>> Welcome to the Cypress College Math Review on applications of linear equations.
To solve an application problem, you first want to read the problem carefully several times
until you really understand what's going on.
Sometimes, depending upon the problem, it might be helpful to draw a picture.
Then you want to determine the unknown and assign a variable to it.
Any other unknowns in the problem should be represented in terms of that same variable.
Then you want to set up your equation, solve the equation, be sure to write your answer
as a full sentence using the appropriate units.
In this video, the emphasis is on understanding how to set
up the equation that models the word problems.
In order to accomplish this, many word problems need to be shown.
To save time, I will not spend time showing how to solve the equations.
The solutions will be shown.
Please pause the video if you'd like to go through the steps carefully.
Objective one, perimeter and area problems.
Here is a table of geometric formulas that you need to learn.
The ones that arise the most with applications of linear equations are the perimeter formulas
for the square, the rectangle, and the triangle.
For now, you'll see these formulas come up the most.
Example. "It will take 380 feet of fencing to fence in Gabriel's backyard.
The length of his backyard is 50 feet less than twice the width.
Find the dimensions of Gabriel's backyard.
So we have a rectangular backyard.
And we have 380 feet of fencing.
Well that's the distance around.
Any time you have a fencing problem, that's the distance around the backyard.
The formula for perimeter of a rectangle is 2 times the length plus 2 times the width,
write down your formula.
Now we need to set up the variable.
We know that the length...
Is, so "is" is equal, the verb is the "equals," 50 feet less than, so that's something minus 50,
50 less than is always something minus 50, twice the width.
So length is solved for and width is over with the other expressions,
so we would make the width the variable.
And the length is 2W minus 50.
So we have our variables set up.
Now we replace length and width in our formula with these expressions.
So 2 times the quantity 2W minus 50 plus 2 times W equals 380.
Next we solve this equation.
Here's the solution, pause the video if you want to go through the steps.
And the answer is a sentence, something like,
"The dimensions of Gabriel's backyard are 80 feet by 110 feet."
"The perimeter of a triangle is 170 meters.
One side is 40 more than the second side.
The third side is 15 less than the first side.
Find the lengths of the sides of the triangle.
So we have a triangle and we're working with its perimeter.
So the formula is P equals A plus B plus C, where A, B, and C are the sides of the triangle.
So let's rewrite that as the first side plus the second side plus the third side equals 170.
Now let's set up our variables.
It says at one side is 40 more than the second side.
So the one side I'm going to call the first side.
So the first side equals 40 plus the second side.
The third side is 15 less than the first side.
So third equals something minus 15, because 15 less than, so first side minus 15.
Looking at these two equations, we notice that the first side is in both of the equations.
That's what we want to set up for our variable, then.
The first side will be X. So the first side is X. The third side is X minus 15.
And in this equation, we're going to have to solve for the second side.
The second side would be X minus 40.
Now, let's substitute in those expressions for the first side, the second side,
and the third side, and that gives us our equation.
Next we solve the equation, pause the video if you'd like to go through the steps.
Our answer is a sentence something like,
"The sides of the triangle are 75 meters, 35 meters, and 60 meters."
It's always a good idea to check your answer to word problems.
Add up the three sides of that triangle, and yes, you do get the perimeter, which is 170.
Objective two, distance, rate, and time problems.
For these problems you really want to understand the units.
For distance, your different options are miles, feet, meters, yards, something like that.
So when you see 100 miles, that's a distance.
So recognize which is a distance, which is a rate, and which is a time.
Rates. What does mph stand for?
Mph stands for "miles per hour."
And what does that look like in terms of mathematics?
It's miles on top and hours on the bottom.
"Per" means divide.
So miles per hour looks like this.
Other options for rates are feet per second, kilometers per hour, meters per second.
Those are all rates.
Time is the easiest.
Hours, minutes, seconds, or other options.
So what does our formula look like?
Distance equals rate times time.
And why is that true?
Because distance could be in miles.
Rate, miles per hour, time in hours, and you can see that the units of hours cancel off.
Miles equals miles.
You can always use the units to help yourself remember the formula.
Put in the units and see if the units match.
Here's something that will help you set up a lot of this type of word problems.
Here's distance, rate, and time.
And often, only one of these three things will be given in terms of numbers
for both of the different participants.
So maybe both the distances are given, or both the rates, or both the times.
So mark which one is given.
One of them will be asked for in the question.
How far did they go?
How fast did they go?
How long did it take them?
So mark which one was asked for in the question and whichever is left of those three, distance,
rate, or time, that's the easiest way to set up the word problem.
For linear word problems, it's usually distance.
It's almost always distance as the setup for the linear equations,
which is what we're working on here.
Example. Andy and Tommy start from the same point and walk in opposite directions.
Andy walks twice as fast as Tommy does.
In seven hours, they're 42 miles apart.
How fast was Andy walking?
How fast was Tommy walking?"
So we can tell that this is a distance, rate,
time problem because we have a distance, 42 miles, a time, seven hours.
And they're asked how fast you're going.
Let's take a look at each of the pieces of the puzzle.
First of all, they're starting at the same point and walking in opposite directions.
So drawing a picture of what's going on should help us.
Next, let's take a look at each of the different pieces that we're given.
Andy walks twice as fast as Tommy does.
That does not tell us how fast either of them are walking.
So that's just a comparison of the rates, but we are not given the rates.
In seven hours, that's a time.
In fact, since they start at the same time and end at the same time,
that is the time for each of them.
So we are given the time for each of them.
42 miles apart, that's a distance.
But that's the total distance, that's not the distance for each one.
Distance, rate, and time.
We are given the time for each one.
We are asked how fast is each walking, so we're asked for the rates.
So distance is the easiest way to set this problem up.
And what do we have?
So we read the problem again and we say, "Okay, what do we know about the distance?"
Their total distance is 42 miles.
That doesn't mean each one walked 21 miles.
Because one was walking faster than the other.
But that is the total distance that they traveled.
So the distance that Andy walked plus the distance
that Tommy walked is equal to the total distance.
What's the formula for distance?
Rate times time.
So we have rate times time plus rate times time equals 42.
The first rate times time deals with Andy, and the second rate times time deals with Tommy.
Now let's fill in those four sets of the parentheses.
Well the time for each one is seven.
Seven hours each one traveled.
Now what about the rates?
We know a comparison of the two rates.
Andy's rate is twice Tommy's rate.
So if we let R be Tommy's rate, then 2R is Andy's rate.
Substitute those into the parentheses.
And now we have our equation.
Here's the solution to the equation.
And our answer would be a sentence, something like, "Tommy is walking at 2 miles an hour
and Andy is walking at 4 miles an hour."
Carla and Joaquin went for a bike ride from their house to the beach.
Carla rode at 15 miles an hour while Joaquin rode at 11 miles an hour.
It took Joaquin one hour longer to make it to the beach.
How long did it take each of them to ride to the beach?"
We recognize that this is a distance, rate,
time problem because we're given the two rates and we're asked for the times.
So we're given Carla's rate and we're given Joaquin's rate.
Those are given specifically in terms of a number for both of the different participants.
One hour longer, that's a comparison of the times.
We're not given the times of each.
How long did it take each of them to ride to the beach?
So that's what's asked for in the question.
So the rates are given, the time is what's asked for,
which means distance is the easiest way to set up this problem.
What kind of a picture can we draw?
Well, they started at the same place, their house.
And they ended up at the same place, the beach.
So here's our picture.
The same distance.
So since we're setting this up in terms of a distance,
we have to come up with a model for distance.
Well they went the same distance, so the distance that Carla rode is the same
as the distance that Joaquin rode.
What's the formula for distance?
Rate times time.
So now we have these four sets of parentheses and we have to figure out what goes in each.
Well we're given the two rates.
Carla's rate was 15 miles an hour and Joaquin's rate was 11 miles an hour.
What about the times?
Well we have a comparison of the times.
Joaquin took one hour longer to get to the beach.
So Joaquin's time equals one hour plus Carla's time.
If we let X be Carla's time, then Joaquin's time is 1 plus X. We substitute these two expressions
in for the times.
And we have our equation.
Here's the solution to the equation.
And the answer would be a sentence, something like, "It took Carla two
and three quarters hours while it took Joaquin three and three quarters hours."
"On a 340 mile trip, a car traveled at an average of 55 miles an hour
and then reduced the speed to 40 miles an hour for the remainder of the trip.
The trip took a total of seven hours.
For how long did the car travel at each speed?"
We recognize that this is a distance, rate, time problem because we're given some rates.
We're given something about the time.
And we're given a distance.
What's our picture?
Well, the car traveled long at 55 miles an hour and then at 40 miles an hour
and the total distance was 340 miles.
So here's our picture.
We're given the total distance, not the individual distances.
We are given the individual rates though.
The 55 miles an hour is one rate, and the other rate is 40 miles an hour.
The seven hours is not the individual times.
The seven hours is the total time.
Distance, rate, time.
The rates are given.
What's the question?
For how long did the car travel at each speed?
That's time.
So time is asked for.
Which means distance is the best way to set this up.
Well we know the title distance is 340 miles.
So the distance on the first leg, when he's going faster,
plus the distance on the second leg, when you're going slower, is equal to the total distance.
The formula for distance is rate times time.
And the total distance is 340 miles.
Now we have four sets of parentheses that we need to fill in.
We're given the two rates, so we can put in the 55 miles an hour and the 40 miles an hour.
Now we have to figure out what to put in for the times.
You can't assume that it's three and a half hours each.
In other words, we can't just take the seven hours and split it by two.
You cannot put X in for both.
You cannot assume they're both the same.
So, we know that the time when traveling faster plus the time
when traveling slower is equal to seven hours.
If there are two unknowns in a problem that are supposed to add up to seven,
then you can represent one as X and the other as 7 minus X. If we substitute in X for the time
when he was going faster, plus 7 minus X for the time
when he's going slower, what does that add up to?
Well we have X minus X, which is zero.
And then 7 is equal to 7.
It works. So when two things add up to 7, and you don't know either one,
substitute X in for one of them and 7 minus X in for the other.
We do that in our parentheses.
And we have our equation.
Here's the solution to our equation.
And the answer is a sentence, something like, "The car traveled at 55 miles an hour
for 4 hours and at 40 miles an hour for 3 hours," which of course is 7 minus 4.
Please note that there's other ways to do a lot of these problems.
A lot of instructors use a tabular method for these word problems and that's fine, also.
This is just another suggestion using a verbal model.
Pause the video and try these problems.
Objective three, value mixture problems.
If you have three gallons and you add two gallons to it,
then you're going to end up with five gallons.
If you have three gallons and you're going to add an unknown quantity,
let's call it X gallons, then you're going to end
up with a mix that has 3 plus X gallons in it.
Let's say you wanted to end up with nine gallons in your final mix.
Then you need two things that add up to nine.
Well, let's represent one as X, the other, then, would be what's left, which is 9 minus X.
And if you take X plus 9 minus X, you get 9.
The formula for a value mixture problem comes from the fact that if you take two pounds
of hamburger that cost $3.00 per pound, let's see, what's dollars per pound?
That's dollar divided by pounds.
That "per" means divide.
So if we have $3.00 a pound and we multiply it times two pounds,
we end up with $6.00 because the pounds cancel.
So the units help us figure out the formula.
So the formula that you're going to use on both sides of the equation
for a value mixture problem is dollars per pound times pounds.
That'll give you the dollars, or the cost,
of each of the different quantities you're working with in the mixture problem.
And to recognize a mixture problem, you're looking for the word "mix" or "mixture."
"How many pounds of walnuts valued at $7.50 per pound should be mixed with 10 pounds
of cashews valued at $9.00 per pound to create a trail mix valued at $8.00 per pound?"
So this is a mixture problem because we see the word "mix."
The cost of the walnuts plus the cost of the cashews equals the cost of the mix.
Now looking at the three different prices, if you take something that's expensive and add it
to something that's cheaper, you're going to get something that's in between.
So the price of the mix is always the price that's in between the other two prices.
So $8.00 is between $7.50 and $9.00, so that must be the price of our mix.
We put the $8.00 on the right.
And we put the other two prices on the left.
And now we're going to look at the amounts.
Well we have 10 pounds of the cashews, so we put the 10 pounds in.
And we don't know the other amount on the left hand side,
so we'll let X be the amount of walnuts.
And the amounts on the left must add up to the amounts on the right, so X plus 10,
that must be the amount of the mix.
So that's the pounds of the mix.
So we have our equation set up.
Here's our solution.
And our answer.
A chocolate shop manager mixes dark chocolate, worth $5.00 per pound,
with white chocolate, worth $9.00 per pound.
Find how many pounds of each she should use to get 24 pounds
of an assortment mix, worth $7.50 per pound.
So we see the word "mix," so we know that this is a mixture problem.
It's a value mixture problem.
So here we have our equation because we have the cost of all the dark chocolate plus the cost
of the white chocolate, and that will give us the cost of the mix.
What do we know?
We know the different costs.
$7.50 is between $5.00 and $9.00, so $7.50 must be the cost of the mix.
We put that in.
We put the other two prices over on the left.
Which amount were we given?
This time we're given the amount on the right hand side,
so we put the 24 pounds next to the $7.50.
So we know that our two amounts on the left hand side have to add up to 24.
So we need two expressions that add up to 24.
Well let's let one be X and the other would be what's left from the 24
or 24 minus X. So we have our equation.
Here's our solution.
And our answer.
Concentration mixture problems, that's the other type of mixture problem.
Let's say you have a bucket that has one gallon of a mix and it's 50% antifreeze.
So how much pure antifreeze is in the bucket?
Well that'd be a half a gallon.
How do we get that?
Well 50% of one gallon is a half of a gallon.
So you'd take the concentration, which is .5, or 50%, and multiply it times the amount,
or one gallon, to get the half a gallon.
Each term of your equation in a concentration mixture problem is going
to be the amount of pure substance.
So the formula, concentration times amount,
will be used for every single term on both sides of the equation.
"A chemist wants to make 50 milliliters of a 16% acid solution
by mixing a 13% solution with an 18% solution.
How many milliliters of each should be used?"
So each term of our equation in this mixture problem will be the amount of pure substance.
And the amount of pure substance is always equal to the concentration times the amount.
That formula will be used for every single term.
The concentrations are 16, 13, and 18.
Obviously the one in the middle between those two numbers must be the one on the right.
You're going to take something that's weak, add something that's more concentrated,
and get something that's in between.
So between 16, 13, and 18, 16 must be the mixture that goes over on the right hand side.
So we put in the other concentrations.
And now we look at the amount.
We're given 50 milliliters of the 16%.
So we put that in for the amount.
So we need two expressions that add up to 50 because we know the amounts on the left have
to add up to the amount on the right.
So we need two expressions that add up to 50.
So one would be X and one would be 50 minus X. Now we have our equation.
Here's our solution.
And here's our answer.
"Maya has 3 gallons of a solution that's 30% antifreeze,
which she wants to use to winterize her car.
How much pure antifreeze should she add to this solution
so that the new solution will be 65% antifreeze?"
So she's starting off with a 30% solution, she's ending up with a 65% solution,
she's adding in pure antifreeze, so what would the concentration of pure antifreeze be?
Well that would be 100%.
100% as a decimal would be 1.
You're given three gallons of the 30%.
So we're going to let X be the number of gallons of pure antifreeze.
We know the amounts on the left must add to the amount on the right, so we have 3 plus X
for the amount on the right hand side.
Here's our equation.
Here's our solution.
And here's our answer.
This problem's different.
"Denise has 75 liters of a 30% salt solution."
We'll go ahead and put that information in.
How many liters does she need to evaporate off--
Oh, so instead of adding two things together, we're taking something away.
So we have subtraction instead.
"-- to get a 35% solution?"
Okay, so we've put in what we know.
Now, you're evaporating off what?
Well, water evaporates off.
The salt does not disappear, the water is what disappears.
So the concentration of salt in pure water is of course zero.
So we put a zero in there for the concentration of the water that's evaporating off.
How much water?
Well, we don't know.
So we put X in for the amount of water on the left hand side.
This time, the amounts on the left hand side will subtract,
because once again, you are taking something away.
We're starting off with 75 liters, we are taking away X liters.
So at the end, your amount will be 75 minus X. Here's our equation.
Here's our solution.
And our answer.
"Bonnie has 50 gallons of a 20% solution."
Let's put that in already.
"How many gallons must she drain off?"
So we're not evaporating this time, we're actually draining the same solution off.
So the concentration is going to be 20%, and the unknown amount of X. Then we're going
to replace it with the same amount.
So the amount that we replace is also going to be X. And we're replacing it with pure chemical,
so that would be 100% for the concentration, or 1 as a decimal.
And at the end, we want a mix that's 35% pure.
The only thing left to figure out is how many gallons is going to be in our new mix?
Well we started off with 50 gallons, we subtracted X gallons,
then we added the X gallons right back in.
Which means our final solution will have 50 gallons in it.
Here's our equation.
Our solution.
And our answer.
Pause the video and try these problems.
Objective five, consecutive integer problems.
Well what are consecutive integers?
Well, they're like four, five, six.
Those are consecutive integers.
How do we represent them?
Well we choose the smallest one and let that be X. And how do we get to five from four?
Well we add one.
So consecutive integers can be represented as X, X plus 1, X plus 2, and so on so forth.
What about consecutive odd integers?
Like three, five, seven?
So the first one will be X, the smallest is X. How do we get from three to five?
Well we add two.
And then we add two again.
So consecutive odd integers can be represented as X, X plus 2,
X plus 4, and so on and so forth.
Consecutive even integers like six, eight, ten...
So if we let X be the smallest one, how do we get from six to eight?
We also add two.
So consecutive even integers can be set up the same way as odd integers can.
X, X plus 2, X plus 4, and so on and so forth.
Example. "The sum of three consecutive integers is 48 more
than twice the smallest, find the integers."
So it said consecutive integers, not even or odd.
So we're going to set them up as X, X plus 1, and X plus 2.
So the verbal model is their sum is 48 more than twice the smallest.
Their sum, I'm going to add X, and X plus 1, and X plus 2.
And the smallest one is X. So here's our equation.
Here's our solution.
And our answer.
Notice you can always check word problems.
This is a nice one to check.
So we took the sum, we made sure that was equal to 48 more than twice the smallest, which is 45.
And we get 138 on both sides.
It checks.
"Find three consecutive odd integers such that the sum
of 3 times the smaller and twice the largest is 143."
So consecutive odd integers, we'll represent them as X, X plus 2, and X plus 4.
Our verbal model is the sum of 3 times the smaller plus twice the largest is 143.
The smaller is X, the largest of the three is X plus 4.
And now we have our equation.
Here's our solution.
And our answer.
Problems involving fixed costs and variable costs.
To find the total cost of an item, that would equal the fixed cost plus the variable cost.
To find out the variable cost, you'll have
to multiply the cost per item times the number of items.
Our problem involves the charge for a phone call.
"One phone company charges 45 cents for the first minute and 20 cents per minute after that.
If a phone call costs $2.85, how many minutes did the people talk?
And don't forget to add in the first minute to your answer."
So the fixed cost is the 45 cents that you have to pay when you just pick up the phone,
you're going the be charged for that first minute.
That's the fixed cost of 45 cents or .45 dollars.
The variable cost is the charge per minute times the number of minutes.
So we're going to have .2 dollars per minute times the number
of minutes that you were on the phone.
But that is just the minutes after that first minute.
And we just put an X in for it anyhow.
So here's our equation.
We solve it and we find out that the person talked
for 12 additional minutes beyond the first minute that they were charged the 45 cents for.
Which means that the phone call was 13 minutes long.
Pause the video and try these problems.
Problems involving complementary and supplementary angles.
Angles that are complementary add up to 90 degrees.
Angles that are supplementary add up to 180 degrees.
If you're working with two angles that are complementary, then you can represent one
of them as X and the other as 90 minus X because they have to add up to 90 degrees.
If you're working with two angles that are supplementary, then you can represent them as X
and 180 minus X, because they need to add up to 180.
"Find the measure of an angle if the measure of the angle is 8 degrees less
than 3 times the measure of its supplement."
So we're talking about an angle and its supplement.
So we'll have X and 180 minus X for the supplement.
Reading the sentence again, we translate that into a verbal model.
The measure of the angle, well that's X, is, equals, 8 less than,
so that's something minus 8, so 3 times the supplement, 3 times parentheses,
the supplement is 180 minus X. We have our equation.
Here's our solution.
And our answer.
"Find the measure of an angle whose supplement measures 6 degrees more
than 7 times its complement."
So here we're working with an angle and its supplement and its complement.
So the angle we'll represent as X. Its supplement would be 180 minus X.
And its complement would be 90 minus X. Now we'll translate the sentence into an equation.
Its supplement, so supplement, measures, so that's the equals, 6 more than, 6 plus,
7 times, 7 times, the complement.
So supplement equals 6 plus 7 times the complement.
For the supplement of the angle we'll put in 180 minus X. For the complement we put
in 90 minus X. Here's our solution.
And our answer.
Investment problems.
The formula for simple interest is principal times rate times time.
For our example, we have that Tyson invested 10,000 dollars more at 8% than he did at 6%.
His annual interest was 3,390 dollars from the two investments.
Find out how much he invested in each account.
So we have two different accounts and we have a total annual interest.
So the interest from the first account plus the interest
from the second account equals the total interest.
So on the left hand side we're using our formula, principal times rate times time.
We know the interest rates are 8% and 6%.
And we know the total interest was $3,390.00.
Now how much did he invest in that one year?
So 1 for time.
Well 10,000 more at 8%.
10,000 more, so the amount that he invested at 6%, we'll let that be X. And the amount
that he invested at 8% must be 10,000 more than X, so 10,000 plus X. Here's our equation.
Here's our solution.
And our answer.
And once again, it's a good idea to check your answers when you can.
So here's the check.
Pause the video and try these problems.