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Contents

Problem Solving

In the last section, we learned how to solve equations. We're going to use this knowledge to help us solve some word problems. We're going to learn some strategies to help us solve these word problems. We're going to identify unknowns, draw diagrams, and try and create a plan to solve. By the end of this section, I hope you will gain some insights into writing expressions and solving equations. Let's get started.

number changes

Let's look at a problem involving changing a number. We decrese a number by 7, multiply the result by 3, and divide the final amount by 2. Let's get started by finding out what these words mean. Let's translate them into an expression. Notice that we have some unknown number, we're going to call that a. What do you think? What's the expression that could represent the operations we performed? Choose the best answer.

Number Changes

We first took the number and decreased it by 7, so a minus 7. We multiplied that result by 3, so I say 3 times the quantity a minus 7. Finally, we divide the amount by 2. Notice I put the entire quantity over the number 2. I need to divide this entire number by 2. You chose the last one, nice work. We know it's not the first one because we know a minus 7 times 3 is really a minus 21. And I didn't subtract 21 from my original number, so this one's out. Here I would have multiplied by 3 first. which we didn't do. so this one's out as well. the same is true for the third case.

More Number Changes

So, notice, this case, we were only able to write an expression. Now, I have the same problem but if added the sentence, the result is negative 18. Before we knew that the end result would be negative 18, we could only have an expression. But now that I have this extra piece of information, I can set this equal to negative 18. Now, we have an equation. We have an expression on the left and a number on the right. Now, you can use this equation to figure out the value of a. Try and find a and put your answer in the box. Good luck.

More Number Changes

To solve this equation, we want to undo the division by 2. We want to clear the fraction, so we multiply by 2. Notice this is one whole fraction. I can reduce the 2's that appear in the numerator and the denominator. I'm left with 3 times the quantity a minus 7. Now, we want to clear the grouping so we can distribute this positive 3 to each term. 3 times a is 3 a, and 3 times negative 7 is negative 21. Adding 21 to both sides, we get 3 a equals negative 15. So, for the solution, a is equal to negative 5. Remember, just like all the other equations, we can check our answer. I can plug in negative 5 for a, and this result should be negative 18. Negative 5 minus 7 is negative 12, 3 times negative 12 is negative 36. And at negative 36 divided by 2 is negative 18. So our answer, a equals negative 5, does check.

Challenge Question

Here's a challenge question. Divide a number by 6 and then increase the amount by 1 8th. The result is 1 4th. Do you think you can find the original number? Try your best. And I'm not going to post the solution for this one. So, work with others in the forum or with a friend.

Perimeter

Before we get to our next word problem, let's learn about perimeter. You might already know what this word means, but if you don't, I'm sure you'll be able to figure it out. For example, for this triangle, I know that the perimeter is 15 centimeters. For this rectangle, I have sides of 2 units in length, and a width of w. I don't know the width. The expression for this perimeter is 4 plus 2 w. And I don't know the units for this perimeter, so I'm going to write the word units. It could be meters, centimers, or even inches. So, what do you think perimeter means? Try and find the expressions for the perimeters for these two shapes. You can write your answers in the boxes.

Perimeter

You might have guessed that the perimeter is the distance around an object. So if I just add up the distances around an object, I can find the perimeter. For this one, I have 2 d's and, and 2 3's, so the perimeter is 2d plus 6. The d's have the same length, so we can add them together. They're like terms. The perimeter of this shape is 2a plus 3b. We don't know the length of these two sides, but we can add them together because they're both a, they have the same length. The same is true for the b's. Be careful that you don't write 5ab here. We can't combine these two terms. a is one number, and b is another number. I'm not multiplying, I'm really adding the distances around the object. If you got these two right, nice work.

Perimeter Equation

Let's use our idea of perimeter to solve this problem. The length of a rectangle is 6 feet more than its width, and the perimeter of that rectangle is 52 feet. What are the dimensions, length and width, of the rectangle? To start this problem, let's draw a diagram. Here's a rectangle, with a width and a length. Notice the opposite sides of the rectangle are the same. This is true for all rectangles. Before we answer this question, let's break our problem down into a smaller step. Let's start with an equation. So I want you to write an equation for the perimeter of the rectangle. Use p for the perimeter, l for the length, and w for the width. Write your equation in this box. Remember, equations have an equal sign, expressions don't.

Perimeter Equation

To find the perimeter of our rectangle, we simply add up the sides around the shape. So we have two l's, and two w's. You might have written this equation, which also is correct. We want to combine the two l's, however, because they're like terms. The same is true for the w's.

Perimeter Quiz

Here's the equation we found. I'm going to rewrite this equation by dividing by of the equation by 2, I need to be careful on this side. I need to take the 2 and divide it into each term, both the 2l and the 2w. I get p divided by 2 equals l plus w. This p divided by 2 might look funny. But we know dividing by a number is the same as multiplying by the reciprocal. So, I can rewrite this as p times 1 half. So, I have p times one half. This should connect to your drawing. I know if I go half way around the parameter I should only be at l plus w. Notice, it is l plus w, that's half way around the rectangle. We know the perimeter of the rectangle is 52 feet so I am going to write that here. The length of the rectangle is 6 feet more than the width. So, I can write the length being equal to w plus 6. The length is 6 more than the width. Now, we can make a substitution. Wherever I see the letter l, I can replace it with w plus half times 52 equals w plus 6 plus w. We made the substitution so we could work with one equation and one variable. Now, we can actually solve for this one variable. I'm going to leave that up to you. Enter the width here and the length here. And don't worry about writing units, I've got those for you.

Perimeter Quiz

any multiplication here, this is addition. I'm just adding w to its like term. Subtracting 6 from both sides, we get 20 equals 2w, then we divide both sides by in mind that we still need to find the length. Let's check in with our diagram. I can replace these w's with 10, so each side is 10 feet. Now, you could have reasoned your way and figured out the length of l since you know the perimeter has to add up to 52. Or you could be a little bit more clever. You could just say the length of the rectangle is 6 more than the width. So, to find the length, I just have to take the width and add 6, so the length needs to be 16. And you could, of course, check your answer by finding the perimeter. Add up all the sides and your perimeter should be 52. Hopefully, you use your equation skills and got that one right.

Ratios

Another type of word problem involves ratios. A ratio is a way of comparing one quantity to another. For example, the number of dimes to the number of quarters in my hand is two to three. This means that for every two dimes in my hand I have three quarters. So try and answer this question. How many coins are in my hand in total? Choose the best answer.

Ratios

So let's start off really simple. Let's try creating a table to help us out. If I started with two dimes, then I'd have to have three quarters. For ever two dimes there's three quarters, so five coins in total. If I add two more dimes, then I have to add three more quarters. So I'd get ten coins in total. The ration is a way of representing this constant increase. For every increase of two in dimes I increase, and three in quarters. I could also have 6 dimes and 9 quarters, which is 15 coins in total. But I don't know which case I'm in. I could have had any number of coins. So, it's actually impossible to know. And in fact, I had 10 coins in my hand. For every two dimes, I have three quarters. Two more dimes, three more quarters. This ratio is constant. I know that last question wasn't obvious, so don't worry if you didn't get it right. I just wanted to get you thinking about ratios.

Ratios with Coins

Let's investigate this ratio more closely. >> A ratio is a comparison of the quantity of two objects, like the number of dimes to the number of quarters. >> So, let's find the ratio in each of these rows. >> Take the number of dimes and divide it by the number of quarters. >> Enter your simplified fractions here.

Ratios with Coins

For the first row, I just have 2 dimes divided by 3 quarters, or 2 3rd, 2 to numerator and the denominator so, I'm left with 2 3rd. >> And, for 6 to 9 I'm also left with 2 3rd. >> I can remove a 3 as the common factor. >> Notice that in every case the ratio stayed the same. >> As we gain 2 coins here, we always gain 3 here. >> That's why the ratio stays constant, 2 to 3. >> Even though our amounts of dimes and quarters is increasing, the ratio, or that fraction comparing their quantities, is always the same. >> Pretty neat.

Writing Ratios

We can write ratios in many different ways. >> For example, if I want to write the number of dimes to the numbers of quarters, I could write it as 2 to using a fraction bar. >> All of these express a ratio. >> You might see this symbol later. >> I'm just using it to represent a number. >> This pound sign or hash tag is representing the word number. >> I just wanted to save some room when I was writing.

Practice Writing Ratios

So, how can you write these ratios? The number of quarters to the number of dimes, and the total number of dimes to the total number of coins. >> Write each ratio in three different ways in these boxes. >> Be sure you use the correct form.

Practice Writing Ratios

The number of dimes to the number of quarters is 2 to 3. So, if I reverse that ratio, the number of quarters to the number of dimes must be 3 to 2. I can write that with words, with a colon, or as a fraction. The second one was really tricky. I don't know the total number of dimes and I don't know the total number of coins So, what can I do? Well, let's try some cases. If I started out with two dimes, I'd have to have 3 quarters, which would be 5 coins in total. But I didn't necessarily have to start with 2 dimes. What if I started with 4 dimes? I doubled the number of dimes that I started with. So, I also need to double my number of quarters. So, 4 dimes would have 6 quarters, which would be 10 coins in total. Notice that 2 divides into 4 and 10. 2 is the common factor, so I can simplify this fraction. And even if I had 6 dimes, I would have to have 9 quarters, or 15 coins in total. And this fraction still reduces to 2 5ths. I removed a 3 as the common factor. That's pretty incredible. I didn't even know the total number of coins, or the total number of dimes, but I knew the ratio. It was 2 to 5. If you didn't get the second part of the problem right, that's okay. I just wanted to get you thinking about ratios.

Proportions

Let's keep our same ratio and this time look at using proportion and a table. So, the number of dimes to the number of quarters is still two to three. But this time I have 10 dimes in my hand. How many quarters am I holding? We could solve this problem by extending our table. I could keep adding two dimes and three dimes in each case. So I'd have 8 dimes, 12 quarters, and 20 coins in total. And then 10 dimes, 15 quarters, and 25 in total. So if I add 10 dimes, then that means I have 15 quarters. But, maybe there was a different way to see this. If I take two dimes and if I multiply it by 5, then I also have to multiply my quarters by 5. The same is true for the number of coins. I would have 5 times 5, which is 25. Ratios often have a multiplier. We take our original ratio and multiply it by a number. So to get from 2 dimes to 10 dimes, we multiply by 5. So to get a 3 quarters to 15 quarters we also multiply by 5. The number of the objects continue to increase, so I can have an n multiplier, any number, and this would be 2n. 3n and 5n. These expressions represent the number of coins for our dimes, our quarters, and our total, depending on our multiplier.

Proportions and Cross Multiplying

We could also solve this problem by setting up a proportion. We can compare the number of dimes and quarters in the ratio to the number of dimes and quarters in our hand. When we set two ratios equal, we get a proportion. The number of dimes in our ratio was 2. For our ratio we had 2 to 3, or 2 3rd. And in our hand, we had 10 dimes, and I didn't know the number of quarters, so I'm going to use a variable, q. Once the proportion is set up, we can solve for unknown amounts using cross-multiplication. This method comes from multiplying both sides of the equation by the denominators. Notice 3 divided by 3 makes 1, so I'm left with 2 q on this side. And q divided by q makes one as well, so I'm left with 3 times would have just cross multiplied 3 times 10 is 30, and q times 2 is 2 q, we would get this equation. We want to be careful when we cross multiply. We can only do it when there's one entire fraction on the left and on the right. If there's other sums or differences, you should try something else. And then to find the number of quarters, we divide by 2. So, the number of quarters is 15.

Proportion Check

So, why the mood of out of proportion when conditions are table? Consider this problem. >> We're going to keep our ratio with the number of dimes to the number of quarter's the same. >> How many dimes are there if there are 231 quarters? Now here, I wouldn't want to list out a long table of numbers, that would take too much time. >> Instead, I could set up a proportion and solve. >> You try setting up a proportion and solving. >> Put your final answer here.

Proportion Check

We know for our ratio the number of dimes to quarters is 2 to 3 or 2 3rd. 2 is the number of dimes, so I need the number of dimes is in the numerator on the other side. I need to make sure the units match. I don't know the number of dimes, so I'm going to call that d. And 231 would go in the denominator, for the number of quarters. Again, notice that the quarters are both in the denominator. There's other ways to set up this proportion. Just be sure you're consistent with your units. Now, we can solve this proportion using cross multiplication. I'm going to multiply both sides of the equation by the denominators. The 3s will reduce to 1, and the 231s will also reduce to 1. This leaves me with 231 times 2, which is 462 and 3 times d, or 3d. Finally, we divide by 3 and we get or wrong, try solving this a different way. See if you can set up other proportions and get the same answer.

Convenience Store Proportions

Okay. Try this one with proportions. >> A convenience store charges 3 packs of gum for 87 cents. >> How much should be charged for 10 packs of gum, assuming each pack of gum costs the same amount? >> You can put yours in this box here, and ignore the dollar sign.

Convenience Store Proportions

We can write two ratios and set up a proportion. We can set them equal to each other. I know that for 3 packs of gum, I have to pay 87 cents. For 10 packs of gum, I don't know the cost. I'm going to use a variable to represent that number, I'm going to use t. We have one equation and one unknown. So, let's solve for this variable. We multiply both sides of the equation by the denominators. 87 hundredths times t. The 87 hundredth is reduce to 1, because they are a factor in the numerator and the denominator, and the t is reduce to just move the decimal one place to the right. I know t represents an amount of money, so I know 8 and 7 tenths is really 8 dollars and 70 cents. I going to divide that amount by 3, so 1t is equal to $2.90, this was the cost of our ten pack of gum. If you've gotten at least one of these right, nice work. We're going to solve similar problems involving expressions and equations in the next sections. Keep up the great work.

Angles

In all the problems so far, we've been setting up equations with words, and then solving them. >> We're going to try that same approach, but this time with geometry. >> We're going to look at angles. >> An angle is used to represent a certain amount of rotation. >> You might have heard doing a 360. >> If you're looking out in one direction and turn all the way around, you've gone 360 degrees. >> So, some angle is only part of 360 degrees. >> We're not going to look at just any angle, we're going to look at some special angles that have unique relationships.

Complementary Angles

Here I have angle A and angle B. We call the angles by their vertex. The point right here at the corner of each angle is the vertex. But we're not so concerned with that. We're going to look at the angle measures. Angle A is going to measure 30 degrees and angle B is going to measure 60 degrees. A and B are said to be complementary angles. Why do you think that is? What do you think this work complementary means? Let's see if you can figure it out. So if C and D are complementary angles, and E and F are complementary angles, what do you think D and F measure? Type your answer in here and don't worry about the degree symbol, I've already put it for you.

Complementary Angles

Well, we said A and B were complimentary angles and if we notice, they sum to 90 degrees or a right angle. You usually draw a box in the corner to indicate a right angle. If I want to find angle D, I know C and D have to add up to 90 degrees. I know C is 55 degrees. So, if I just subtract 55 from both sides, angle D is 35 degrees. We can use the exact same process for angle E and angle F. If I subtract 12 degrees from 90, I get 78 degrees. If you figured out what complementary meant, nice work. If not, that's okay. Hopefully, you'll learn along the way.

Complement of Any Angle

Complementary angles sum to 90 degrees. So, if I know the compliment of 30 degrees is 60 degrees, what do you think the complement of x is? How do we find the complement of any unknown angle? Write an expression in this box here. Think about how you found the complement in the last problems.

Complement of Any Angle

To find the compliment of 30 degrees, we took 90 degrees and subtracted that angle. So, we got 60. We can use the same reasoning to find the compliment angle of this unknown angle. We know that angles add up to 90 degrees, so we start with 90 degrees, then we subtract off angle x. We know we start with 90 degrees, a right angle, and then we just subtract off angle x. This will leave us with just the compliment angle.

Supplementary Angles

We've worked with complementary angles. Let's see if we can figure out what supplementary angles are. Here's angle R, which measures 60 degrees, and here's angle Q, which measures 120 degrees. These angles are considered to be supplementary angles. Let's see if you know what this word means. Here are two questions. Here, angles S and T are supplementary, and angles U and V are supplementary. What do you think are the missing angles, T and V? Find their measures. And don't worry about entering the degree symbol. Good luck.

Supplementary Angles

Supplementary must mean that the two angles add up to 180 degrees. We can see that because angle R and angle Q add up to 180. This should also make sense, because a circle has 360 degrees. So, if I take half of it or draw a straight line through it, I know that's only 180 degrees, just like in this case. So, if angle S and angle T are supplementary, they must add up to 180 degrees. I knew angle S measures 55 degrees so I just need to find the measure of angle T. You can either think about it in your head or just subtract 55 degrees from both sides. So, T is equal to 125 degrees. You might be saying, whoa, this doesn't make sense. This angle is smaller than this angle and you're right. Well, it turns out that this diagram isn't to scale. This would be a more accurate picture of our angle measures. We could find angle V in a similar fashion. We know U and V have to add up to 180 degrees and then, we subtract 12 degrees from both sides. So, V must be 168. If you got those two right, nice work.

Supplement of Any Angle

From the last problems, we could find out the supplement of an angle by subtracting it from 180 degrees. >> So let's write an expression for the supplement of any angle. >> If I have some angle x, I don't know what it is, how do I find the supplement? Write your expression here, and don't worry about putting the degree symbol.

Supplement of Any Angle

We know x in the supplement add up to 180, >> So we know the supplement is just 180 minus x. >> If you put 180 minus x, nice work.

Complement vs. Supplement

A lot of people have trouble remembering what complement and supplement mean. Here are some tips that you can use to help you remenber. >> Complement angles form a corner, or a 90 degree angle. >> So the C stands for corner. >> Supplementary angles form a straight line. >> So the S, for straight. >> And, if you remember to list the letters alphabetically, you can also list the degrees numerically in order, increasing. >> C goes with 90, >> S goes with 180. >> Complimentary angles add to 90. >> Supplemetary angles add to 180.

Complement and Supplement Equation

Let's put the idea of complement and supplement to use. We want to find the measure of an angle whose supplement is 10 more degrees than twice its complement. This problem can seem kind of confusing. So let's start with what we don't know. We know we need to find the measure of an angle. So that's my unknown, I'm going to call that x. The second part that I've underlined in blue represents an equation. The supplement is, or equals, 10 more than twice it's complement. So notice how I've taken the words and translated them into an equation. Let's put the ideas of complement and supplement to use. We know the complement for any angle is 90 minus x. And the supplement for any angle, is 180 minus x. So, what equation can you write using only x to replace these words? Enter your answer here.

Complement and Supplement Equation

Well, I just need to make a substitution. I know the supplement is 180 minus x, so that takes this spot. And the complement is 90 minus x. That's the angle that partners with x to make 90 degrees. And the complement is 90 minus x, so that goes in this spot. Here's the equation and one variable.

Find the Angle

So, now that we have an equation to solve, I want you to find the unknown angle. >> Find x.

Find the Angle

We can solve this problem like our other equations. First, we distribute the 2. like terms, we add 2 x to both sides and we get a 180 plus x equals a 190. So, x is equal to 10, or 10 degrees.

Checking Our Work

We found out that the measure of the angle was 10 degrees. We really should make sure that this is correct. Let's make sure we can check this with our ideas of supplement and complement. We know the complement of angle x has to be 80 degrees, since 80 and 10 make 90 degrees. And the supplement of angle x has to be 170 degrees, since these two add up to 180. So, in my equation, I can replace the complement with 80 degrees and the supplement with 170 degrees. This allows me to check my work. I want to know if this answer is really true. Was the supplement 10 more than two times the complement? And yes, 170 degrees is equal to 170 degrees. Our supplement was double the complement plus 10. So we can be assured that our answer was correct. The unknown angle was 10 degrees.

Mastering Complements and Supplements

Let's see if you've got these ideas down. >> Try answering this question. >> We'll call the unknown angle x, >> And you can write your answer in this box. >> Good luck.

Mastering Complements and Supplements

Remember, when we start this problem we always start this by identifying our unknown and in this case it was the angle. We don't know the angle, so we call it x. Next we want to identify an equation in words. The supplement of an angle is 40 more than twice its complement. These words actually represents our equation. We have the supplement on the left and 40 more than twice the complement on the right. This word is often serves as the word equals. This allows us to set the supplement equal to 40 more than twice the compliment. The supplement of our unknown angle would be 180 minus x, and the compliment of our unknown angle would be 90 minus x. Now we solve this equation like before. We distribute the positive 2 to get 180 minus 2x. Then we combine like terms to get rewritten this statement over here and my last step, I just subtract 180 from both sides. So x is 40 degrees. If you got that one right, great work. You've really come a long way in understanding equations and expressions. This stuff is pretty tough. The last thing we want to do is make sure our answer is correct. The supplement of 40 degrees is 140 degrees, since these two add up to 180. And the compliment of our angle is 50 degrees. So let's compare these results with what we've started with. Is this complement 40 more than twice our supplement? If I double the complement, I get 100 degrees. And if I add 40 to it, I get 140 degrees. So, yeah, that's my supplement.

Vertical Angles

For this last problem, we're going to learn about one more angle relationship, vertical angles. Vertical angles are directly across from one another. They share a central vertex, and they form Vs. That's how we can remember them. It turns out vertical angles are equal. So if T measures 30 degrees, M must also measure 30 degrees. Degrees. This also forces some unique relationships in this diagram. I'm going to see if you can figure them out. If angle M measures 32 degrees, I want you to find the measures of all the other angles. Enter their measures here.

Vertical Angles

We know angle M and angle T are vertical angles, so they must be equal. T is 32 degrees. This last part is a little trick. Maybe you saw it. This is a straight line. So, angles M and A are supplementary, they add to 180 degrees. Notice, angle A and T are also supplementary. They both add up to 180 degrees. This also is a result of the vertical angles. There's only one possible value for A. We know supplementary angles add to 180 degrees. So, if I subtract off 32 degrees, I get angle A. So, angle A is 148 degrees. Angle H also shares this same supplementary relationships. So, it's forced to have the same value. Pretty interesting, huh?

Practice 1

Well, Chris, there's a lot to know about word problems, isn't there? Let's practice doing some word problems. >> The first problem we're going to do is a number problem. >> If twice a number added to 7 is equal to 5 less than 5 times the number, what is the number? You can put your answer in this box.

Practice 1

Let's see how we got 4. If we know twice a number added to 7, we know how to write twice a number. Twice a number means 2 times the number. I'm going to use n as the variable this time. So if I had 2 times the number, or twice the number, added to 7 means plus 7. And it's equal to 5 less When I see 5 less, that means I'm going to be subtracting 5 from whatever comes after, is equal to equal to 5 less than 5 times the number. Now that we've translated our words into an equation, we can simply solve this equation. First thing I'm going to do is subtract 2 n from both sides, and I get 7 is equal to 5 n minus 2 n is 3 n minus 5. And now, I just add 5 to the both sides and I get 12 is equal to 3 n. Now I have 12 equals 3 n. So all I need to do is divide both sides by 3 and I get n is equal to 12 divided by 3, which is 4. That wasn't too bad was it? Let's try another one.

Practice 2

Now that we've tried a number problem, let's try a geometry problem. Here we have the width of a rectangle is 9 inches less than its length. If the perimeter of the rectangle is 58 inches, find the dimensions. You can put one dimension here and one dimension here.

Practice 2

If you got 19 inches by 10 inches, great work. If you didn't get that answer, let's see how we got that answer. First, I know that the width of a rectangle is less then its length. So that's going to be l minus 9. The width is 9 inches less than its length. We also know that the perimeter is 58 inches. We know the equation for perimeter is 2 times the length plus 2 times the width, or 2l plus with what w equals which is l minus 9. And we know that the perimeter is 58 inches. So, I have 2l plus 2 times w, but w equals l minus 9, and that equals have 2l plus 2l. 2 times negative 9 is negative 18 and that equals 58. Now, I can combine like terms, 2l plus 2l is 4l, add 18 to both sides, so now, I have I get that l equals 19. So, I have that 4l equals 76, I divide both sides by 4, which gives me, l equals 19. So, here's one of the dimensions of my rectangle. length minus 9. So, the width is equal to the length, which is 19, minus 9 which gives me 10. So, the dimensions of the rectangle are 19 inches by 10 inches. Let's try a ratio problem.

Practice 3

Now that you've tried a number problem and a geometry problem, let's try a ratio and proportion problem. This time we have a recipe that calls for 8 ounces of chocolate chip cookies for each two dozen cookies. We want to know how many ounces of chips you would need to make 42 cookies. You can put your answer in this box.

Practice 3

How'd you do? If you got 14 ounces, great work. Let's see how we got that answer. When we do ratio problems, it's important to put the ratio in words before we put numbers. So, the ratio I need to write is the amount of chocolate chips over the number of cookies. So, the ratio we want to write here is how many ounces of chocolate chips we need for how many cookies. We know that there's 8 ounces of chocolate chips, so we know the amount of chocolate chips. But we don't know the number of cookies because this tells us we have 2 dozen cookies. For those of you that don't know, one dozen is equal to 12. For example, when you buy a dozen eggs, you get 12 eggs. So, if we have 2 dozen cookies, that's going to be 24 cookies. So we need 8 ounces of chocolate chips for 24 cookies. The question says, how many ounces of chips we need to make 42 cookies? So, the number of cookies is 42, and we want to know how many ounces of chocolate chips we need. We know the number of cookies we need to make is 42, and we have to find out how many ounces of chocolate chips. So, this is what we're looking for. I'm going to use the variable c in this case. We have two ratios. When we set them equal to each other, that creates a proportion. To solve a proportion, we simply cross-multiply. When we cross-multiply, we get 24 times c is equal to 8 times 42. So, we get 24 c equals 8 times 42 is 336. To solve this, all we have to do is divide both sides by 24. 336 divided by 24 is make 42 cookies. Are you ready to try another one?

Practice 4

Let's try an angle problem. >> The supplement of an angle measures 30 degrees more than twice its complement. >> Find the measure of the angle. >> You can put your answer in this box.

Practice 4

First, we have to keep a couple things in mind. I have to decide what variable I'm going to use for the angle. I'm going to let a equal the measure of the angle. Next, I have to remember what supplement and complement mean. If you recall, the supplement of an angle, I'll write it as s, is equal to 180 minus the measure of the angle. So here, we have the supplement as 180 minus the measure of the angle. The other thing we have to remember is the complement. The complement of an angle is 90 degrees minus the measure of the angle. So, I have the measure of the angle is a. The supplement is 180 minus a, and the complement is 90 minus a. Now, we can try to turn the words into an equation. The supplement of the angle s measures, which means equals, 30 degrees more. So that's 30 plus twice it's compliment. I know that s equals 180 minus a, and I know that the compliment is 90 minus a. Now I can solve the equation for a. The first thing we need to do to solve this equation is to distribute this 2. So I have 180 minus a equals 30 plus 2 times 90 is 180. And 2 times negative a is negative 2 a. The next thing I want to do combine these like terms. 180 plus 30 is 210. I need to get all my variables onto one side of the equation, and I'm going to do that by adding 2 a to both sides of the equation. When I add 2 a to both sides of the equation, the left side I have 180. Negative a plus 2 a gives me a. And that's going to equal 210. The next thing I need to do is subtract 180 from both sides. And I get a equals 210 minus 180 which gives me 30. So, the measure of an angle who's supplement measures 30 degrees more than twice it's complement is 30 degrees. I think we've had enough of word problems for now. Let's move on to something else.