Download Solving Systems Using Substitution

Solving Systems Using
Solve each equation.Substitution
ALGEBRA 1 LESSON 9-2
(For help, go to Lessons2-4 and 7-1.)
1. m – 6 = 4m + 8
2. 4n = 9 – 2n
3. 1 t + 5 = 10
3
For each system, is the ordered pair a solution of both equations?
4. (5, 1)
y = –x + 4
5. (2, 2.4) 4x + 5y = 20
y=x–6
2x + 6y = 10
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Solving Systems Using
Substitution
ALGEBRA 1 LESSON 9-2
Solutions
1.
m – 6 = 4m + 8
m – m – 6 = 4m – m + 8
–6 = 3m + 8
2.
n = 11
–14 = 3m
–4 2 = m
2
3
3.
4n = 9 – 2n
4n + 2n = 9 – 2n + 2n
6n = 9
1 t + 5 = 10
3
1t =5
3
t = 15
9-2
Solving Systems Using
Substitution
Solutions (continued)
ALGEBRA 1 LESSON 9-2
4. (5, 1) in y = –x + 4
1
–5 + 4
1 =/ –1
(5, 1) in y = x – 6
1
5–6
1 =/ –1
no
no
No, (5, 1) is not a solution of both equations.
5. (2, 2.4) in 4x + 5y = 20
4(2) + 5(2.4)
20
(2, 2.4) in 2x + 6y = 10
2(2) + 6(2.4) 10
8 + 12
20
20 = 20
yes
4 + 14.4 10
18.4 =/ 10
no
No, (2, 2.4) is not a solution of both equations.
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Solving Systems Using
Substitution
Solve using substitution. y = 2x + 2
ALGEBRA 1 LESSON 9-2
y = –3x + 4
Step 1: Write an equation containing only one variable and solve.
y = 2x + 2
–3x + 4 = 2x + 2
4 = 5x + 2
2 = 5x
0.4 = x
Start with one equation.
Substitute –3x + 4 for y in that equation.
Add 3x to each side.
Subtract 2 from each side.
Divide each side by 5.
Step 2: Solve for the other variable.
y = 2(0.4) + 2
y = 0.8 + 2
y = 2.8
Substitute 0.4 for x in either equation.
Simplify.
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Solving Systems Using
Substitution
ALGEBRA 1 LESSON 9-2
(continued)
Since x = 0.4 and y = 2.8, the solution is (0.4, 2.8).
Check: See if (0.4, 2.8) satisfies y = –3x + 4 since y = 2x + 2 was used in
Step 2.
2.8
–3(0.4) + 4
2.8
–1.2 + 4
2.8 = 2.8
Substitute (0.4, 2.8) for (x, y) in the equation.
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Solving Systems Using
Substitution
Solve using substitution.
ALGEBRA 1 LESSON 9-2
–2x + y = –1
4x + 2y = 12
Step 1: Solve the first equation for y because it has a coefficient of 1.
–2x + y = –1
y = 2x –1
Add 2x to each side.
Step 2: Write an equation containing only one variable and solve.
4x + 2y = 12
4x + 2(2x –1) = 12
4x + 4x –2 = 12
8x = 14
x = 1.75
Start with the other equation.
Substitute 2x –1 for y in that equation.
Use the Distributive Property.
Combine like terms and add 2 to each side.
Divide each side by 8.
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Solving Systems Using
Substitution
ALGEBRA 1 LESSON 9-2
(continued)
Step 3: Solve for y in the other equation.
–2(1.75) + y = 1
–3.5 + y = –1
y = 2.5
Substitute 1.75 for x.
Simplify.
Add 3.5 to each side.
Since x = 1.75 and y = 2.5, the solution is (1.75, 2.5).
9-2
Solving Systems Using
Substitution
A youth group with 26 members is going to the beach. There
ALGEBRA 1 LESSON 9-2
will also be five chaperones that will each drive a van or a car. Each
van seats 7 persons, including the driver. Each car seats 5 persons,
including the driver. How many vans and cars will be needed?
Let v = number of vans and c = number of cars.
Drivers
Persons
v
7 v
+
+
c
5 c
= 5
= 31
Solve using substitution.
Step 1: Write an equation containing only one variable.
v+c=5
Solve the first equation for c.
c = –v + 5
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Solving Systems Using
Substitution
ALGEBRA 1 LESSON 9-2
(continued)
Step 2: Write and solve an equation containing the variable v.
7v + 5c = 31
7v + 5(–v + 5) = 31
7v – 5v + 25 = 31
2v + 25 = 31
2v = 6
v =3
Substitute –v + 5 for c in the second
equation.
Solve for v.
Step 3: Solve for c in either equation.
3+c=5
c=2
Substitute 3 for v in the first equation.
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Solving Systems Using
Substitution
ALGEBRA 1 LESSON 9-2
(continued)
Three vans and two cars are needed to transport 31 persons.
Check: Is the answer reasonable? Three vans each transporting 7
persons is 3(7), of 21 persons. Two cars each transporting 5 persons is
2(5), or 10 persons. The total number of persons transported by vans
and cars is 21 + 10, or 31. The answer is correct.
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Solving Systems Using
Substitution
Solve each system using
substitution.
ALGEBRA 1 LESSON 9-2
1. 5x + 4y = 5
2. 3x + y = 4
3. 6m – 2n = 7
y = 5x
2x – y = 6
3m + n = 4
(0.2, 1)
(2, 2)
(1.25, 0.25)
9-2