In this unit, we will look at solving equations and inequalities. We'll consider how many solutions or answers exist for a given problem, and we'll work on writing expressions with variables. By the end of this unit, I hope you'll have a greater understanding of solving for unknowns, and how to be confident in knowing you have the right answer, and checking your work. Alright, let's get started.
Let's start out by looking at a scale and some weights. A scale can determine if the weights on the left side and the right side are equal, or if one side is heavier than the other. I've added some weights to the scale. The left side of the scale has a total of 8 pounds. The right side of the scale also has a total of 8 pounds. We can write an equation that represents the weight of these scales on each side. I have 6 pounds and 2 pounds is equal to 5 pounds and 3 pounds. We know they both make 8. Now, let's do something crazy. Let's add a ton of weight to both sides, like 100 pounds. Think about what'd happen to the scale, and try and answer this question. If 100 pounds of weight is added to both sides, which side weighs the most? The left, the right, or they weigh the same.
We know each side weighed 8 pounds each. Then we added 100 pounds to both sides, so we have 108 pounds on both sides, so they must weigh the same. Not only can I add 100 pounds to both sides of the scale, I could also think about removing 100 pounds from both sides of the scale. If I remove 100 pounds from both sides of the scale, then I'd be back at 8 pounds. 8 pounds on the left and 8 pounds on the right. This is one of the most important concepts in algebra. We want to be able to add a number to both sides of an equation and subtract a number from both sides of an equation. It's the Properties of Equality. If you chose the last answer, nice work.
So, we're back to our 8 pounds on both sides of our scale. They weigh the same. What if we removed half of the weight from both sides? What do you think would happen? Would the left weigh more, the right weigh more, or would they weigh the same?
If you said they'd weigh the same, great work. Dividing by 2 on both sides gives me 4 pounds. Notice that when I've been solving these problems, I keep my equals sign lined up in the center. And after every time I perform an operation on both sides, I draw a line. This helps me organize my work and my thoughts. Make sure that you can do this or organize your thoughts in some manner when you go to solve problems.
It turns out that when we add, subtract, multiply or divide, we can do it to both sides of the scale. This also works for equations. These properties have a special name, and they're called the properties of equality. Let's use these properties of equality to solve a more challenging problem. We know equations work like scales, so when we write an equation, we know the side on the left, or that value, matches the value on the right. Equations work just like scales. We're saying that this number, or this value on the left, is the same as this number, or the value on the right. So, using what you know about the properties of equality, what do you think the unknown weight m is? Can you find it's value?
If you got 12 pounds, nice work. When approaching this problem, we can think about combining the like terms or the like weights on one side. I can combine the m's as like terms, so I'll have 3 m on this side and 2 m on this side. I can also combine the 200 and the 50. These are both like terms since they're just numbers. 200 and 38 are also like terms. I can add those weights together. So, I have 238 plus 3 m is equal to 250 plus 2 m. I have 238 plus 3 m is equal to 250 plus 2 m. We know we can remove weights from both sides so we can take off a 2 m, from here and here. I remove 2 m from the weights and 2 m from the weights over here. So, I'm just left with 238 plus m equals 250. I want to think of about getting m alone. I want to isolate this variable. I want to figure out how much this weighs. So, I should remove 238 pounds from both sides. We could easily think about removing 200 and 200, and then removing 38, and removing 38 from 50. So, we know one way m is equal to 12 pounds. When looking back at our work, we should always remember that we're trying to isolate m. You want to remove the variable from both sides and remove constants from both sides. Eventually, we want the variable on one side and constants on the other. We could have done this in a reverse order and maybe you did it a little bit differently. That's okay. We just want to make sure we are performing the same steps and reasoning through our results. When solving the equation, we always want to start off by combining like terms. This helps us know what to remove from both sides of the equation. We could have removed the numbers from both sides of the equation first. That's okay. We just need to be consistent and always remove or perform the same operation on both sides of the equation at each step.
Here's another equation, and this time it looks harder. We have this negative 4 sitting outside these parentheses. Well, we only need to multiply this negative 4 to the terms inside the parentheses. This is called distribution. When distributing the negative 4, I do negative 4 times x, and negative 4 times negative 4. You want to make sure that you carry this sign in front of the second term with it. I've also been careful to write each statement out. I don't want to make an error with negative signs. Finish this problem now. What's the solution? What does x equal?
When solving the equation, we want to multiply these terms, and then combine like terms on the same sides of the equation. Negative 4 times x is negative 4x, and negative 4 times negative 4 is positive 16. This is one of the most common mistakes in algebra, forgetting to multiply a negative by another negative when distributing. Be careful with that. Next we combine like terms, so, negative 4x positive 2x make negative 2x. Here I'm going to try and keep my variable positive. I'm going to add 6x to both sides. Adding 6x to both sides leaves me with 8 equals 4x plus 16. Subtracting 16 from both sides gives me negative 8 equals 4x. And then my last step is just to divide by the coefficient, or the number in front of the variable, which is 4. This leaves me with x equals negative 2. Great work if you got negative 2.
When we solved this equation, we got x equals negative 2. What's awesome about math is that we can know we are right. We can substitute negative 2 in for x and check to see if the equation is true. It's like if we are seeing if the scales are still balanced. To check to see if we are right, we take the value negative substituting values, we don't want to make mistakes with negative signs. Now that we've plugged in negative 2 into our equation, we know that the left side should be equal to the right side. The equation should be balanced. So, here's your skills check. What number should go into these boxes to check the answer? Remember, the last boxes should be equal if we've done it right.
Negative 6 times negative 2 is positive 12. Negative 2 minus 4 makes negative 6. And, I know I have multiplication here, negative 4 times negative 6, because I had multiplication up here, with this larger parentheses. 12 plus 8 makes 20, and negative 4 times negative 6 makes positive 24. And whoa, look at that. Our answers actually match. 20 does equal 20. So, we know x equals negative 2 was definitely correct.
We've solved the equations with integers, now let's try something more difficult. Here's an equation with fractions. Let's use our knowledge of fractions to help us out. When solving an equation involving fractions it's best to clear the denominators first. We can actually undo the division of these denominators by multiplying through by the least common denominator. So, here's your question, what is the least common denominator for these fractions? As a hint, you don't need to worry about the 4, since the 4 is already a factor of 8.
The least common denominator is going to be 40. I dont' need to worry about the and the 8 in the denominator. The only factors in common between 5 and 8 are 1. So that leaves 5 times 1 times 8, which is 40. That quiz is pretty tough, so don't worry if you didn't get it right on the first try.
Let's clear these fractions by multiplying through by 40. I'm going to multiply the left by 40 and the right by 40. I've multiplied this side by 40 and notice I didn't need to multiply these inside terms because eventually, this will get distributed to each of them. I've put parentheses around the right side because I needed to distribute the 40 to both these terms. 40 times a 4th is 40 4th or I can figure out 40 over 1, and then divide 40 and 4 both by 4. And I'll get 10 over 1, or 10. Now, I'm going to distribute or multiply this 10 into this inside parentheses. I have 10 times negative 2 5th y here, and I have 10 times 1, here. I can reduce a common factor of 5 in the numerator and the denominator, so I get negative 4 over 1 or negative 4y. And then, I have my 10 on the end. Notice how both times I reduced before I multiplied. Usually, it's easier to reduce common factors before you multiply. Try doing that on this side. If you carry out the distribution on the right side, what terms would you get down here? Be sure you enter a number and a variable in the first box and a number in the second box. You'll also want to include both of their signs. If the term is negative, be sure to type a negative sign in front of your answer. Good luck.
If I distribute the positive 40 to 3 8th y, I'll get 40 times 3 8th y. And if I distribute the 40 to the negative 17 8th, I'll get 40 times negative 17 8th. Be sure you carry the minus sign. Reducing a common factor of 8 gives us 5 here and So, I just have 15, and then y. On the right side, I have 5 times negative 17, which is negative 85. This is amazing if you got this right. If you remember how to work your fractions and reduce common factors, great work.
We know equations are like scales. They need to remain balanced. The left side needs to equal the right. There are some choices we can make along the way to help us with our math. Let's see what those are. I'm going to solve this problem two ways. When thinking about solving this equation, I'm going to combine my like terms on one side of the equation. So here, I'll have 3b plus 32 and then on the right side, I'll have 2b plus 47. I've listed my equation twice, so now that I can solve it two different ways. First, we can subtract 2b from both sides. 3b and negative 2b make positive 1b. I have 1 times b here, but I don't really need to write the 1. I can leave it off. The positive 32 comes straight down, positive 2b and negative 2b sum to 0, so I don't list anything here. And then, I have positive 47. In the end, b equals positive 15. But let's see another way to solve this. I subtracted 2b first, but what if I didn't subtract the 2b, what if I subtracted the 3b? Positive 3b and negative 3b makes 0, the 32 comes down, the 47 comes down and then 2b and negative 3b make negative 1b. And subtracting 47, we get negative 1b is equal to negative 15. So, I'm not interested in the negative value of b, I want to know the positive value of b. So, I know if negative b is negative 15, then positive b must be positive 15. that could also think about multiplying both sides by a negative 1, That would give us our result, b equals positive 15. So, when you're solving equations, it's best to look out for the variables. Do you want to keep your variable positive? Or do you want to make it negative? It's probably easier to work with a positive b variable here. Sometimes, we make errors with negative signs, so it's best to avoid those.
Let's finish this problem now. What's the answer to this equation? You can put your answer in this box
If you said 5, way to go. First, I'm going to add 4 y to both sides. So, I'll have 0 here, 10, 19 y, and negative 85, with my equal sign in the middle. The left is equal to the right. Adding down, I get 95 is equal to 19 y. And, negative 85 and 85 make 0. And for the last step, I divide both sides by 19, the coefficient of my variable. This seems like really tough multiplication. But, let's take a second and stop and think about it. Let's try not using a calculator for this. I know 20 times 5 is 100. Here, I have it 19 times y. Well, short of 100. So, 19 times 5 must be 95. So, this y has to be 5.
We've solved the equations with fractions, now let's see how we can use a similar approach to work with decimals. Here I have 2 and 3 10th times n plus 4 values as 10th and this as 100th. If I want to remove the decimals from my equation I would multiply by a number that moves the decimal point to the right. We can multiply both the left by a 100 and the right by a 100. If we remember back from scientific notation multiplying by 100 moves the decimal 2 places to the right. Now I have an equation just made up of integers. See if you can solve the rest of this problem. And, notice too about the last step that you'll do. Think about why that happens.
Now that we've seen how to clear decimals from an equation, let's see you put this to use. Here's a question. To clear the decimals from this equation, what number would you multiply both sides by?
We should multiply both sides by 10. I know I should multiply by 10 because I have 5 and 8 10ths and 2 and 1 10th. Tenth is the smallest place value I have. And I know multiplying by 10 moves the decimal one place to the right. So that will clear those decimals.
If we do multiply the left and the right side by 10, which of the following equations would we get? Choose your best guess. Good luck, and take your time. Think about what numbers you actually need to multiply by 10. One of these choices will be correct.
If I multiply the left side by 10, I get 58a. >> I notice 58a is on the left side in all my answer choices, >> So there's something else going on. >> Notice that I use brackets around the entire expression. That's because I already had parenthesis here. >> I only need to distribute the 10 to the 2 and 1 10th and the 2, because when we distribute the 10 will be applied to the 3a and the 2. >> And then of course on the end, I have 2 times 20. >> So we know this one must be correct. >> This was a really tricky quiz, so don't worry if you didn't get it right on the first try. >> Hopefully you learned something about decimals and the distributive property along the way.
Try solving this equation. >> You can put your answer in this box up here ... And remember to check your answer. That way, you'll know if you're right.
Here's another equation you can try out. >> Remember, you can clear the fractions to help yourself get started.
The solution to this problem is x equals 9 7th. Well, let's see how we got 9 want to do is get rid of the fractions. Because I don't want to have to add, subtract, multiply and divide fractions all the way through solving the problem. To clear the fractions out of an equation, I look at all the denominators, which I have 6, 2 and 3. And I have to find the least common denominator, because I want to multiply the entire equation through by the least common denominator. The least common denominator of 3, 2 and 6 is 6. So, I'm going to multiply the entire equation through by 6. I have to multiply 6 by each term that is in the equation. I'm going to write the steps out here. But once you get comfortable, you can feel free to skip this step. So, I have 6 times 1 3rd, and 1 3rd is being multiplied by 2 x minus 1. On the right side of the equation, I have 6 times negative 1 half x plus 6 times 7 6th. When I multiply 6 times 1 3rd, I get by 2 x minus 1. On the right side of the equation I have 6 times negative 1 half x. 6 over 2 is 3, so I have negative 3 x. When I multiply 6 times 7 6th, the 6th cancel out and I'm left with a 7. Once I clear the fractions from the equation, I need to distribute wherever needed. Here, I have 2 times 2 x minus 1, so I need 2 times 2 x, and I need 2 times negative 1. 2 times 2 x is 4 x, and 2 times negative 1 is negative 2. And I just write down the other side of my equation. Once I've cleared the grouping symbols and distributed where needed, I would combine like terms on each side of the equation. However, here, there's only one x term and one constant term on the left side, and one x term and one constant term on the right side. The next step is to move all my constants to one side of the equation. We learned that whatever you do to one side of the equation, you have to do it to the other side of the equation. So, if I add 3 x on this side, I have to add 3 x on this side. When I add 3 x to both sides, I have 4 x plus 3 x gives me 7 x minus 2 equals negative 3 x plus 3 x is 0. So, I just have 7. Now that I have the variable on one side of the equation, I need to move the constant to the other side of the equation. When I add 2 to both sides of the equation, I get 7 x on the left side. Negative 2 plus 2 is zero. And I get 7 plus 2 which equals 9 on the right side. So, I have 7 x equals 9. The last step is to divide both sides of the equation through by the coefficient of the variable. The coefficient of the variable is 7, so I divide both sides by 7. 7 over 7 is 1, so I have x equals 9 7th. Solving an equation with fractions in it is a little bit more difficult. But if we simply multiply through by the least common denominator, all the fractions dissappear and you're left with a simpler problem to solve.
Here's the second to last practice problem. >> Good luck.
Here's your last practice problem. >> Be careful with these fractions, especially on this side.
The solution to this equation is x equals 3 over 26. If you got the answer right, you are well on your way. Let's see how we got that answer. Once again, this is an equation that has fractions in it. Since I don't want to have to add and subtract, multiply and divide a lot of fractions, I can clear the fractions by multiplying through by the least common denominator. The denominator are 6, through by 6. I'm going to skip this step of showing the individual multiplication of the terms. When I multiply 6 times x plus 7 over 6, the six is cancelled out, and I'm left with x plus 7. When I multiply 6 times 1 half x, 6 times 1 half is 3, and I have to multiply that by x. When I multiply 6 times 2 least common denominator, I have to multiply the least common denominator by every term. And that's every term whether there's a fraction in the term or not. So, I also have to multiply 6 times 4x, which gives me 24x. Now that I've eliminated all the fractions from my equation, I can combine my terms. On the left side, I only have x plus 7. And on the right side, I have 3x plus 24x is going to give me 27x. And then, I have the plus 4. The next thing I want to do since I like to deal with positive coefficients is to subtract x from both sides. Even though there's not a coefficient in front of this x, we know that there's a 1 there. So, 1x minus 1x gives me 0 and the only thing I have left on the left side of the equation is 7. On the right side of the equation, I have equation, I want to isolate it by subtracting 4 from both sides of the equation. the equation by the coefficient of the variable, which is 26. When I do that, I know 26 over 26 is 1, so I have x on the right side of the equation and 3 over have a number equals x or x equals a number. It's the same thing. Congratulations on completing solving linear equations. Now that you've learned how to solve linear equations, let's do something a little bit different. Chris, what do you have up for us next?