Holt McDougal Algebra 1 5-2 Solving Systems by Substitution Warm Up Solve each equation for x. 1. y...
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Transcript of Holt McDougal Algebra 1 5-2 Solving Systems by Substitution Warm Up Solve each equation for x. 1. y...
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Warm UpSolve each equation for x.
1. y = x + 3
2. y = 3x – 4
x = y – 3
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Solve systems of linear equations in two variables by substitution.
Objective
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Sometimes it is difficult to identify the exact solution to a system by graphing. In this case, you can use a method called substitution.
The goal when using substitution is to reduce the system to one equation that has only one variable. Then you can solve this equation by the methods taught in Chapter 2.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Solving Systems of Equations by Substitution
Step 2
Step 3
Step 4
Step 5
Step 1Solve for one variable in at least one equation, if necessary.
Substitute the resulting expression into the other equation.
Solve that equation to get the value of the first variable.
Substitute that value into one of the original equations and solve.
Write the values from steps 3 and 4 as an ordered pair, (x, y), and check.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Solve the system by substitution.
Example 1A: Solving a System of Linear Equations by Substitution
y = 3x
y = x – 2
Step 1 y = 3xy = x – 2
Both equations are solved for y.
Step 2 y = x – 23x = x – 2
Substitute 3x for y in the second equation.
Solve for x. Subtract x from both sides and then divide by 2.
Step 3 –x –x2x = –22x = –22 2x = –1
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Solve the system by substitution.
Example 1A Continued
Step 4 y = 3x Write one of the original equations.
Substitute –1 for x. y = 3(–1)y = –3
Step 5 (–1, –3)
Check Substitute (–1, –3) into both equations in the system.
Write the solution as an ordered pair.
y = 3x–3 3(–1)
–3 –3
y = x – 2–3 –1 – 2
–3 –3
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Solve the system by substitution.
Example 1B: Solving a System of Linear Equations by Substitution
y = x + 1
4x + y = 6
Step 1 y = x + 1 The first equation is solved for y.
Step 2 4x + y = 64x + (x + 1) = 6
Substitute x + 1 for y in the second equation.
Subtract 1 from both sides. Step 3 –1 –1
5x = 5 5 5
x = 1
5x = 5
5x + 1 = 6 Simplify. Solve for x.
Divide both sides by 5.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Solve the system by substitution.
Example1B Continued
Step 4 y = x + 1 Write one of the original equations.
Substitute 1 for x. y = 1 + 1y = 2
Step 5 (1, 2)
Check Substitute (1, 2) into both equations in the system.
Write the solution as an ordered pair.
y = x + 1
2 1 + 12 2
4x + y = 6
4(1) + 2 66 6
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Solve the system by substitution.
Example 1C: Solving a System of Linear Equations by Substitution
x + 2y = –1
x – y = 5
Step 1 x + 2y = –1 Solve the first equation for x by subtracting 2y from both sides.
Step 2 x – y = 5(–2y – 1) – y = 5
Substitute –2y – 1 for x in the second equation.
–3y – 1 = 5 Simplify.
−2y −2yx = –2y – 1
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Example 1C Continued
Step 3 –3y – 1 = 5Add 1 to both sides. +1 +1
–3y = 6
–3y = 6–3 –3
y = –2
Solve for y.
Divide both sides by –3.
Step 4 x – y = 5 x – (–2) = 5
x + 2 = 5 –2 –2
x = 3
Step 5 (3, –2)
Write one of the original equations.
Substitute –2 for y.
Subtract 2 from both sides.Write the solution as an
ordered pair.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Check It Out! Example 1a
Solve the system by substitution.
y = x + 3
y = 2x + 5
Both equations are solved for y.Step 1 y = x + 3 y = 2x + 5
Substitute 2x + 5 for y in the first equation.
Solve for x. Subtract x and 5 from both sides. –x – 5 –x – 5
x = –2
Step 3 2x + 5 = x + 3
Step 22x + 5 = x + 3
y = x + 3
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Check It Out! Example 1a Continued
Solve the system by substitution.
Step 4 y = x + 3 Write one of the original equations.
Substitute –2 for x. y = –2 + 3
y = 1
Step 5 (–2, 1) Write the solution as an ordered pair.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Check It Out! Example 1b
Solve the system by substitution.x = 2y – 4
x + 8y = 16
The first equation is solved for x.Step 1 x = 2y – 4
Substitute 2y – 4 for x in the second equation.
Simplify. Then solve for y.
(2y – 4) + 8y = 16
x + 8y = 16 Step 2
Step 3 10y – 4 = 16Add 4 to both sides. +4 +4
10y = 20
y = 2
10y 2010 10= Divide both sides by 10.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Check It Out! Example 1b Continued
Solve the system by substitution.
Step 4 x + 8y = 16 Write one of the original equations.
Substitute 2 for y. x + 8(2) = 16
x + 16 = 16
x = 0 – 16 –16
Simplify.Subtract 16 from both
sides.
Step 5 (0, 2) Write the solution as an ordered pair.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Check It Out! Example 1c
Solve the system by substitution.
2x + y = –4
x + y = –7 Solve the second equation for x
by subtracting y from each side.
Substitute –y – 7 for x in the first equation. 2(–y – 7) + y = –4
x = –y – 7 Step 2
Step 1 x + y = –7– y – y
x = –y – 7
2(–y – 7) + y = –4
–2y – 14 + y = –4Distribute 2.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Combine like terms. Step 3
+14 +14
–y = 10
Check It Out! Example 1c Continued
Solve the system by substitution.
–2y – 14 + y = –4
Add 14 to each side.
–y – 14 = –4
y = –10
Step 4 x + y = –7 Write one of the original equations.
Substitute –10 for y. x + (–10) = –7
x – 10 = – 7
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Check It Out! Example 1c Continued
Solve the system by substitution.
x – 10 = –7 Step 5+10 +10
x = 3
Add 10 to both sides.
Step 6 (3, –10) Write the solution as an ordered pair.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Sometimes you substitute an expression for a variable that has a coefficient. When solving for the second variable in this situation, you can use the Distributive Property.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
When you solve one equation for a variable, you must substitute the value or expression into the other original equation, not the one that had just been solved.
Caution
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Example 2: Using the Distributive Property
y + 6x = 11
3x + 2y = –5 Solve by substitution.
Solve the first equation for y by subtracting 6x from each side.
Step 1 y + 6x = 11– 6x – 6x
y = –6x + 11
Substitute –6x + 11 for y in the second equation. 3x + 2(–6x + 11) = –5
3x + 2y = –5 Step 2
3x + 2(–6x + 11) = –5 Distribute 2 to the expression in parentheses.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Step 3
Example 2 Continued
y + 6x = 11
3x + 2y = –5 Solve by substitution.
3x + 2(–6x) + 2(11) = –5
–9x + 22 = –5
Simplify. Solve for x.
Subtract 22 from both sides.–9x = –27
– 22 –22
Divide both sides by –9.
–9x = –27–9 –9
x = 3
3x – 12x + 22 = –5
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Step 4 y + 6x = 11
Substitute 3 for x.y + 6(3) = 11
Subtract 18 from each side.y + 18 = 11
–18 –18
y = –7
Step 5 (3, –7) Write the solution as an ordered pair.
Simplify.
Example 2 Continued
y + 6x = 11
3x + 2y = –5 Solve by substitution.
Write one of the original equations.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Check It Out! Example 2
–2x + y = 8
3x + 2y = 9Solve by substitution.
Solve the first equation for y by adding 2x to each side.
Step 1 –2x + y = 8+ 2x +2x
y = 2x + 8
Substitute 2x + 8 for y in the second equation. 3x + 2(2x + 8) = 9
3x + 2y = 9 Step 2
3x + 2(2x + 8) = 9 Distribute 2 to the expression in parentheses.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Step 3 3x + 2(2x) + 2(8) = 9
7x + 16 = 9
Simplify. Solve for x.
Subtract 16 from both sides.7x = –7
–16 –16
Divide both sides by 7.
7x = –77 7x = –1
Check It Out! Example 2 Continued
–2x + y = 8
3x + 2y = 9Solve by substitution.
3x + 4x + 16 = 9
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Step 4 –2x + y = 8
Substitute –1 for x.–2(–1) + y = 8
y + 2 = 8
–2 –2
y = 6
Step 5 (–1, 6) Write the solution as an ordered pair.
Check It Out! Example 2 Continued
–2x + y = 8
3x + 2y = 9Solve by substitution.
Subtract 2 from each side.
Simplify.
Write one of the original equations.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Example 3: Consumer Economics Application
Jenna is deciding between two cell-phone plans. The first plan has a $50 sign-up fee and costs $20 per month. The second plan has a $30 sign-up fee and costs $25 per month. After how many months will the total costs be the same? What will the costs be? If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.
Write an equation for each option. Let t represent the total amount paid and m represent the number of months.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Example 3 Continued
Total paid is
sign-up fee plus
paymentamount
for eachmonth.
Option 1 t = $50 + $20 m
Option 2 t = $30 + $25 m
Step 1 t = 50 + 20mt = 30 + 25m
Both equations are solved for t.
Step 2 50 + 20m =30 + 25m Substitute 50 + 20m for t in the second equation.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Step 3 50 + 20m = 30 + 25m Solve for m. Subtract 20m from both sides.–20m – 20m
50 = 30 + 5m Subtract 30 from both sides. –30 –30
20 = 5m Divide both sides by 5.
Write one of the original equations.
Step 4 t = 30 + 25m
t = 30 + 25(4)
t = 30 + 100
t = 130
Substitute 4 for m.Simplify.
Example 3 Continued
5 5
m = 4
20 = 5m
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Step 5 (4, 130)Write the solution as an
ordered pair.
In 4 months, the total cost for each option would be the same $130.
Jenna should choose the first plan because it costs $290 for the year and the second plan costs $330.
Example 3 Continued
Option 1: t = 50 + 20(12) = 290
Option 2: t = 30 + 25(12) = 330
If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Check It Out! Example 3
One cable television provider has a $60 setup fee and charges $80 per month, and the second has a $160 equipment fee and charges $70 per month.
a. In how many months will the cost be the same? What will that cost be.
Write an equation for each option. Let t represent the total amount paid and m represent the number of months.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Total paid is fee plus
paymentamount
for eachmonth.
Option 1 t = $60 + $80 m
Option 2 t = $160 + $70 m
Check It Out! Example 3 Continued
Step 1 t = 60 + 80mt = 160 + 70m
Both equations are solved for t.
Step 2 60 + 80m = 160 + 70m Substitute 60 + 80m for t in the second equation.
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Step 3 60 + 80m = 160 + 70m Solve for m. Subtract 70m from both sides. –70m –70m
60 + 10m = 160 Subtract 60 from both sides.
Divide both sides by 10.
–60 –6010m = 10010 10
m = 10
Write one of the original equations.
Step 4 t = 160 + 70m
t = 160 + 70(10)
t = 160 + 700
t = 860
Substitute 10 for m.Simplify.
Check It Out! Example 3 Continued
Holt McDougal Algebra 1
5-2 Solving Systems by Substitution
Step 5 (10, 860) Write the solution as an ordered pair.
In 10 months, the total cost for each option would be the same, $860.
The first option is cheaper for the first six months.
Check It Out! Example 3 Continued
Option 1: t = 60 + 80(6) = 540
Option 2: t = 160 + 70(6) = 580
b. If you plan to move in 6 months, which is the cheaper option? Explain.