Systems of Linear Equations and Word Problems

This section covers Introduction to SystemsSolving Systems by Graphing; Solving Systems with Substitution Solving Systems with Linear Combination or EliminationTypes of Equations;  Algebra Word Problems with SystemsInvestment Word ProblemMixture Word ProblemDistance Word Problem, Which Plumber ProblemGeometry Word Problem, Work Problem, and Three Variable Word Problem.

Introduction to Systems

“Systems of equations” just means that we are dealing with more than one equation and variable.  So far, we’ve basically just played around with the equation for a line, which is y = mx + b.

But let’s say we have the following situation.  You’re going to the mall with your friends and you have $200 to spend from your recent birthday money. You discover a store that has all jeans for $25 and all dresses for $50.  You really, really want to take home 6 items of clothing because you “need” that many new things.

Wouldn’t it be clever to find out how many pairs of jeans and how many dresses you can buy so you use the whole $200 (tax not included – your parents promised to pay the tax)?

Now, you can always do “guess and check” to see what would work, but you might as well use algebra!   It’s much better to learn the algebra way, because even though this problem is fairly simple to solve, the algebra way will let you solve any algebra problem – even the really complicated ones.

The first trick in problems like this is to figure out what we want to know.  This will help us decide what variables (unknowns) to use.  So what we want to know is how many pairs of jeans we want to buy (let’s say “j”) and how many dresses we want to buy (let’s say “d”).  So always write down what your variables will be:

  Let j = the number of jeans you will buy

Let d = the number of dresses you’ll buy

Like we did before, let’s translate word-for-word from math to English.  Always write down what your variables are in the following way:

Now we have the 2 equations as shown below.  Notice that the j variable is just like the x variable and the d variable is just like the y.  It’s easier to put in j and d so we can remember what they stand for when we get the answers.

This is what we call a system, since we have to solve for more than one variable – we have to solve for 2 here.  The cool thing is to solve for 2 variables, you typically need 2 equations, to solve for 3 variables, you need 3 equations, and so on.  That’s easy to remember, right?

We need to get an answer that works in both equations; this is what we’re doing when we’re solving; this is called solving simultaneous systems, or solving system simultaneously.

There are several ways to solve systems; we’ll talk about graphing first.

Solving Systems by Graphing

Remember that when you graph a line, you see all the different coordinates (or x/y combinations) that make the equation work.  In systems, you have to make both equations work, so the intersection of the two lines shows the point that fits both equations (assuming the lines do in fact intersect; we’ll talk about that later).  So the points of intersections satisfy both equations simultaneously. 

We’ll need to put these equations into the y = mx + b (d = mj + b) format, by solving for the d (which is like the y):

Now let’s graph:

We can see the two graphs intercept at the point (4, 2).  This means that the numbers that work for both equations is 4 pairs of jeans and 2 dresses!

We can also use our graphing calculator to solve the systems of equations:

(Note that with non-linear equations, there will most likely be more than one intersection; an example of how to get more than one solution via the Graphing Calculator can be found in the Exponents and Roots in Algebra section.)

Solving Systems with Substitution

Substitution is the favorite way to solve for many students!  It involves exactly what it says: substituting one variable in another equation so that you only have one variable in that equation.  So below are our two equations, and let’s solve for “d” in terms of “j” in the first equation.  Then, let’s substitute what we got for “d” into the next equation.

Even though it doesn’t matter which equation you start with, remember to always pick the “easiest” equation first (one that we can easily solve for a variable) to get a variable by itself.

So we could buy 4 pairs of jeans and 2 dresses.

Note that we could have also solved for “j” first; it really doesn’t matter.  You’ll want to pick the variable that’s most easily solved for.

Let’s try another substitution problem that’s a little bit different:

Solving Systems with Linear Combination or Elimination

Probably the most useful way to solve systems is using linear combination, or linear elimination.   The reason it’s most useful is that usually in real life we don’t have one variable in terms of another (in other words, a “y =” situation).

The main purpose of the linear combination method is to add or subtract the equations so that one variable is eliminated.  Now let’s see why we can add, subtract, or multiply both sides of equations by the same numbers – let’s use real numbers as shown below.  Remember these are because of the Additive Property of Equality, Subtraction Property of Equality, Multiplicative Property of Equality, and Division Property of Equality:

So now if we have a set of 2 equations with 2 unknowns, we can manipulate them by adding, multiplying or subtracting (we usually prefer adding) so that we get one equation with one variable.  For, example, let’s use our previous problem:

So we could buy 4 pairs of jeans and 2 dresses.

Here’s another example:

Types of equations

In the example above, we found one unique solution to the set of equations.  Sometimes, however, there are no solutions (when lines are parallel) or an infinite number of solutions (when the two lines are actually the same line, and one is just a “multiple” of the other) to a set of equations.

When there is at least one solution, the equations are consistent equations, since they have a solution.  When there is only one solution, the system is called independent, since they cross at only one point.  When equations have infinite solutions, they are the same equation, are consistent, and are called dependent or coincident (think of one just sitting on top of the other).

When equations have no solutions, they are called inconsistent equations, since we can never get a solution

Here are graphs of inconsistent and dependent equations that were created on the graphing calculator:

Inconsistent Equations Dependent Equations Consistent Equations

Systems with Three Equations

Let’s get a little more complicated with systems; in real life, we rarely just have two unknowns with two equations.

So let’s say at the same store, they also had pairs of shoes for $20 and we managed to get $60 more from our parents since our parents are so great!

Now we have a new problem: to spend the even $260, how many pairs of jeans, dresses, and pairs of shoes should we get if want say exactly 10 total items?

Let’s let j = the number of pair of jeans, d = the number of dresses, and s = the number of pairs of shoes we should buy.

So far we’ll have the following equations:

We’ll need another equation, since for three variables, we need three equations (otherwise, we’d theoretically have infinite ways to solve the problem).  In this type of problem, you would also have/need something like this:  we want twice as many pairs of jeans as pairs of shoes.  Now, since we have the same number of equations as variables, we can potentially get one solution for the system.

So, again, now we have three equations and three unknowns (variables).  We’ll learn later how to put these in our calculator to easily solve using matrices (see the Matrices and Solving Systems with Matrices section) , but for now we need to first use two of the equations to eliminate one of the variables, and then use two other equations to eliminate the same variable:

Now this gets more difficult to solve, but remember that in “real life”, there are computers to do all this work!

The trick to do these problems “by hand” is to keep working on the equations using either substitution or elimination until we get the answers.

Remember again, that if we ever get to a point where we end up with something like this, it means there are an infinite number of solutions:

     4 = 4     (variables are gone and a number equals another number and they are the same)

And if we up with something like this, it means there are no solutions:

     5 = 2     (variables are gone and two numbers are left and they don’t equal each other)

So let’s go for it and solve   :

So we could buy 6 pairs of jeans, 1 dress, and 3 pairs of shoes.

Here’s one more example of a three variable system of equations, where we’ll only use linear elimination:

I know – this is really difficult stuff!  But if you do it step-by-step and keep using the equations you need with the right variables, you can do it.  Think of it like a puzzle – you may not know exactly where you’re going, but do what you can in baby steps, and you’ll get there (sort of like life!).

And we’ll learn much easier ways to do these types of problems.

Also – note that equations with three variables are represented by planes, not lines (you’ll learn about this in Geometry).  They could have 1 solution (if all the planes crossed in only one point), no solution (if say two of them were parallel), or an infinite number of solutions (say if two or three of them crossed in a line).  OK, enough Geometry for now!

Algebra Word Problems with Systems

Let’s do more word problems; you’ll notice that many of these are the same type that we did earlier in the Algebra Word Problems section, but now we can use more than one variable.  This will actually make the problems easier!

Again, when doing these word problems:

  • If you’re wondering what the variable (or unknown) should be when working on a word problem, look at what the problem is asking.  This is usually what your variable is!
  • If you’re not sure how to set up the equations, use regular numbers (simple ones!) and see what you’re doing.  Then put the variables back in!

Investment Word Problem:

Suppose Lindsay’s mom invests $10000, part at 3%, and the rest at 2.5%, in interest bearing accounts.   The totally yearly investment income (interest) is $283.  How much did Lindsay’s mom invest at each rate?

Solution:

So we always have to define a variable, and we can look at what they are asking.   Since we’ve learned about systems, let’s use two variables:   let x = the amount of money invested at 3%, and y = the amount of money invested at 2.5%.

Remember that the yearly investment income or interest is the amount that we get from the yearly percentages.  (This is the amount of money that the bank gives us for keeping our money there.)  To get the interest, we have to multiply each percentage by the amount invested at that rate.  We can add these amounts up to get the total interest.

So we have two equations and two unknowns.  We know that the total amount (x + y) must equal 10000, and we also know that the interest (.03x + .025y) must equal 283:

We also could have set up this problem with a table:

Mixture Word Problem:

Two types of milk, one that has 1% butterfat, and the other that has 3.5% of butterfat, are mixed.  How many liters of these two different kinds of milk are to be mixed together to produce 10 liters of low-fat milk, which has 2% butterfat?

Solution:

(Note that we did a similar mixture problem using only one variable here in the Algebra Word Problems section.)

Let’s first define two variables for the number of liters of each type of milk.  Let x = the number of liters of the 1% milk, and y = the number of liters of the 3.5% milk.  Let’s use a table again:

We can also set up mixture problems with the type of figure below.  We add up the terms inside the box, and then multiply the amounts in the boxes by the percentages above the boxes, and then add across.  This will give us the two equations.

Now let’s do the math!

Distance Word Problem:

Lia walks to the mall from her house at 5 mph.  10 minutes later, Lia’s sister Megan starts riding her bike at 15 mph (from the same house) to the mall to meet Lia.  They arrive at the mall the same time.  How far is the mall from the sisters’ house?  How long did it take Megan to get there?

Solution:

OK, this is another tough one.  Remember always that distance = rate x time.  It’s difficult to know how to define the variables, but usually in these types of distance problems, we want to set the variables to time, since we have rates, and we’ll want to set distances equal to each other (the house is always the same distance from the mall).

Let’s let L equal the how long (in hours) it will take Lia to get to the mall, and equal to how long (in hours) it will take Megan to get to the mall.  (Sometimes we’ll need to add the distances together instead of setting them equal to each other.)

We must use the distance formula for each of them separately, and then we can set their distances equal, since they are both traveling the same distance (house to mall).

Let’s draw a picture and work the problem:

Which Plumber Problem:

Many word problems you’ll have to solve have to do with an initial charge or setup charge, and a charge or rate per time period.  In these cases, the initial charge will be the y-intercept, and the rate will be the slope.  Here is an example:

Michaela’s mom is trying to decide between two plumber companies to fix her sink.  The first company charges $50 for a service call, plus an additional $36 per hour for labor.  The second company charges $35 for a service call, plus an additional $39 per hour of labor. At how many hours will the two companies charge the same amount of money?

In these cases, the money spent depends on the plumber’s set up charge and number of hours, so let y = total cost of the plumber, and x = number of hours of labor.  And again, set up charges are typical y-intercepts, and rates per hour are slopes.

Solution:

To get the number of hours when the two companies charge the same amount of money, we just put the two y’s together and solve for x (substitution, right?):

Plumber Problem

Geometry Word Problem:

Many times we’ll have a geometry problem as an algebra word problem; these might involve perimeter, area, or sometimes angle measurements (so don’t forget these things!).  Let’s do one involving angle measurements.

Two angles are supplementary.  The measure of one angle is 30 degrees smaller than twice the other.  Find the measure of each angle.

Solution:

We have to know that two angles are supplementary if their angle measurements add up to 180 degrees (and remember also that two angles are complementary if their angle measurements add up to 90 degrees, in case you see that).

Let’s define the variables and turn English into Math.  Let x = the first angle, and y = the second angle; we really don’t need to worry at this point about which angle is bigger; the math will take care of itself.

Then we know that x plus y must equal 180 degrees by definition, and also x = 2y – 30. (Remember the English-to-Math chart?)  Let’s solve:

See – these are getting easier!  

Here’s one that’s a little tricky though:
Work Problem

8 women and 12 girls can paint a large mural in 10 hours.  6 women and 8 girls can paint it in 14 hours.  Find the time to paint the mural, by 1 woman alone, and 1 girl alone.

Solution:

(This is a “work problem” that is typically seen when studying Rational Equations – fraction with variables in them –  and can be found here in the Rational Expressions and Functions section.)   But let’s solve it with using systems.  (There’s also a simpler version of this problem here in the Direct, Inverse, Joint and Combined Variation section).

Let’s let w = the part of the job by 1 woman in 1 hour, and g = the part of the job by 1 girl in 1 hour.  Let’s set up and solve:

Work Problems with SystemsLet’s do one more with three equations and three unknowns:

Three Variable Word Problem:

A florist is making 5 identical bridesmaid bouquets for a wedding.  She has $610 to spend (including tax) and wants 24 flowers for each bouquet.  Roses cost $6 each, tulips cost $4 each, and lilies cost $3 each.  She wants to have twice as many roses as the other 2 flowers combined in each bouquet.  How many roses, tulips, and lilies are in each bouquet?

Solution:

Let’s look at the question that is being asked and define our variables:  Let r  = the number of roses, t  = the number of tulips, and l  = the number of lilies.  So let’s put the money terms together, and also the counting terms together:

Now let’s do the math:

Systems of Equations Three Equations

The “Candy” Problem

Sometimes we get lucky and can solve a system of equations where we have more unknowns (variables) then equations.   (Actually, I think it’s not so much luck, but having good problem writers!)    Here’s one like that:

Sarah buys 1 lb of jelly beans and 2 lb of chocolates for $2.00.  She then buys 1 lb of jelly beans and 2 lbs of caramels for $3.00.  She also buys 1 lb of jelly beans, 3lbs of licorice and 2 lbs of caramels for $1.50.   How much will it cost to buy 1lb of each of the four candies?

Solution:

Let’s look at the question that is being asked and define our variables:  Let j = the cost of 1 lb of jelly beans, o = the cost of 1 lb of chocolates, l = the cost of 1 lb of licorice, and c = the cost of 1 lb of caramels.  So we have this system of equations:

Candy Problem

Now let’s try to do the math:

Candy Problem Math

Understand these problems, and practice, practice, practice!

On to Introduction to Functions – you’re ready! 

57 thoughts on “Systems of Linear Equations and Word Problems

  1. this did not help me at all… i asked for help to help me with my homework and thhis makes no sense… the problem i have is:
    The yellow pages identify two different electrical businesses. Business A charges $50 for a service call, plus an additional $36 per hour for labor. Business B charges $35 for a service call, plus an additional $39 per hour of labor. At how many hours will the two companies charge the same amount of money?
    this site did not help with this question

      • i have one question

        8 men and 12 boys finish apiece of work in 10 hours .And 6 men and 10 boys finishes the same work in 14 hours what is the speed of one man and that of boy yo finsh the work

        • This is a great problem – I think I’ll add it to my site. Let m = part of job by one man in one hour, and b = the part of the job by 1 boy in one hour. So we have 10 hours with 8 men and 12 boys that do 1 job, and 14 hours with 6 men and 8 (I think you mean 8, not 10 – 10 will not work) boys that do 1 job. (sort of like rate * time = distance, but distance = 1).
          So 10(8m + 12b) = 1, and 14(6m + 8b) = 1, so solving this, m = 1/140 and b = 1/280. So it would take 140 hours for one man to do the job, and 280 for one boy.

          I went ahead and added to my site here and here.

  2. Can you help me solve this problem?

    An executive in an engineering firm earns a monthly salary plus a Christmas bonus of 6100 dollars. If she earns a total of 90600 dollars per year, what is her monthly salary in dollars?

    • Yes! Let x = the first integer; 2x + 3 = the second integer. x + 2x + 3 = 51, so 3x = 48, so x = 16. The other integer is 35. Does that make sense?
      Lisa

    • Good question, Meera!!
      Take a look at it again – I tried to add more steps to make it more clear. We are using the middle equation to substitute into the first and third equations. Does that makes sense? Lisa

  3. I completely understand thanks so much. See I have been studying all summer on algebra 1 because I want to take test that lets me skip it so this website let me move past the systems of equations unit!

    Thanks

  4. I have a doubt. In a school cricket tournament was held. Sachin hits only fours and sixes.He scored 112 runs. Number of fours were twice the number of sixes. How many fours and sixes he scored? Please answer me for dis as soon as poss. Thank you.

    • I don’t know cricket at all, but here goes ;) Let f = the number of fours that he scores and s = the number of sixes that he scores. 4f + 6s = 112. f = 2s. Solve by substitution: 4(2s) + 6s = 112; 14s = 112; s = 8. f = 2s = 16. So he scores 8 sixes and 16 fours. Does that make sense? Lisa

  5. can you help me lisa?
    given two numbers. The second number equal to six times the first number after the minus one. The second number is also equal to the first number squared and added three. determine the number?

    • Sure – I can try! Let x = the first number and y = the second number. y = 6(x – 1); y = x^2 + 3 Is that what you meant? Then put the two y’s together and get 6x-6=x^2+3. Then put everything to one side, and get x^2 -6x +9 = 0. Then factor to get (x-3)(x-3). So the first number is 3 and the second is 12. Does that make sense?

  6. algebra word problems. need helps..question is as follows..
    A marketing director notices that the sales level for a certain product and amount spent on television advertising are linearly related. When $6000 is spent on television advertising, sales for the product are $255,000, and when $8000 is spent on television advertising, sales for the product are $305,000. Find an equation that gives the sales level for the product in terms of the amount spent on television advertising. Very urgent need..

    • Hi! Thanks for writing. I put the numbers in my calculator and did a linear regression (is that what you guys are doing?) and got
      Y= 25x + 105000

      Hope that helps.
      Lisa

  7. Can any help mi to solve this problem!!!!
    A small garden is divided by a fence into two parts, a square and an isosceles triangle, as given in the following diagram (not to scale). The perimeter of the whole garden is 5 metres longer than the perimeter of the square part. The perimeter of the triangular part is equal to the perimeter of the square part. Find the width w in metres.

  8. I got 2 equations and 2 unknowns : t is side of triangle and w is side of square (width?).

    2t + 3w = 3w + 5
    2t = 3w
    W= 5/3. T= 5/2
    Hope this helps!
    Lisa

  9. On your example Solving Systems with Substitution, where did the 300 come from? This makes it very confusing to follow. My question is similar. Where do I pull out a number that is not in the equation?

    A hospital has a budget of $1500 to purchase aspirin and penicillin. The pharmaceutical company sells aspirin for $50 a box and penicillin for $75 a box. How many boxes of each can they purchase? The answer is simple but it is confusing putting it into a formula according to your example of Solving Systems with Substitution.

    • Hi! Thanks so much for writing!
      I think you mean the 300 that comes from “pushing through” the 50 and multiplying it first by -j and then by 6. This is the distributive property – when you get rid of parentheses like that, you have to multiply the outside number by both things inside. Does that make sense?
      In your example, I don’t think I can solve the problem until I know how many total boxes of aspirin and penicillin they need to purchase? If we have two unknowns, we need two equations. Do you have more information from that problem?
      Thanks,
      Lisa

  10. Just stopping by to say thanks. Your site is a big help when it comes to puzzling through these problems and getting them to make sense. So…

    Thank you!

  11. Daniel buys 1 lb of jelly beans and 2 lb of chocolates for $2.00. He then buys 1lb of jelly beans and 2 lbs of caramels for $3.00. He also buys 1 lb of jelly beans, 3lbs licorice and 2 lbs caramels for $1.50. How much will it cost to buy 1lb of each of the four candies?

    1j+2x=2.00
    1j+2c= 3.00
    1j+3l+2c=1.5

    • Hi! Thanks for writing. This is a strange problem (I think I’ll add it!) since there are less equations than unknowns. BUT it works out that all when we got to get j + x + l + c, but substituting all variables in terms of j, all the j’s cross out, and we just end up with 2. So 2 is the answer.

      Try that – put the other variables in terms of j and then see what you get for j + x + l + c. Let me know if you need more help for now on this.
      Lisa

      • I have a better way to solve the problem.
        Add equation 1 and 2
        I get 2j+2o+2c=5, then I know j+o+c=2.5.
        now the problem becomes finding the value of l.
        from equation 2, i know j+2c=3, then equation 3
        j+3l+2c=3+3l=1.5, 3l=-1.5, then l=-0.5.
        so j+o+c+l will be 2.5-0.5=2.

  12. Could you help me to solve this problem:

    A sports apparel manufacturer plans to sell a new set of products. Cost charged to the retailer is RM 33 per set. What is the price to be charged by the manufacturer to the retailer so that retailers can reduce the price by 20% and still earn profit of 15% of the cost?

    Thank you

    • Could you explain this problem a little better? You say “cost charged to the retailer is RM 33″ and then say what is the price to be charged by the manufacturer to the retailer? What is the difference of these 2 things? Thanks! Lisa

  13. Can you please give me three examples of word problems of linear equation in one variable about work? Please, I just need those for my project. Please help me.

    • Sure! 1) A company hired Ann with a $5K bonus and $35K a year. How much money will she have earned in x years? F(x) = 35000x + 5000
      2) A jewelry store pays Laura $40K a year plus 2% of every sale she makes. How much will she make in a year if she sells x dollars in jewelry? F(x) = 40000 + .02x
      3) A clothing store has a huge sale with everything 75% off. How much will Julie spend if she buys clothes that are worth x dollars at the regular price? F(x) = .25x

      Hope that helps – Lisa

  14. My daughter needs help with problem number 13.
    Thank you for your help.

    Mrs. Travis wants to have a clown deliver balloons to her secretary’s office. Clowns R Fun charges $1.25 per balloon and $6 delivery. Singing Balloons charges $1.95 per balloon and $2 for delivery. What is the minimum number of balloons Mrs. Travis needs to purchase in order for Clowns R Fun to have a lower price than Singing Balloons?

    • Thanks for writing. Here’s how I would solve the problem:
      You want to find where the two equations intersect. Let b = the number of balloons that Mrs. Travis would need to purchase – so it would be 1.25b + 6 = 1.95b + 2. Solve for b and get around 5.7. But plugging in 5.7, you’d still have Clowns R Run costing more. So they would need to purchase 6 balloons.

      • “Sarah buys 1 lb of jelly beans and 2 lb of chocolates for $2.00. She then buys 1 lb of jelly beans and 2 lbs of caramels for $3.00. She also buys 1 lb of jelly beans, 3lbs of licorice and 2 lbs of caramels for $1.50. How much will it cost to buy 1lb of each of the four candies?”
        This is an absurd problem as i get the price for 1lb licorice is -$0.5. What does it mean? When I buy 1lb licorice, the business owner gives me $0.5? I think the better way is change $1.5 to $4.5. That will make the problem sense!

        • Yes – I did this problem after someone asked me earlier:
          Daniel buys 1 lb of jelly beans and 2 lb of chocolates for $2.00. He then buys 1lb of jelly beans and 2 lbs of caramels for $3.00. He also buys 1 lb of jelly beans, 3lbs licorice and 2 lbs caramels for $1.50. How much will it cost to buy 1lb of each of the four candies?

          1j+2x=2.00
          1j+2c= 3.00
          1j+3l+2c=1.5

          Reply ↓

          lisa
          on December 9, 2013 at 10:23 pm said: Edit
          Hi! Thanks for writing. This is a strange problem (I think I’ll add it!) since there are less equations than unknowns. BUT it works out that all when we got to get j + x + l + c, but substituting all variables in terms of j, all the j’s cross out, and we just end up with 2. So 2 is the answer.

          Try that – put the other variables in terms of j and then see what you get for j + x + l + c. Let me know if you need more help for now on this.
          Lisa

          Reply ↓

  15. To fill a tank from two tabs getting 15 days. Two tabs start till 12 days and close one tabe. To fill the getting more 8 days then find days fill tank to seprate tabs plz send me answer
    .

    • Here’s what I got: Since it takes 2 tabs to fill a tank in 15 days, after 12 days of both tabs, the “job” will be 12/15 done. Then with one tab on for the next 8 hours, we can solve the equation 12/15 + 8/x = 1 to get x = 40 days for one of the tabs to fill the job. Then put it back in the equation “together/alone + together/alone = 1″, 15/40 + 15/x = 1, we get the other tab will take 24 days to fill the tank. I’m not sure if this is correct – what do you think? Lisa

  16. Two families go to a hockey game. One family purchases two adult and four youth tickets for $28. Another family purchases four adult tickets for $45.50. Let x represent the cost of one adult ticket and y for the cost of one youth ticket. Write a linear system that represents this situation.

    I having issues solving it can someone help, please?

    • OK, let’s try to translate almost word for word from english to math: 2x + 2y = 28 and 4x = 45. Solve this last equation and get x = 11.25. Then plug this in for x in the first equation: 2(11.25) + 4y = 28; 4y = 28 – 22.50; 4y = 5.5; y = 1.375, or round to 1.38. So x = $11.25 and y = $1.38. Hope that makes sense :)

  17. I have this problem that I cannot figure out. I hope you can help me, please.

    A municipal committee decides to build 50 housing units for low income families. These housing units come in three types, A, B, and C. Type A units will cost $500K each to build and will provide the city with revenue from rental payments of the amount $25K per year. The corresponding numbers for type B units are $625K and $30K per year, and the numbers for type C units are $400K and $22.5K per year.

    If the total cost to build the 50 units was $24,750,000; and the yearly revenue from rental payments is $1,262,500; how many of each type of unit was built by the city?

    • Thanks for writing. I set up a system of equations with 3 equations and 3 unknowns: A + B + C = 50; 500A + 625B + 400C = 24750 (since the other numbers are in thousands), and 25A + 30B + 22.C = 1262.5.
      When I put the matrices in the calculator and solved, I got A = 25, B = 10, and C = 15.
      Let me know if you have any more questions ;)
      Lisa

  18. I am trying to learn how to use a TI 84 plus silver edition algebraic calculator. I keep getting “error” when I try to graph these two equations for the intersecting point.
    Please help.
    y = x – 4
    y = -x + 10

    • Hi and thanks for writing! Are you using the “minus” sign (under the X sign) for the first equation, and the “negative” sign (under the 3)? When are you getting the error? Sorry you are having trouble ;( Lisa

  19. A 500 g jar of mixed nuts contain 30% cashews, 20% almonds, and 50% peanuts.
    a. How many grams of cashews must you add to increase the percentage of cashews to 40%? What is the new percentage of almonds and peanuts?
    b. How many grams of almonds must you add to the original mixture to make the percentage of almonds and cashews the same? Now what is the percentage of each type of nut?

    c + a + p = 500
    .3c + .2a + .5p = 1(500)

    ? then what?

    • Thanks for writing! This is a tricky one; what I did for a) was to find out how many grams of cashews there were to begin with (.3 x 500 = 150 g. cashews out of 500g). Then I set up a table and came up with the following formula: 150 + x = .4(500 + x), and got about 83.3 grams of cashews to add to the 150g to get to about 233.3g (out of a total 583.3). I also see that originally there were 100g of almonds (20% of 500g) and 250g of peanuts (50% of 500g). So after adding the 83.3g of cashews, we’d have 100/583.3=17.1438% of almonds and 250/583.3g of peanuts or 42.859%. Add up all the percentages now and you get 100% – yeah!
      For b), I just used the equation (100 + x)/(500 + x) = 150/(500 + x), and got 50 g of almonds to add – then the % of almonds and cashews will both be 27.27% and the % of peanuts will go down to (250/550) or 45.45%.
      Hope this helps!
      Lisa

  20. I would greatly appreciate your help solving this word problem. Thanks in advance!

    You have $100 to spend on a barbeque where you want to serve chicken and steak. Chicken costs $1.29 per pound and steak costs $3.49 per pound. Find a function that relates the amount of chicken and the amount of steak you can buy. Then graph the function. What is the meaning of the slope in this context? Use this ( and any other information represented by the equation or graph) to discuss what your options are for the amounts of chicken and amount of steak you can buy for the barbeque.

    • Thanks for writing! Good problem! If you let x = lbs of chicken you can buy and y = amount of steak you can buy, you get 1.29x + 3.49y = 100. Solve for y to get the function: y = 100/3.49 – (1.29/3.49)x. The slope is -1.29/3.49, or about -.37. It’s negative, so the positive amount of the slope would represent how much less steak (lbs) you could buy if you were to buy 1 more lb of chicken. You also have to remember that the domain (how much chicken you can buy) has to start at 0 and end at around 77.52, so you don’t have negative amounts of chicken or steak. I could see that by graphing the function. For the steak, you can buy 0 lbs up to about 28.65 (the y intercept). Does this make sense? Can you see how to graph it? Lisa

  21. It’s starting to make sense,many thanks to you, but I’m still not sure about how to graph it. Would the Y axis represent cost/dollars, and the X axis represent pounds? I’m not sure what increments to use for the cost, if this is correct. Your help is so much appreciated! Wendy

    • I would make x the number of lbs of chicken (you can make x go from 1 to say 80 by 10′s) and y the number of lbs of steak (make y go from 0 to 30 and count by 10′s). You will have a line coming down and sort of creating a triangle – a negative slope. Every point on the line represents how much steak you can buy (the y), given that much chicken bought (the x). Does that make sense?

  22. Can you please give me five examples of word problems of system of linear equation and quadratic equation. i need it for my project. pls. w/ sol.

  23. Maximizing the Yield for Stock Investments Using a System of Linear Inequalities with a Geometric Approach

    The financial manager of a company has $17,000 to invest in low-risk, medium-risk, and high-risk stocks. The amount invested in low-risk stocks, will be at most $3000 more than the amount invested in medium-risk stocks. At least $7,000 will be invested in low-risk and medium-risk stocks. No more than $13,000 can be invested in medium risk and high-risk stocks. The expected yields on the stocks are 6% for low-risk, 7% for medium-risk, and 8% for high-risk. The financial manager will choose the amount to invest in each type of stock in order to maximize the yield on the company’s investment.

    1.Name in words the three quantities that must be determined.

    2.Write the three quantities algebraically using only the variables x and y. Use the variables x and y to represent the first two quantities. Then use the variables x and y to write an expression representing the third quantity using the fact that the total amount to be invested is $17,000.

    Write the complete set of inequalities needed to solve the problem using the given information. Simplify the expressions in the inequalities by combining like terms. Include inequalities specifying that the amount of each type of investment must be greater than or equal to 0.

    3.Write the objective function to compute the total yield from the three types of investments. Simplify the expression by multiplying to remove parentheses, and by combining like terms.

    4.Use scratch paper to graph the system of inequalities to determine the points of intersection on the boundary of the feasible set. Compute the coordinates of the points of intersection on this boundary by solving each set of 2 equations that intersect on the boundary of the feasible set. List the resulting points.

    5.Substitute the coordinates from each of the points computed in question 5 in the total yield equation from question 4 to test for the maximum total yield. Identify the point that results in the maximum total yield. What is the yield for this choice?

    6.Use the point identified in question 6 to compute the amounts to be invested in each type of stock in order to maximize the yield on the company’s investment.

    I am having the hardest time with this assignment I have been trying to figure it out for a week! Please help!

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