2 for y. x = –2 Simplify.**?**–6=–6 –2– 2(2) = –6 4=4 4(–2) +6 (2) = 4 ? EXAMPLE 2 Use the substitution method GUIDED PRACTICE CHECK Substitute –2for xand2 for y in each of the original equations. Equation 1 Equation 2 4x+6y= 4 x– 2y = –6**1.**y =2x +5 (1, 7) ANSWER EXAMPLE 1 for Examples 1 and 2 Use the substitution method GUIDED PRACTICE Solve the linear system using the substitution method. 3x + y = 10**x – y = 3**2. (0, –3) ANSWER EXAMPLE 2 for Examples 1 and 2 Use the substitution method GUIDED PRACTICE Solve the linear system using the substitution method. x + 2y = –6**3x + y = –7**3. (–2, –1) ANSWER EXAMPLE 2 for Examples 1 and 2 Use the substitution method GUIDED PRACTICE Solve the linear system using the substitution method. –2x + 4y = 0**EXAMPLE 3**Solve a multi-step problem WEBSITES Many businesses pay website hosting companies to store and maintain the computer files that make up their websites. Internet service providers also offer website hosting. The costs for website hosting offered by a website hosting company and an Internet service provider are shown in the table. Find the number of months after which the total cost for website hosting will be the same for both companies.**y= 10+ 21.95 x**EXAMPLE 3 Solve a multi-step problem SOLUTION STEP 1 Write a system of equations. Let ybe the total cost after xmonths. Equation 1: Internet service provider**y=22.45x**EXAMPLE 3 Solve a multi-step problem Equation 2: Website hosting company The system of equations is: y=10+21.95x Equation 1 y=22.45x Equation 2**ANSWER**The total cost will be the same for both companies after 20 months. EXAMPLE 3 Solve a multi-step problem STEP 2 Substitute 22.45xfor yin Equation 1 and solve for x. y= 10 + 21.95x Write Equation1. 22.45x= 10 + 21.95x Substitute 22.45xfor y. 0.5x=10 Subtract 21.95xfrom each side. x=20 Divide each side by 0.5.**In Example 3, what is the total cost for website hosting for**each company after 20 months? 4. ANSWER $449 for Example 3 GUIDED PRACTICE**WHAT IF? In Example 3, suppose the Internet service provider**offers $5 off the set-up fee. After how many months will the total cost for website hosting be the same for both companies? 5. ANSWER 10 mo for Example 3 GUIDED PRACTICE**EXAMPLE 4**Solve a mixture problem ANTIFREEZE For extremely cold temperatures, an automobile manufacturer recommends that a 70% antifreeze and 30% water mix be used in the cooling system of a car. How many quarts of pure (100%) antifreeze and a 50% antifreeze and 50% water mix should be combined to make 11 quarts of a 70% antifreeze and 30% water mix?**EXAMPLE 4**Solve a mixture problem SOLUTION STEP1 Write an equation for the total number of quarts and an equation for the number of quarts of antifreeze. Let xbe the number of quarts of 100% antifreeze, and let y be the number of quarts of a 50% antifreeze and 50% water mix.**yquarts of**50%–50% mix 11 quarts of 70%–30% mix xquarts of 100% antifreeze 1x+0.5y=0.7(11) EXAMPLE 4 Solve a mixture problem Equation 1: Total number of quarts x + y = 11 Equation 2: Number of quarts of antifreeze x+ 0.5y = 7.7**EXAMPLE 4**Solve a mixture problem x+y=11 The system of equations is: Equation 1 x + 0.5y = 7.7 Equation 2 STEP2 Solve Equation1 forx. x+ y=11 Write Equation 1 x=11–y Revised Equation1 STEP3 Substitute11–yforxin Equation2 and solve fory. x+0.5y=7.7 Write Equation2.**ANSWER**Mix 4.4 quarts of 100% antifreeze and 6.6 quarts of a 50% antifreeze and 50% water mix to get 11 quarts of a 70% antifreeze and 30% water mix. EXAMPLE 4 Solve a mixture problem Substitute11–yforx. (11 – y) + 0.5y = 7.7 Solve fory. y=6.6 STEP4 Substitute 6.6 for yin the revised Equation 1 to find the value of x. x = 11 –y= 11 –6.6= 4.4**WHAT IF? How many quarts of 100% antifreeze and a 50%**antifreeze and 50% water mix should be combined to make 16 quarts of a 70% antifreeze and 30% water mix? 6. ANSWER 6.4 quarts of 100% antifreeze and 9.6 quarts of a 50% antifreezeand 50% water mix for Example 4 GUIDED PRACTICE**Add the equations to**2x + 3y = 11 eliminate one variable. –2x + 5y = 13 EXAMPLE 1 Use addition to eliminate a variable Solve the linear system: 2x + 3y = 11 Equation 1 –2x + 5y = 13 Equation 2 SOLUTION STEP 1 STEP 2 Solve fory. 8y =24 y =3**ANSWER**The solution is(1, 3). EXAMPLE 1 Use addition to eliminate a variable STEP 3 Substitute 3 for yin either equation and solve for x. 2x + 3y = 11 Write Equation 1 2x + 3(3)= 11 Substitute3for y. x = 1 Solve for x.**?**? 2(1)+ 3(3)= 11 2(1)+ 5(3)= 13 EXAMPLE 1 Use addition to eliminate a variable CHECK Substitute 1 for xand 3 foryin each of the original equations. 2x + 3y= 11 2x + 5y= 13 11= 11 13= 13**Subtract the equations to**4x + 3y = 2 eliminate one variable. 5x + 3y = –2 EXAMPLE 2 Use subtraction to eliminate a variable Solve the linear system: 4x + 3y = 2 Equation 1 5x + 3y = –2 Equation 2 SOLUTION STEP 1 STEP 2 Solve forx. – x = 4 x = 4**ANSWER**The solution is (–4, 6). EXAMPLE 2 Use subtraction to eliminate a variable STEP 3 Substitute4 for x in either equation and solve for y. 4x+ 3y = 2 Write Equation 1. 4(–4)+ 3y = 2 Substitute –4 for x. y = 6 Solve fory.