Word Problems Involving Systems of Linear Equations

Word Problems Involving Systems of Linear EquationsMany word problems will give rise to systems of equations --- that is, a pair of equations like this:

You can solve a system of equations in various ways. In many of the examples below, I'll use thewhole equation approach. To review how this works, in the system above, I could multiply the first equation by 2 to get the y-numbers to match, then add the resulting equations:

If I pluginto, I can solve for y:

In some cases, the whole equation method isn't necessary, because you can just do a substitution. You'll see this happen in a few of the examples.The first few problems will involve items (coins, stamps, tickets) with different prices. If I have 6 tickets which cost $15 each, the total cost is

If I have 8 dimes, the total value is

This is common sense, and is probably familiar to you from your experience with coins and buying things. But notice that these examples tell me what the general equation should be: The number of items times the cost (or value) per item gives the total cost (or value). This is where I get the headings on the tables below.You'll see that the same idea is used to set up the tables for all of these examples: Figure out what you'd do in a particular case, and the equation will say how to do this in general.Example.Calvin has $8.80 in pennies and nickels. If there are twice as many nickels as pennies, how many pennies does Calvin have? How many nickels?In this kind of problem, it's good to do everything in cents to avoid having to work with decimals. So Calvin has 880 cents total.Let p be the number of pennies. There are twice as many nickels as pennies, so there arenickels. I'll arrange the information in a table.

Be sure you understandwhythe equations in the pennies and nickels rows are the way they are: The number of coins times the value per coin is the total value. If the words seem too abstract to grasp, try some examples:If you have 3 nickels, they're worthcents.If you have 4 nickels, they're worthcents.If you have 5 nickels, they're worthcents.So if you havenickels, they're worthcents.The total value of the coins (880) is the value of the penniesplusthe value of the nickels. So I add the first two numbers in the last column, then solve the resulting equation for p:

Calvin has 80 pennies.Therefore, he hasnickels.

Tables for problems.I'll often arrange the equations for word problems in atable, as I did above. Roughly:1. Thenumberof things will go in the first column. This might be the number of tickets, the time it takes to make a trip, the amount of money invested in an account, and so on.2. Thevalue per itemorratewill go in the second column. This might be the price per ticket, the speed of a plane, the interest rate (in percent) earned by an investment, and so on.3. Thetotal valueortotal amountwill go in the third column. This might be the total cost of a number of tickets, the distance travelled by a car or a plane, the total interest earned by an investment, and so on.With this arrangement:

There are many correct ways of doing math problems, and you don't have to use tables to do these problems. But they are convenient for organizing information --- and they give you a pattern to get started with problems of a given kind (e.g. interest problems, or time-speed-distance problems).In some cases, youaddthe numbers in some of the columns in a table. In other cases, you set two of the numbers in a column equal, or subtract one number from another. There is no general rule for telling which of these things to do: You have to think about what the problem is telling you.Example.A total of 78 seats for a concert are sold, producing a total revenue of $483. If seats cost either $2.50 or $10.50, how many $2.50 seats and how many $10.50 seats were sold?Suppose x of the $2.50 seats and y of the $10.50 seats were sold.

The first and third columns give the equations

Multiply the second equation by 10 to clear decimals:

Solve the equations by multiplying the first equation by 25 and subtracting it from the second:

Then, so. Thus, 42 of the $2.50 seats and 36 of the $10.50 seats were sold.

Example.Tickets to a concert cost either $12 or $15. A total of 300 tickets are sold, and the total receipts were $4140. How many of each kind of ticket were sold?


Multiply the first equation by 15 and subtract equations:

Then

There were 120 tickets sold for $12 each and 180 tickets sold for $15 each.

Example.An investor buys a total of 360 shares of two stocks. The price of one stock is $35 per share, while the price of the other stock is $45 per share. The investor spends a total of $15000. How many shares of each stock did the investor buy?Let x be the number of shares of the $35 stock and let y be the number of shares of the $45 stock.

The first and third columns give

Multiply the first equation by 45, then subtract the second equation:

Since, I have. The investor bought 120 shares of the $35 stock and 240 shares of the $45 stock.

The next problem is more complicated than the others, since it involves solving a system ofthreeequations withthreevariables. You'll see that I do it by substitution. If you take more advanced courses (such aslinear algebra), you'll learn methods for solving systems like these which are like the whole equation method. They involve representing the equations usingmatrices.Example.Phoebe has some 32-cent stamps, some 29-cent stamps, and some 3-cent stamps. The number of 29-cent stamps is 10 less than the number of 32-cent stamps, while the number of 3-cent stamps is 5 less than the number of 29-cent stamps. The total value of the stamps is $9.45. How many of each stamp does she have?I will do everything in cents.I'll let x be the number of 32-cent stamps, let y be the number of 29-cent stamps, and let z be the number of 3-cent stamps. Here's the table.

The last column says

The number of 29-cent stamps is 10 less than the number of 32-cent stamps, so

The number of 3-cent stamps is 5 less than the number of 29-cent stamps, so

I want to get everything in terms of one variable, so I have to pick a variable to use. Since the last two equations both involve y, I'll do everything in terms of y.z is already solved for in terms of y, since. I'll solve for x in terms of y:

Plugandintoand solve for y:

Then

Phoebe has 20 32-cent stamps, 10 29-cent stamps, and 5 3-cent stamps.

The next problem is about numbers. The setup will give two equations, but I don't need to solve them using the whole equation approach as I did in other problems. Since one variable is already solved for in the second equation, I can just substitute for it in the first equation.Example.The sum of two numbers is 90. The larger number is 14 more than 3 times the smaller number. Find the numbers.Let L be the larger number and let S be the smaller number. The sum is 90:

The larger number is 14 more than 3 times the smaller number:

Pluginto the first equation and solve:

Then. The numbers are 19 and 71.

The next set of examples involvesimple interest. Here's how it works. Suppose you invest $600 (theprincipal) in an account which payssimple interest. At the end of one interest period, the interest you earn is

You now havedollars in your account.Notice that you multiply the amount invested (the principal) by the interest rate (in percent) to get the amount of interest earned.By the way --- How does "percent" fit the pattern of the earlier problems, where I had things like "dollars per ticket" or "cents per nickel"? In fact, "percent" is short for "per centum", andcentumis the Latin word for a hundred. So "4 percent" means "4 per 100". Since "per" translates to division, I get, as you probably know from earlier math courses.Example.$6000 is divided between two accounts, one payinginterest and the other payinginterest. At the end of one interest period, the interest earned by theaccount exceeds the interest earned by theaccount by $65. How much was invested in each account?

I have

Rewrite the equations:

Clear the decimals by multiplying the second equation by 100:

Multiply the first equation by 3 and subtract equations to solve for y:

Then

$2500 was invested atand $3500 was invested at.

Example.Bonzo invests some money atinterest. He also invests $1700 more than twice that amount atinterest. At the end of one interest period, the total interest earned was $278. How much was invested at each rate?

The last column gives an equation which can be solved for x:

Then, so $2100 was invested atand $5900 was invested at.

There are various kinds ofmixture problems. The first few involve mixtures of different things which cost different amounts per pound. For instance, if you have 4 pounds of candy which costs $2 per pound, the total cost of the candy is

In other words, the number of pounds times the price per pound is the total cost.Example.Calvin mixes candy that sells for $2.00 per pound with candy that costs $3.60 per pound to make 50 pounds of candy selling for $2.16 per pound. How many pounds of each kind of candy did he use in the mix?

The first and third columns give two equations:


Then

He used 45 pounds of the $2 candy and 5 pounds of the $3.60 candy.

Example.Phoebe wants to mix raisins worth $1.60 per pounds with nuts worth $2.45 per pound to make 17 pounds of a mixture worth $2 per pound. How many pounds of raisins and how many pounds of nuts should she use?Suppose she uses x pounds of raisins and y pounds of dried fruit.


Multiply the second equation by 100 to clear the decimals. This gives

Solve the equations by multiplying the first equation by 160 and subtracting it from the second:

Hence,and. She needs 8 pounds of raisins and 9 pounds of nuts.

Mixture problems do not always wind up with two equations to solve. Here's an example where the setup gives a single equation.Example.How many pounds of coffee worth $4 per pound must be mixed with 20 pounds of Elmer's Glueworth $2 per pound to obtain a mixture worth $3.60 per pound?Let x be the number of pounds of coffee. Set up a table:

The last line says. Solve for x:

You will need 80 pounds of coffee.

Analloyis a mixture of different kinds of metals. Suppose you have 50 pounds of an alloy which issilver. Then the number of pounds of (pure) silver in the 50 pounds is

That is, the 50 pounds of alloy consists of 10 pounds of pure silver andpounds of other metals.Notice that you multiply the number of pounds of alloy by the percentage of silver to get the number of pounds of (pure) silver.Example.Phoebe mixes an alloy containingsilver with an alloy containingsilver to make 100 pounds of an alloy withsilver. How many pounds of each kind of alloy did she use?

The first and third columns give two equations:



Then

She used 60 pounds of thealloy and 40 pounds of thealloy.

Other mixture problems involvesolutions. For instance, a solution may beacid, oralcohol. What does this mean? Suppose you have 80 gallons of a solution which isacid. Then the number of gallons of (pure) acid in the solution is

So you can think of the 80 gallons of solution as being made of 16 gallons of pure acid andgallons of pure water.Notice that you multiply the gallons of solution by the percentage of acid to get the number of gallons of (pure) acid.Example.How many gallons of each of aacid solution and anacid solution must be mixed to produce 50 gallons of aacid solution?


Multiply the second equation by 10 to clear the decimals:


Then

Use 15 gallons of thesolution and 35 gallons of thesolution.

Example.Amounts of aalcohol solution and aalcohol solution are to be mixed to produce 24 gallons of aalcohol solution. How many gallons of thealcohol solution and how many gallons of thealcohol solution should be used?Suppose x gallons of thealcohol solution and y gallons of thealcohol solution are used.



Solve the equations by multiplying the first equation by 65 and subtracting the second:

Then, so. Thus, 16 gallons of thesolution and 8 gallons of thesolution must be used.

Two angles arecomplementaryif their sum is--- that is, if they add up to a right angle.For example,andare complementary, because

Example.Two angles are complementary, and the larger one ismore than 3 time the smaller one. Find the angles.Let L be the larger angle, and let S be the smaller angle. The angles are complementary:

The larger one ismore than 3 time the smaller one:

Plugintoand solve for S:

Then. The smaller angle isand the larger angle is.

Example.Two angles are complementary. One angle isless than twice the other. Find the two angles.The sum is:

One angle isless than twice the other:

Pluginto the first equation and solve:

Then. The angles areand.

"Age" Word Problems(page 1 of 2)In January of the year 2000, I was one more than eleven times as old as my son William. In January of 2009, I was seven more than three times as old as him. How old was my son in January of 2000?Obviously, in "real life" you'd have walked up to my kid and and asked him how old he was, and he'd have proudly held up three grubby fingers, but that won't help you on your homework. Here's how you'd figure out his age for class:First,namethings andtranslatethe English into math: Let "E" stand for my age in 2000, and let "W" stand for William's age. ThenE= 11W+ 1in the year 2000 (from "eleven times as much, plus another one"). In the year 2009 (nine years after the year 2000), William and I will each be nine years older, so our ages will beE+ 9andW+ 9. Also, I was seven more than three times as old as William was, soE+ 9 = 3(W+ 9) + 7 = 3W+ 27 + 7= 3W+ 34. This gives you two equations, each having two variables:ADVERTISEMENT

E= 11W+ 1E+ 9 = 3W+ 34If you know how tosolve systems of equations, you can proceed with those techniques. Otherwise, you can use the first equation to simplify the second: sinceE= 11W+ 1, plug "11W+ 1" in for "E" in the second equation:E+ 9 = 3W+ 34(11W+ 1) + 9 = 3W+ 3411W 3W= 34 9 18W= 24W= 3

Remember that the problem did not ask for the value of the variableW; it asked for the age of a person. So the answer is:William was three years old in January of 2000.

The important steps above were to set up thevariables, labelling them all clearly with theirdefinitions, and then to increment the variables by the required amount (in this case, by 9) to reflect the passage of time. Don't try to use the same expression to stand for two different things. If "E" stands for my age in 2000, then "E" can not alsostand for my age in 2009. Make sure that you are very explicit about this when you set up your equations; write down the two sets of information (our ages at the first time, and then our ages at the second time) as two distinct situations.

In three more years, Miguel's grandfather will be six times as old as Miguel was last year. When Miguel's present age is added to his grandfather's present age, the total is68. How old is each one now?Copyright Elizabeth Stapel 2000-2011 All Rights ReservedThis exercise refers not only to their present ages, but also to both their ages last year and their ages in three years, so labelling will be very important. I will label Miguel's present age as "m" and his grandfather's present age as "g". Thenm+g= 68. Miguel's age "last year" wasm 1. His grandfather's age "in three more years" will beg+ 3. The grandfather's "age three years from now" is six times Miguel's "age last year" or, in math:g+ 3 = 6(m 1)This gives me two equations with two variables:m+g= 68g+ 3 = 6(m 1)Solving the first equation, I getm= 68 g. (Note: It's okay to solve for "g= 68 m", too. The problem will work out a bit differently in the middle, but the answer will be the same at the end.) I'll plug "68 g" into the second equation in place of "m":g+ 3 = 6m 6g+ 3 = 6(68 g) 6g+ 3 = 408 6g 6g+ 3 = 402 6gg+ 6g= 402 37g= 399g= 57Since "g" stands for the grandfather's current age, thenthe grandfather is 57 years old. Sincem+g= 68, thenm= 11, andMiguel is presently eleven years old. One-half of Heather's age two years from now plus one-third of her age three years ago is twenty years. How old is she now?This problem refers to Heather's age two years in the future and three years in the past. So I'll pick a variable and label everything clearly:age now:Hage two years from now:H+ 2age three years ago:H 3Now I need certain fractions of these ages:one-half of age two years from now: (1/2)(H+ 2) =H/2+ 1one-third of age three years ago: (1/3)(H 3) =H/3 1The sum of these two numbers is twenty, so I'll add them and set this equal to20:H/2+ 1 +H/3 1 = 20H/2+H/3= 203H+ 2H= 1205H= 120H= 24Heather is24years old.Note: Remember that you can always check your answer to any "solving" exercise by plugging that answer back into the original problem. If Heather is24now, then she will be26in two years, half of which is13, and she was21three years ago, a third of which is7. Adding, I get13 + 7 = 20, so the solution works.

The following puzzler is an old one, but it keeps cropping up in various forms...

"Age" Word Problems(page 2 of 2)"Here lies Diophantus," the wonder behold . . .Through art algebraic, the stone tells how old:"God gave him his boyhood one-sixth of his life,One twelfth more as youth while whiskers grew rife;And then yet one-seventh ere marriage begun;In five years there came a bouncing new son.Alas, the dear child of master and sageAfter attaining half the measure of his fathers lifechill fate took him.After consoling his fate by this science of numbersfor four years, he ended his life."Find Diophantus' age at death.My first task is to "translate" the poetry from the headstone into practical terms: "Boyhood" can stand for pre-adolscent childhood; he spent one-sixth of his life in this period.Copyright Elizabeth Stapel 2000-2011 All Rights Reserved "Youth while whiskers grew" can stand for pubescence (the teenage years, into young adulthood); he spent one-twelfth of his life in this period. "Ere marriage began" can stand for "unmarried adulthood" or "bachelorhood"; he spent one-seventh of his life in this period. He had five years between the wedding and the time his first child was born. Tragically, this child died young, living only half as long as his father eventually would; looked at the other way, half of Diophantus' life was spent while his child was alive. Diophantus died four years after burying his child.

I will letdstand for Diophantus' age at death. Then:childhood:d/6adolescence:d/12bachelorhood:d/7childless marriage:5age of child at death:d/2life after child's death: 4His whole life had been divided into intervals which, when added together, give the sum of his life. So I'll add the lengths of those periods, set their sum equal to his (as-yet unknown) total age, and solve:

.d/6+d/12+d/7+ 5 +d/2+ 4 =d(25/28)d+ 9 =d9 =d (25/28)d9 = (3/28)d84 =dDiophantus lived to be84years old.You can check the answer if you like, by plugging "84" into the original problem. If he lived to be84, then one-sixth of his life is14years, one-twelfth of his life is7years (so he'd be21, and he certainly should have a beard by this age), one-seventh of his life is12years (so he didn't marry until he was33years old), his child was born when he was38, the boy died at42(when Diophantus was80), and then Diophantus died four years later.

Always try to label your variables and expressions clearly, as this will go a long way toward helping you get your equations set up correctly. And remember that you can always check your answers (like I did on the last example above); checking your answers is an especially good idea on tests.

Question 126539:If a boat goes downstream 72 miles in 3 hours and upstream 60 miles in 6 hours, the rate of the river and the rate of the boat in still water respectively are ________?

The basic formula to remember here isordistance/time = rate1. First we need a rate of speed for travel both downstream and upstream.a. Downstream:The boat travelled 72 miles in 3 hours ormiles per hour.b. Upstream:The boat travelled 60 miles in 6 hours ormiles per hour.2. Now that we have these rates we can establish some variablesLet x = the rate of the boat in still waterLet y = the rate of the riverLet x + y = 24 (downstream rate of boat)Let x - y = 10 (upstream rate of boat)3. Add the 2 equations:

+

3. Substite and solve for y:

4. Check using the other equation:

This is true so our answers are correct.The boat's speed in still water is 17 mph.The rate of the river is 7 mph.

Question 285932:Upstream, downstream. Juniors boat will go 15 miles per hour in still water. If he can go 12 miles downstream in the same amount of time as it takes to go 9 miles upstream, then what is the speed of the current?

D = R*T, where D=distance, R=rate of speed, and T=time.S = speed in calm water = 15C = current speedS-C = speed going upstreamS+C = speed going downstream.T = D/R.Upstream...T = 9/(S-C).Downstream...T = 12/S.Transitive property lets us eliminate T...which is good since we don't know what T is...9/(S-C) = 12/S.Cross multiply...9S = 12*(S-C) = 12S - 12C.Substitute S=15...

9(15) = 12(15) - 12C.135 = 180 - 15C.Add 15C to both sides...15C + 135 = 180.Subtract 135 from both sides...15C = 45.Divide both sides by 15C = 3 = current speed.Checking the answer by checking the time up and downstream...T = 9/(S-C) = 9/9 = 1 hr.T = 15/(S+C) = 15/(12+3) = 15/15 = 1 hr.Correct..Answer:The current is 3 mph.

A boat travels 18 km downstream in 2 hours. It requires 6 hours to travel back to its original starting point upstream. What is the rate of the boat in still water and what is the rate of the current?

Given: Upstream Time = 6 HOURSGiven: Downstream Time = 2 HOURSGiven: Distance = 18 KILOMETERS DistanceFind V Speed in still waterFind C = Speed of Current(V+C) * 2 = 18 Downstream Equation(V-C) * 6 = 18 Upstream Equation V+C = 18/2V-C = 18/6 2V = 18/2 + 18/6V = 6 KM/H 6 + C = 9C = 3 KM/H

C = 3 KM/H

Algebra Help: How to Solve Boat-in-the-River Word Problems

Another common math word problem that the Algebra student should know how to solve are the "boat-in-the-river" word problems. These are actually just a variation of the dreaded uniform motion word problems which I've discussed in two previous articles. With the boat in the river problems, we assume that the boat has a uniform speed in still water and that the speed of the water (or the speed of the river current) is constant. The Algebra student will be presented with a math word problem in which he must solve for the speed of the boat (in still water, implied), or the speed of the river current, or the time spent going upstream or the time going downstream or some combination of these variables.With these problems, Saxon uses the following variables:B = speed of the boat in still waterW = the speed of the water (or the rate of the river current)Td= time spent going downstreamTu= time spent going upstreamDd= distance gone downstreamDu= distance gone upstreamAs with other uniform motion word problems, an important algebraic equation to remember is:Distance = Rate * TimeWith the boat-in-the-river math problem, we modify this slightly when we are going downstream to get:Dd= (B + W) * TdThat is, the speed (or rate) of the boat going downstream is the speed of the boat in still water plus the speed of the water (or river current). That is because when one is going downstream, the river works with you.When a boat is going upstream, the river works against you so we have to change our distance equation to reflect this by subtracting the speed of the water from the speed of the boat to get our rate:Du= (B - W) * TuThe best way to understand how to solve these is to work through a couple examples.Sample Boat-in-the-River Word Problem #1:Hannah and Fred can go 45 miles downstream in their boat in the same time it takes them to go 15 miles upstream. If the speed of the current was 3 mph, what was the speed of their boat in still water and how long did it take for them to go 45 miles downstream?With these kinds of word problems I recommend to the Algebra student that you always start out by writing down the two essential distance equations, before anything else:Dd= (B + W) * TdDu= (B - W) * TuSecondly we need to write down the information we know from dissecting the text of the word problem.Dd= 45Td= Tu= T(Since we know that the two times are the same, let's just use T for our time variable)Du= 15W = 3Now we can plug in our known values to get our system of two equations:45 = (B + 3) * T15 = (B - 3) * TSimplify and rearrange the equations so that they are easier to solve:B*T + 3T = 45B*T - 3T = 15We see that we can easily add the two equations to get a value for B*T:2B*T = 60B*T = 30We can now substitute B*T into the first equation (or the second, either would work) to solve for T:30 + 3T = 453T = 15T = 5Which then tells us that B = 6.The speed of the boat is 6 mph and the time spent going downstream (which is the same as the time spent going upstream) was 5 hours.Sample Boat-in-the-River Word Problem #2:The boat Murphy's Law could go 91 miles downstream in 7 hours, but required 12 hours to go 84 miles upstream. What were the speed of the current and the speed of the boat in still water?Our essential equations:Dd= (B + W) * TdDu= (B - W) * TuKnown values from the word problem:Dd= 91Td= 7Tu= 12Du= 84Plug in the values to get our equations:91 = (B + W) * 784 = (B - W) * 12Simplify and rearrange the equations:7B + 7W = 9112B - 12W = 84We now must modify these two equations so that we can add them together to eliminate the W variable. We can do this by multiplying the top equation by 12 and the bottom equation by 7:84B + 84W = 109284B - 84W = 588Adding the two equations gives us:168B = 1680B = 10Plug in the B to the original first equation to solve for W:7(10) + 7W = 917W = 21W = 3The speed of the boat was 10 mph and the speed of the current (water) was 3 mph.

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Word Problems Involving Systems of Linear Equations