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Algebra 1 Solving Formulas

Name: Algebra 1: Solving Formulas
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Section 2: Linear Equations: Lecture 3 | 12:02 min

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Lecture Comments (11)

	1 answer Last reply by: Professor Eric Smith Mon Dec 30, 2019 10:39 AM Post by Yousef Jad on December 21, 2019 X square - 12X+a=(X+b) square find the value of a and the value of b
	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?
	4 answers Last reply by: Rain Zhang Sun Jun 16, 2019 2:05 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 âˆ’ 5 ) = 2 âˆ’ ( 6 âˆ’ 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.

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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 + 9

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

n = − 7.5

5x + 8 = 13x − 9

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

x = [17/8]

7( 3x + 5 ) = 2( 12x − 8 ) + 14

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

x = [37/3]

2( 13x − 5 ) = 2 − ( 6 − 5x )

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]

x = [2/3]

2( 8y + 11 ) / 4 = 6( y − 5 )

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]

y = [71/4]

4( 6s + 2 ) = 8( 3 − 2s )

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]

s = [2/5]

8n − 14 = 5n − 20

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

n = − 2

3(5f − 6) = 2(12f − 4) + 13

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)]

f = − [23/9]

[(5(4x − 8))/10] = 2(15x + 5)

[(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]

x = − [1/2]

4(4k − 8) = 4(3k − 6) − 18 − 2k

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]

k = − [5/3]

Solve for y:4x − 11y = 15

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

y = − [(15 − 4x)/11]

Solve for a:11a + 12b − 6 = 14

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

a = [(20 − 12b)/11]

Solve for t:s − 8st = 3t − 7

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))]

t = [( − 7 − s)/(( − 8s − 3))]

Solve for v:4u + 3uv = v − 1

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)]

v = [( − 1 − 4u)/(3u − 1)]

Solve for n:[(n + m)/(n − m)] = 4m

(n − m) ×[(n + m)/(n − m)] = 4m(n − m)
n + m = 4m(n − m)
n + m = 4mn − 4m²
n + m − 4mn = 4mn − 4m² − 4mn
n + m − 4mn = − 4²
n + m − 4mn − m = − 4m² − m
n − 4mn = − 4m² − m
n(1 − 4m) = − 4m² − m
[(n(1 − 4m))/((1 − 4m))] = [( − 4m² − m)/((1 − 4m))]

n = [( − 4m² − m)/((1 − 4m))]

Solve for c:
[(c − d)/2c] = 5d

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

c = 10cd + d

Solve for k:
[(j − k)/(j + k)] = 3j

(j + k)( [(j − k)/(j + k)] ) = 3j(j + k)
j - k = 3j(j + k)
j − k = 3j² + 3k
j − k − j = 3j² + 3k − j
− k = 3j² + 3k − j
k = − (3j² + 3k − j)

k = − 3j² − 3k + j

Solve for r:
16r − 12s = 48

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

r = [(12 + 3s)/4]

Solve for u:2u − 7uv = 5v + 9

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

u = [(5v + 9)/((2 − 7v))]

Solve for x:6xy + y = 3x − 2y

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))]

x = [( − 3y)/((6y − 3))]

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*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

Objectives

Solving Formulas

Example 1

Example 2

Example 3

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

Section 1: Properties of Real Numbers
	Basic Types of Numbers	30:41
	Operations on Numbers	19:26
	Order of Operations	12:06
	Properties of Real Numbers	18:52
Section 2: Linear Equations
	The Vocabulary of Linear Equations	12:22
	Solving Linear Equations in One Variable	28:52
	Solving Formulas	12:02
	Applications of Linear Equations	28:41
	Applications of Linear Equations, Motion & Mixtures	24:26
Section 3: Graphing
	Rectangular Coordinate System	22:55
	Slope & Graphing	27:58
	Linear Equations in Two Variables	20:36
Section 4: Functions
	Introduction to Functions	21:24
	Graphing Functions	16:12
Section 5: Systems of Linear Equations
	Systems of Linear Equations	25:54
	Solving a System Using Substitution	20:01
	Solving a System Using Elimination	19:40
	Applications of Systems of Equations	24:34
Section 6: Inequalities
	Solving Linear Inequalities in One Variable	17:13
	Compound Inequalities	16:13
	Solving Equations with Absolute Values	14:12
	Inequalities with Absolute Values	17:07
	Graphing Inequalities in Two Variables	15:33
	Systems of Inequalities	21:13
Section 7: Polynomials
	Integer Exponents	44:51
	Adding & Subtracting Polynomials	18:33
	Multiplying Polynomials	25:07
	Dividing Polynomials	44:56
Section 8: Factoring Polynomials
	Greatest Common Factor & Factor by Grouping	28:27
	Factoring Trinomials	21:44
	Factoring Trinomials Using the AC Method	30:09
	Special Factoring Techniques	30:14
Section 9: Quadratic Equations
	Solving Quadratic Equations by Factoring	23:38
	Solving Quadratic Equations	29:27
	Equations in Quadratic Form	16:47
	Quadratic Formulas & Applications	29:04
	Graphs of Quadratics	26:53
	Polynomial Inequalities	21:42
Section 10: Rational Equations
	Multiply & Divide Rational Expressions	26:41
	Adding & Subtracting Rational Expressions	20:24
	Complex Fractions	18:23
	Solving Rational Equations	16:24
	Rational Inequalities	18:54
	Applications of Rational Expressions	20:20
	Variation & Proportion	27:04
Section 11: Radical Equations
	Rational Exponents	14:32
	Simplify Rational Exponents	15:12
	Adding & Subtracting Radicals	17:22
	Multiply & Divide Radicals	19:24
	Solving Radical Equations	15:05
	Complex Numbers	29:16

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 × R × 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 × 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 I’m going to do is I’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 done0328

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 × 6.0422

I guess I better do some scratch work.0434

37 × 6 I have 42, 3 × 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 – 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 V’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 × 5V, on the left side those W’s would take care of each other.0569

I will be left with RV + Q = W × 5V.0585

In this point I do not have to deal with the fractions but notice we have not got those V’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 V’s, I’m going to make them into one.0627

The way I’m going to do this is I’m going to think of my distributive property but I’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 – 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 V’s be multiplied by 5W – R, that entire thing inside the parentheses.0682

We will divide both sides by that and then it should be all alone 5W – 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 V’s running around in there.0717

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

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Eric Smith

Solving Formulas

Slide Duration:

Table of Contents

Section 1: Properties of Real Numbers

Basic Types of Numbers

30m 41s

Intro