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Solve for the specified variable. See Example 8. …

01:39

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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 Problem 105 Problem 106 Problem 107 Problem 108 Problem 109 Problem 110 Problem 111 Problem 112 Problem 113 Problem 114 Problem 115 Problem 116 Problem 117 Problem 118 Problem 119 Problem 120

Problem 35 Medium Difficulty

Solve for the specified variable. See Example 8.
$P=a+b+c$ for $b$

Answer

$$b=P-a-c$$

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Algebra

Elementary and Intermediate Algebra

Chapter 2

Equations, Inequalities, and Problem Solving

Section 4

Formulas

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Algebra Topics That are Reviewed at the Start of the Semester

Equations and Inequalities

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Video Transcript

in this problem were asked to take the equation. P equals a plus B plus C, and we're going to solve it for be. In other words, we want to isolate me by itself, meaning everything else on this in the equation that's on the same side as B. We need to move to the other side. Well, if you know this on the right hand side of our equation, the first thing that's happening is we're adding a to B so we can move the aide to the other side by subtracting a Well, if you know this p minus a well, those aren't like terms, so they're going to stay as it's so in the left hand side, we're gonna have P minus A is equal to well, a minus a zero. Those cancel out so we're simply left would be plus C Well, B is still not by itself, however, be is getting added by sea. So to move the sea to the other side of the equation, we're simply going to subtract, See well, if you know this on the left hand side will have p minus a minus. C. Well, none of those air like terms, so we can simply write this as p minus a minus. C is equal to Well, we have B and then we have C minus C. But that's going to cancel each other out. And now we've got him be all by itself. So what we've found is P minus a minus. C is equal to be.

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Alan S. Tussy, R. David Gustafson

Elementary and Intermediate Algebra

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