### Solving Formulas

- A formula is a type of equation that usually conveys some fundamental principle. They usually have many variables that stand in for unknown quantities.
- To solve a formula, we isolate the desired variable on one side of the equal sign. To do this we use the same process we saw with solving linear equations.
- With formulas, the solution may still contain many variables. It can still be substituted into the original to see if it creates a true statement.

### Solving Formulas

- 3n − 6 − 5n = 5n + 9 − 5n
- − 2n − 6 = 9
- − 2n − 6 + 6 = 9 + 6
- − 2n = 15
- [( − 2n)/( − 2)] = [15/( − 2)]

- 5x + 8 − 13x = 13x − 9 − 13x
- − 8x + 8 = − 9
- − 8x + 8 − 8 = − 9 − 8
- − 8x = − 17

- 21x + 35 = 24x − 16 + 14
- 21x + 35 = 24x − 2
- 21x + 35 − 24x = 24x − 2 − 24x
- − 3x + 35 = − 2
- − 3x + 35 − 35 = − 2 − 35
- − 3x = − 37

- 26x − 10 = 2 − 6 + 5x
- 26x − 10 = − 4 + 5x
- 26x − 10 − 5x = − 4 + 5x − 5x
- 21x − 10 = − 4
- 21x − 10 + 10 = − 4 + 10
- 21x = 14
- x = [14/21] = [2/3]

- 16y + 22 / 4 = 6y − 30
- 4( [(16y + 22)/4] ) = ( 6y − 30 )4
- 16y + 22 = 24y − 120
- 16y + 22 − 24y = 24y − 120 − 24y
- − 8y + 22 = − 120
- − 8y + 22 − 22 = − 120 − 22
- − 8y = − 142
- y = [142/8] = [71/4]

- 24s + 8 = 24 − 16s
- 24s + 8 + 16s = 24 − 16s + 16s
- 40s + 8 = 24
- 40s + 8 − 8 = 24 − 8
- 40s = 16
- s = [16/40] = [2/5]

- 8n − 14 − 5n = 5n − 20 − 5n
- 3n − 14 = − 20
- 3n − 14 + 14 = − 20 + 14
- 3n = − 6

- 15f − 18 = 24f − 8 + 13
- 15f − 18 = 24f + 5
- 15f − 18 − 24f = 24f + 5 − 24f
- − 9f − 18 = 5
- − 9f − 18 + 18 = 5 + 18
- − 9f = 23
- [( − 9f)/( − 9)] = [23/( − 9)]

- [(20x − 40)/10] = 30x + 10
- 10( [(20x − 40)/10] ) = (30x + 10)10
- 20x − 40 = 300x + 100
- 20x − 40 − 300x = 300x + 100 − 300x
- − 280x − 40 = 100
- − 280x − 40 + 40 = 100 + 40
- − 280x = 140
- x = − [140/280] = − [1/2]

- 16k − 32 = 12k − 24 − 18 − 2k
- 16k − 32 = 10k − 42
- 16k − 32 − 10k = 10k − 42 − 10k
- 6k − 32 = − 42
- 6k − 32 + 32 = − 42 + 32
- 6k = − 10
- k = − [10/6]

- 4x − 11y − 4x = 15 − 4x
- − 11y = 15 − 4x
- [( − 11y)/( − 11)] = [(15 − 4x)/( − 11)]

- 11a + 12b − 6 + 6 = 14 + 6
- 11a + 12b = 20
- 11a + 12b − 12b = 20 − 12b
- 11a = 20 − 12b
- [11a/11] = [(20 − 12b)/11]

- s − 8st − 3t = 3t − 7 − 3t
- s − 8st − 3t = − 7
- s − 8st − 3t − s = − 7 − s
- − 8st − 3t = − 7 − s
- t( − 8s − 3) = − 7 − s
- [(t( − 8s − 3))/(( − 8s − 3))] = [( − 7 − s)/(( − 8s − 3))]

- 4u + 3uv − v = v − 1 − v
- 4u + 3uv − v = − 1
- 4u + 3uv − v − 4u = − 1 − 4u
- 3uv − v = − 1 − 4u
- v(3u − 1) = − 1 − 4u
- [(v(3u − 1))/((3u − 1))] = [( − 1 − 4u)/(3u − 1)]

- (n − m) ×[(n + m)/(n − m)] = 4m(n − m)
- n + m = 4m(n − m)
- n + m = 4mn − 4m
^{2} - n + m − 4mn = 4mn − 4m
^{2}− 4mn - n + m − 4mn = − 4
^{2} - n + m − 4mn − m = − 4m
^{2}− m - n − 4mn = − 4m
^{2}− m - n(1 − 4m) = − 4m
^{2}− m - [(n(1 − 4m))/((1 − 4m))] = [( − 4m
^{2}− m)/((1 − 4m))]

^{2}− m)/((1 − 4m))]

[(c − d)/2c] = 5d

- 2c( [(c − d)/2c] ) = 5d(2c)
- c − d = 5d(2c)
- c − d = 10dc
- c − d + d = 10cd + d

[(j − k)/(j + k)] = 3j

- (j + k)( [(j − k)/(j + k)] ) = 3j(j + k)
- j - k = 3j(j + k)
- j − k = 3j
^{2}+ 3k - j − k − j = 3j
^{2}+ 3k − j - − k = 3j
^{2}+ 3k − j - k = − (3j
^{2}+ 3k − j)

^{2}− 3k + j

16r − 12s = 48

- 16r − 12s + 12s = 48 + 12s
- 16r = 48 + 12s
- [16r/16] = [(48 + 12s)/16]
- r = [(48 + 12s)/16] = [(12 + 3s)/4]

- u(2 − 7v) = 5v + 9
- [(u(2 − 7v))/((2 − 7v))] = [(5v + 9)/((2 − 7v))]

- 6xy + y − 3x = 3x − 2y − 3x
- 6xy + y − 3x = − 2y
- 6xy + y − 3x − y = − 2y − y
- 6xy − 3x = − 2y − y
- 6xy − 3x = − 3y
- x(6y − 3) = − 3y
- [(x(6y − 3))/((6y − 3))] = [( − 3y)/((6y − 3))]

*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.

Answer

### Solving Formulas

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
- Objectives 0:06
- Solving Formulas 0:18
- Formulas
- Use the Same Properties as Solving Linear Equations
- Addition Property of Equality
- Multiplication Property of Equality
- Steps to Solve Formulas
- Example 1 3:56
- Example 2 6:09
- Example 3 8:39

### Algebra 1 Online Course

### Transcription: Solving Formulas

*Welcome back to www.educator.com.*0000

*In this lesson we will take a look at solving formulas and that is our only objective for our list of things to do.*0002

*But one thing I want you to know while going through this one is how similar formulas to solve the equations.*0011

*If you are curious what exactly is a formula and how they differ from those linear equations that we saw earlier.*0020

*A formula is a type of the equation and usually conveys some sort of fundamental principle.*0025

*Below, I have examples of all kinds of different formulas, D = RT and I = P Ã— R Ã— T.*0031

*What these represents are usually something else.*0039

*For example, D = RT we will look at that a little bit later on because it stands for distance = rate Ã— time.*0042

*I = PRT, that is an interesting formula and it stands for interest = principal, rate, time.*0058

*It is not that these equations are too unusual but they actually have lots of good applications behind them.*0075

*One thing that is a little bit scarier with these formulas is you will notice how they have a lot more variables than what we saw earlier.*0081

*They have a lot of L, W and other could be multiple variables for just a single formula.*0089

*You know we have those multiple variables present in there and usually we are only interested in solving for a single variable in the entire thing.*0097

*The way we go about solving for that variable is we use exactly the same tools that we used for solving several linear equations.*0105

*You have the addition property for equality and you have the multiplication property for equality.*0114

*Both of these say exactly the same thing that they did before.*0120

*One, you can add the same amount to both sides of an equation, keep it nice and balanced.*0124

*And that you can multiply both sides of an equation by the same value and again keep it nice and balanced.*0129

*The only difference here is that we will be usually adding and subtracting or multiplying, dividing by a variable of some sort.*0135

*One thing that we usually thrown in as an assumption is that when we do multiply by something then we make sure that it is not 0.*0142

*If I do multiply it by a variable, multiply by both sides by an X, then I will make the assumption that X is not 0.*0151

*Just to make sure I do not violate the multiplication property of 0.*0158

*You have lots of variables and the very first thing is probably identify what variable you are solving for.*0167

*I usually like to underline it, highlight it in some sort of way just to keep my eye on it.*0174

*If you have multiple copies of this variable, try to work on getting them together so that you can eventually isolate it.*0180

*Some of these formulas do involve some fractions.*0186

*Use the common denominator techniques so you can clear them out and not have to worry about them.*0189

*Simplify each side of your equation as much as possible then isolate that variable that you are looking for.*0196

*Try and get rid of all the rest of the things that are around you by using the addition and multiplication property.*0203

*You can check to make sure that your solution works out by taking your solution for that variable and playing it back into the original.*0212

*When you do this for your formula, what you often see is that a lot of stuff will cancel out and you will be left with a very simple statement.*0221

*Usually we do not worry about that one too much unless we have some more numbers present in there but you can do it.*0229

*Let us actually look at a formula and watch the solving process.*0238

*You will see that it actually flows pretty easily.*0242

*In this first one I have P=2L + 2W.*0245

*This formula stands for the perimeter of a rectangle.*0248

*It goes through and adds up both of the lengths and both of the widths.*0251

*What we want to do with this one is we want to solve it for L.*0257

*Let us identify where L is in our formula right there.*0261

*What Iâ€™m going to do is Iâ€™m trying to get rid of everything else around it.*0265

*The 2W is not part of what I'm interested in so I will subtract that from both sides.*0271

*P and W are not like terms so even though it will end up on the other side of the equation, I will not be able to combine them any further.*0280

*Now you have 2 - 2 or P - 2W = 2L, continuing on.*0291

*I almost have that L completely isolated, let us get rid of that 2 by dividing.*0298

*Notice how we are dividing the entire left side of this equation by 2 so that we can get that L isolated.*0310

*What I have here is (P -2W)/2 = L.*0322

*One weird part about solving a formula is sometimes it is tough to tell when you are done *0328

*because usually with an equation when you are done you will have a number like L = 5 or L = 2.*0334

*Since you have many variables in these formulas that when we get to the end what we developed is another formula right here.*0340

*It is just in a different order or different way of looking at things.*0348

*This formula that we have created could be useful if we were looking for L many times in a row and we had information about the perimeter and W.*0351

*It is solved, it is a good the way that it is and the way we know it is solved is because L is completely isolated.*0362

*Let us look at another one, in this one we want to solve for R.*0371

*I have Q= 76R + 37, a lots of odd ball numbers but let us go ahead and underline what we would be looking for.*0375

*We want to know about that R.*0382

*I think we can get rid of a few fractions and we would have to multiply everything through by 6.*0387

*Let me do that first, I will multiply the left side by 6, multiply the right side by 6.*0395

*Q = 76R + 37, on the right side we better distribute that 6 and then we will see that it actually does take care of our fraction like it should.*0403

*6Q = 7R, I know I have to take care of 37 Ã— 6.*0422

*I guess I better do some scratch work.*0434

*37 Ã— 6 I have 42, 3 Ã— 6 is 18 + 4 = 22, a lot of 2.*0436

*6Q = 7R + 222, I do not have to deal with any more fractions at this point so let us continue isolating the R and get it all by itself.*0452

*I will subtract 222 from both sides, awesome.*0463

*One final step to get R all by itself, we will divide it by 7.*0482

*(6Q â€“ 222)/7 = R and I will consider this one as solved.*0490

*The reason why we can consider this one done is because we have isolated R completely.*0503

*Even though we do have a Q still floating around in there, it is solved.*0508

*Do not get too comfortable with that, it is not equal to just a single number, very nice.*0513

*This one is a little different, we want to solve the following for V.*0521

*The reason why this one is a little bit different is I have a V over here but I also have another V sitting over there.*0525

*When you have more than one copy of the variable like this, you have to work on getting them together before you can get into the isolating process.*0534

*Let us take care of our fractions and see if we can actually get those Vâ€™s together and work on isolating it.*0543

*We only isolate one of them then it is not solved, we still have a V in there.*0551

*To take care of our fraction I will multiply both sides of this one by W, it is my common denominator.*0556

*(RV + Q) / W = W Ã— 5V, on the left side those Wâ€™s would take care of each other.*0569

*I will be left with RV + Q = W Ã— 5V.*0585

*In this point I do not have to deal with the fractions but notice we have not got those Vâ€™s any closer together so let us keep working on that.*0594

*If I'm going to get them together, I at least better get them on the same side of the equation.*0602

*I'm going to subtract say an RV from both sides right.*0609

*Now comes the fun part, I still have two Vâ€™s, Iâ€™m going to make them into one.*0627

*The way Iâ€™m going to do this is Iâ€™m going to think of my distributive property but Iâ€™m going to think of it in the other direction.*0633

*If both of these have a V then I will pull it out front and I will be left with W5 â€“ R.*0641

*This step is a little bit tough when you see it at first but notice how it does work is that I'm taking both of these and moving them out front into a single V.*0653

*I'm sure if it is valid, go ahead and take the V and put it back in using our distributive property and you will see that you be right back at this step.*0664

*It is valid.*0673

*The important part of why we are using it though is now we only have a single V and we can work further by isolating it.*0675

*How do we get it all by itself, these Vâ€™s be multiplied by 5W â€“ R, that entire thing inside the parentheses.*0682

*We will divide both sides by that and then it should be all alone 5W â€“ R = V.*0690

*I literally just took this entire thing right here and divided it on both sides.*0702

*To your left, now I can call this one done because V is completely isolated, it is all alone.*0709

*There is no other Vâ€™s running around in there.*0717

*I have worked hard to get them together and I know it is completely solved.*0719

1 answer

Last reply by: Professor Eric Smith

Tue Apr 22, 2014 8:11 PM

Post by Maria Camila Bernal on April 16, 2014

One of the practice questions says:

2(13x-5) = 2-(6-5x)

26x-10 = -4+5x (How is that the -5x above turned to a +5x?)

1 answer

Last reply by: Mohamed Elnaklawi

Tue Mar 18, 2014 1:35 PM

Post by Mohamed Elnaklawi on March 18, 2014

in example two, why are you done with the problem if you still have a FRACTION in the end?

1 answer

Last reply by: edder villegas

Sun May 18, 2014 10:51 PM

Post by Paul Cassidy on February 23, 2014

does anyone know if these questions go to those who monitor this web site or know how to let them know that there test questions are not

correct. Here is another one. Equation shown: Q. 2( 13x Ã¢Ë†â€™ 5 ) = 2 Ã¢Ë†â€™ ( 6 Ã¢Ë†â€™ 5x ) the answer that is give is 2/3 that is not correct. the answer is 2/7 and proofs out correctly at that. the error is in the fact that they show -4+10 as a positive 14 instead of a 6. I love this site for the lectures but am continuously finding errors with the practice questions. I'm concerned for those home schooling as they need the right answers as well as the fact that they are paying for it.