### Linear systems application word problems

This is the amount of money that the bank gives us for keeping our money there.

## Writing Systems of Linear Equations from Word Problems

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. We have two equations and two unknowns. We also could have set up this problem with a table: 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. Mixture Word Problem with Money: How much of each type of coffee bean should be used to create 50 pounds of the mixture?

See how similar this problem is to the one where we use percentages? Distance Word Problem: Lia walks to the mall from her house at 5 mph. They arrive at the mall the same time. How long did it take Megan to get there? OK, this is another tough one. 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.

- Word Problems Involving Systems of Linear Equations;
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In these cases, the initial charge will be the y -intercept , and the rate will be the slope. Here is an example:. At how many hours will the two companies charge the same amount of money? Two angles are supplementary.

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The measure of one angle is 30 degrees smaller than twice the other. Find the measure of each angle. We have to know that two angles are supplementary if their angle measurements add up to degrees and remember also that two angles are complementary if their angle measurements add up to 90 degrees, in case you see that.

Remember the English-to-Math chart? See — these are getting easier! Find the time to paint the mural, by 1 woman alone, and 1 girl alone. A florist is making 5 identical bridesmaid bouquets for a wedding. 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? Sometimes we get lucky and can solve a system of equations where we have more unknowns variables then equations. How much will it cost to buy 1 lb of each of the four candies?

We have this system of equations:. For Practice: Skip to content. Follow us: Share this page: This section covers: Always write down what your variables will be: Use easier numbers if you need to: We can also use our graphing calculator to solve the systems of equations: If you add up the pairs of jeans and dresses, you want to come up with 6 items. Graphing Calculator Instructions. This one is actually easier: The next problem is more complicated than the others, since it involves solving a system of three equations with three variables.

You'll see that I do it by substitution. If you take more advanced courses such as linear algebra , you'll learn methods for solving systems like these which are like the whole equation method. They involve representing the equations using matrices. Phoebe has some cent stamps, some cent stamps, and some 3-cent stamps. The number of cent stamps is 10 less than the number of cent stamps, while the number of 3-cent stamps is 5 less than the number of cent stamps. How many of each stamp does she have? I'll let x be the number of cent stamps, let y be the number of cent stamps, and let z be the number of 3-cent stamps.

Here's the table. 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. I'll solve for x in terms of y:.

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Plug and into and solve for y:. Phoebe has 20 cent stamps, 10 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.

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The sum of two numbers is The larger number is 14 more than 3 times the smaller number. Find the numbers. Plug into the first equation and solve:. The numbers are 19 and The next set of examples involve simple interest. Here's how it works. At the end of one interest period, the interest you earn is. You now have dollars 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"?

## Algebra - Applications of Linear Equations

In fact, "percent" is short for "per centum", and centum is the Latin word for a hundred. So "4 percent" means "4 per ". Since "per" translates to division, I get , as you probably know from earlier math courses. How much was invested in each account? Bonzo invests some money at interest. How much was invested at each rate?

There are various kinds of mixture problems. The first few involve mixtures of different things which cost different amounts per pound. How many pounds of each kind of candy did he use in the mix? How many pounds of raisins and how many pounds of nuts should she use? Solve the equations by multiplying the first equation by 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. The last line says.

## System-of-Equations Word Problems

Solve for x:. An alloy is a mixture of different kinds of metals. Suppose you have 50 pounds of an alloy which is silver. 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 and pounds 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.

Phoebe mixes an alloy containing silver with an alloy containing silver to make pounds of an alloy with silver. How many pounds of each kind of alloy did she use? She used 60 pounds of the alloy and 40 pounds of the alloy. Other mixture problems involve solutions. For instance, a solution may be acid, or alcohol. What does this mean? Suppose you have 80 gallons of a solution which is acid.

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 and gallons 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. How many gallons of each of a acid solution and an acid solution must be mixed to produce 50 gallons of a acid solution? Use 15 gallons of the solution and 35 gallons of the solution. Amounts of a alcohol solution and a alcohol solution are to be mixed to produce 24 gallons of a alcohol solution.

How many gallons of the alcohol solution and how many gallons of the alcohol solution should be used? Suppose x gallons of the alcohol solution and y gallons of the alcohol solution are used. Thus, 16 gallons of the solution and 8 gallons of the solution must be used.