Problem

Solve the system by either the substitution or th…

02:01

Problem 48 Hard Difficulty

Solve the system by either the substitution or the elimination method.
$$
\left\{\begin{array}{l}
{3 a+4 b=36} \\
{6 a-2 b=-21}
\end{array}\right.
$$

Answer

(-0.4,9.3)

Related Courses

Algebra

Elementary and Intermediate Algebra

Chapter 4

Systems of Linear Equations and Inequalities

Section 3

Solving Systems of Equations by Elimination (Addition)

Discussion

You must be signed in to discuss.

Top Algebra Educators

Watch More Solved Questions in Chapter 4

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Problem 87

Problem 88

Problem 89

Problem 90

Problem 91

Problem 92

Problem 93

Problem 94

Problem 95

Problem 96

Problem 97

Problem 98

Problem 99

Problem 100

Problem 101

Problem 102

Problem 103

Problem 104

Video Transcript

the way this problem is set up, it would be good to use the elimination method. And so I'm gonna multiply the second equation by negative two. Or by positive, too, so that I will have opposite be terms some leaving the first equation alone. And it is still three. A plus four b equals 36 on the second equation, multiplied by two is 12 a minus four b equals negative 42. And now I'm going to add those together and the eight terms edge of 15 a thebe. Terms are eliminated and the Constance add to negative six. Now we divide both sides of the equation by 15 and we have negative 6 15 THS. That reduces to negative 2/5. Now we want to find the value of B. So I'm going to take my A value and substituted into one of the two equations. I'll call these equations one and two. So go ahead and substituted into Equation one. So I have three times negative. 2/5 plus four times be equals 36 three times negative. 2/5 is negative. 6/5 plus four b equals 36. I'm going to add 6/5 to both sides and I have four b equals. Let's see how What's a good way to do this? I'm going to change my 36 into 185th CE and add the 6/5 to it. So 186 5th and then divide both sides by four and we get 186 20th. Now that does reduce. Let's see you do quick little calculation here. Let's see what that reduces too. 93 10th. I have a feeling like maybe I made a mistake, but we're about to check and find out. Okay, well, so now we want to check our answer by substituting both of these numbers into equation, too. I'll do that over here. So six times a would be six times negative 2/5 minus two times be would be minus two times 93 10th equals negative 21. So that would be negative. 12 iss minus 93. This equals negative 21. Let's see what negative 12 minus 93 is You're 12 minus 93 is negative. 105 So I have negative 10 5/5 equals negative. 21. And believe it or not. That is actually correct. Strange answer. Not so user friendly, but it was correct. So our order pair a comma be is negative. 2/5 comma 93 fits. 93 10th. Excuse me. Well, uh

We have video lessons for 94.35% of the questions in this textbook