INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith

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### Solving by Substituting

• To find exact solutions, use algebraic methods like the substitution method.
• Use substitution when at least one of the coefficients in the system is 1 or –1. Solve for this variable in terms of the other one. Then substitute that expression into the other equation.
• If you eventually get an equation that is always true, then the system has an infinite number of solutions.
• If you eventually get an equation that is never true, then the system has no solution.

### Solving by Substituting

j = 6 − 11k3j − 7k = 92
• 3(6 − 11k) − 7k = 92
• 18 − 33k − 7k = 92
• − 40k = 74
• k = − [74/( − 40)]
k = − [37/20]
5m + 6n = − 299m − n = 6
• − n = 6 − 9m
• n = − 6 + 9m
• 5m + 6( − 6 − 9m) = − 29
• 5m − 36 − 54m = − 29
• − 49m = 7
• m = - [1/7]
• 9 ( − [1/7] ) − n = 6
• − [9/7] − n = 6
• − n = 7[2/7]
n = − 7[2/7]
m = − [1/7]
8a + 2b = 104a − 6b = − 23
• Let's start by solving for b in the first equation: 8a + 2b = 10
• 2b = 10 − 8a
• b = 5 − 4a
• Plug b into the other equation 4a − 6(5 − 4a) = − 23
• 4a − 30 + 24a = − 23
• 28a = 7
• a = [7/28] = [1/4]
• 4g( [1/4] ) − 6b = − 23
• 1 − 6b = − 23
• − 6b = − 24
• b = 4
a = [1/4]
b=4
x = 4y + 95x + 3y = 14
• 5(4y + 9) + 3y = 14
• 20y + 45 + 3y = 14
• 23y + 45 = 14
• 23y = − 31
• y = − [31/23]
• x = 4y + 9x = 4( − [31/23] ) + 9
• x = − [124/23] + 9
• x = − 115[9/23] + 9
x = - 106[9/23]
x = 5y + 217x − 3y = 16
• 7(5y + 21) − 3y = 16
• 35y + 147 − 3y = 16
• 32y + 147 = 16
• 32y = − 131
• y = − [131/32]
• x = 5( − [131/32] ) + 21
• x = − 20[15/32] + 21
x = [17/32]
x = 4y − 75x − 3y = 15
• 5(4y − 7) − 3y = 15
• 20y − 35 − 3y = 15
• 17y − 35 = 15
• 17y = 50
• y = [50/17] = 2[16/17]
• x = 4( [50/17] ) − 7
• x = [200/17] − 7
x = 4[13/17]
x − 3y = 62x − 5y = 10
• x = 6 + 3y
• 2(6 + 3y) − 5y = 10
• 12 + 6y − 5y = 10
• 12 + y = 10
• y = − 2
• x = 6 + 3yx = 6 + 3( − 2)
• x = 6 − 6
x = 0
5x − 6y = 12x − 2y = 8
• x = 8 + 2y
• 5(8 + 2y) − 6y = 12
• 40 + 10y − 6y = 12
• 40 + 4y = 12
• 4y = − 28
• y = − 7
• x = 8 + 2yx = 8 + 2( − 7)
• x = 8 − 14
x = − 6
24x − 4y = 603x − y = 12
• − y = − 3x + 12
• y = 3x − 12
• 24x − 4(3x − 12) = 60
• 24x − 12x + 48 = 60
• 12x + 48 = 60
• 12x = 12
• x = 2
• y = 3x − 12y = 3(2) − 12
• y = 6 − 12
y = − 6
8x − 6y = 143x + 3y = 9
• 3x + 3y = 9
• 3x = 9 − 3y
• x = [(9 − 3y)/3]
• x = 3 − y
• 8x − 6y = 148(3 − y) − 6y = 14
• 24 − 8y − 6y = 14
• 24 − 14y = 14
• − 14y = − 10
• y = [( − 10)/( − 14)] = [5/7]
• x = 3 − yx = 3 − [5/7]
x = 2[2/7]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Solving by Substituting

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Substitution 0:09
• Example
• Number of Solutions 2:47
• Infinite Solutions
• No Solutions
• Example 1: Solve by Substitution 5:44
• Example 2: Solve by Substitution 10:01
• Example 3: Solve by Substitution 15:17
• Example 4: Solve by Substitution 19:41