Solving Systems of Equations With 3 Variables & Word Problems

Summary

In this engaging tutorial, The Organic Chemistry Tutor walks us through the process of solving systems of equations with three variables and tackles some intriguing word problems. The video begins with a step-by-step guide on how to solve a system involving three linear equations with variables x, y, and z. The instructor demonstrates how to strategically eliminate variables and solve for the unknowns. Additionally, viewers are guided through practical word problems involving investments and fruit costs, providing real-world applications for these mathematical concepts. The session wraps up with clear solutions and explanations to enhance understanding.

Highlights

• Engage with three-equation system solving: add, subtract, and conquer! 🧮
• Learn to handle investment word problems with ease. 💰
• Crack the cost of fruit conundrum with simple equations. 🍎

Key Takeaways

• Eliminate variables effectively by pairing equations. 🤓
• Transform percentage into decimal to solve equations. 📈
• Word problems make equations fun and practical! 🍏

Overview

Ever found yourself scratching your head over systems of equations? The Organic Chemistry Tutor is here to make it crystal clear. Kicking off with a classic three-variable system, you're invited to practice choosing pairs of equations to eliminate one variable at a time. It's a seamless strategy that involves a bit of rewriting and a sprinkle of arithmetic, leading to the brilliant 'aha!' moment when everything clicks into place.

Once the basics are handled, dive headfirst into an investment problem. Here, you’ll convert percentages into decimals and unravel the mystery of balancing two separate accounts. By transforming the problem into equations and swiftly solving with elimination, you’ll see how much went into each account and how interest works its magic.

To bring it all home, embark on a marketplace adventure with apples and bananas. Facing a real-world problem of determining costs based on known expenses, you’ll employ the same elimination technique and end up calculating the cost of bundles of fruit. This segment reinforces the practicality of mastering equations and wraps up the tutorial in a satisfyingly comprehensive manner.

Chapters

• 00:00 - 00:30: Introduction: Solving Systems of Equations The chapter titled 'Introduction: Solving Systems of Equations' begins by posing the question of how to solve a system of three equations. An example system is presented: (1) 2x + y + z = 7, (2) 2x - y + 2z = 6, (3) x - 2y + z = 0. These equations form the basis for exploring techniques to find solutions to systems of linear equations.
• 00:30 - 01:00: Choosing Equations and Simplification In this chapter, the focus is on the selection and simplification of equations, specifically how to choose two equations out of three to solve a problem. The method demonstrated involves choosing the first two equations, as adding them together allows one to cancel out the variable 'y'. The instructor encourages viewers to pause the video and attempt to solve the problem themselves if they are able. The process starts with rewriting the chosen equations.
• 01:00 - 01:30: Combining Equations to Cancel Variables In this chapter, the focus is on combining equations to eliminate a variable. The example provided starts with two equations: 2x + 2x = 4x and z + 2z = 3z, leading to the sum of 7 + 6 = 13. The goal is to save this equation and then proceed to combine another set of equations to similarly eliminate the same variable, y.
• 01:30 - 02:00: Modifying and Adding Equations In this chapter, the discussion is focused on the necessity of incorporating the third equation in solving a problem where previously only the first two equations were utilized. The strategy involves solving the equations by choosing either the pair of equations one and three or two and three. The chapter illustrates the process by opting for equations one and three to eliminate the variable y. It includes the steps of modifying the first equation by multiplying it by two resulting in new coefficients for x, y, and z, and an updated constant. The chapter concludes with the restatement of the third equation, setting the stage for further manipulation or analysis.
• 02:00 - 02:30: Finding Solution for X and Z The chapter titled 'Finding Solution for X and Z' explains a step in solving a system of equations by elimination or substitution. The transcript describes the process of aligning equations to enable elimination of one variable through addition. In this case, by aligning and adding the equations, the y terms are canceled out, leaving a new equation in terms of x and z: 5x + 3z = 14. This result simplifies the problem to two variables, making it easier to find a solution for x and z.
• 02:30 - 03:00: Solution for Y and Concluding System of Equations This chapter focuses on solving a system of equations. The process involves manipulating the given equations to isolate the variable 'Y' by multiplying the first equation by negative one. The transformed equations are then set up in such a way to allow further simplification and eventual solving of the system.
• 03:00 - 03:30: Introduction to Word Problems on Investments This chapter provides an introduction to solving word problems related to investments. It details a process for solving simultaneous equations where one of them cancels out a variable, in this case, 'z'. The process involves solving equations for 'x' and substituting back into another equation. This specific transcript walks through cancelling terms, simplifying the equations, and solving for 'x' to demonstrate how substitution works in these problems. It emphasizes understanding the methodology of equation manipulation to find solutions in investment contexts.
• 03:30 - 04:00: Setting up Investment Equations The chapter focuses on solving investment equations by demonstrating the step-by-step process. It starts with an equation where the solution requires isolating variables and performing arithmetic operations such as subtraction and division. The process illustrated includes simplifying the equations by combining like terms, specifically solving for variables z and y through a methodical approach. The concluding step involves confirming the results by substituting back into the original equation.
• 04:00 - 04:30: Solving Investment Equations by Elimination In this chapter titled 'Solving Investment Equations by Elimination', the focus is on solving equations by the method of elimination. The example given subtracts numbers from both sides to isolate the variable 'y', resulting in a solution of (1, 2, 3) for x, y, z. The chapter progresses into solving more complex word problems related to investments, where two investments totaling $13,000 are distributed into separate accounts generating returns.
• 04:30 - 05:00: Verifying Investment Solutions The chapter focuses on verifying investment solutions through an example problem. It involves determining the amount invested in two accounts with different interest rates (15% and 14%). The total interest received is given as $1900 for the first year, and the total investment is $13,000. The chapter guides the reader through setting up equations to solve for the amounts invested in each account.
• 05:00 - 05:30: Introduction to Cost Problems with Fruits The chapter introduces cost problems related to fruits, specifically focusing on interest calculations. It explains an example where the total interest amounts to $1900, and discusses two accounts. The first account has an interest rate of 15% which is represented as 0.15 in decimal form. The process of converting a percentage to a decimal is demonstrated, by dividing by 100, as shown with the conversion of 14% to 0.14. The chapter aims to illustrate how total interest earnings from multiple accounts can be calculated.
• 05:30 - 06:00: Setting up Cost Equations for Apples and Bananas The chapter focuses on the development of cost equations for selling apples and bananas. It emphasizes the necessity of having two equations to solve for two variables, which are the cost prices of apples and bananas in this case. To simplify the equations that contain decimals, it suggests multiplying through by 100 to eliminate the decimal places. By doing this, equations are transformed from having decimal components to whole numbers, making them easier to handle, such as converting 0.15 to 15 and 0.14 to 14, and 1900 to 190,000 by adding two zeros.
• 06:00 - 07:00: Eliminating Variables in Cost Equations In this chapter, the focus is on solving equations by the method of elimination. Specifically, the chapter elaborates on eliminating variables in cost equations to simplify the process. For instance, in the given transcript, the y variable is targeted for elimination by multiplying the first equation by negative 14, resulting in a new equation that aids in solving the system of equations by eliminating the y component. The mathematical alteration shows the step where each term of the first equation is multiplied, converting '13000' into '-182000', thereby maintaining balance in the equation while removing the y variable from the equation system.
• 07:00 - 08:30: Finding Cost for Apples and Bananas The chapter discusses solving a system of equations to find the amount of money invested at different interest rates. Initially, two equations are given: one for investments earning 15% interest and another for those earning 14%. By subtracting one equation from the other, the variable y (representing investment at 14%) cancels out, leaving x, which represents investment at 15%. Calculations show that $8,000 was invested in the account with 15% interest. Combined with a total investment amount of $13,000, it is concluded that $5,000 was invested in the account with 14% interest.
• 08:30 - 09:30: Final Calculation of Total Cost for Fruits The chapter discusses the final calculation of the total cost for purchasing fruits. It explains that $8,000 was invested in a first account at a 15% interest rate, resulting in $1,200 earned in interest. Additionally, a second account received a $5,000 investment at a 14% interest rate.

Solving Systems of Equations With 3 Variables & Word Problems Transcription

• 00:00 - 00:30 now what can we do to solve a system of three equations let's say that two x plus y plus z is equal to seven and two x minus y plus two z is equal to six and x minus two y plus z is equal to zero
• 00:30 - 01:00 feel free to pause the video if you know what to do if you know how to solve it so what we need to do is choose two out of the three equations let's choose the first two equations because if we add them notice that we can cancel y so first i'm going to rewrite it
• 01:00 - 01:30 so once we add them 2x plus 2x is 4x z plus 2z is 3z and 7 plus 6 is 13. now i'm going to save this equation now what we need to do next is we need to combine another two equations and cancel the same variable y
• 01:30 - 02:00 but we need to include equation three because we didn't use it we only use the first two equations we can use equation one and three or two and three let's use one and three to cancel y we need to multiply the first equation by two so two x times two is four x y times two is two y z times 2 is 2z 7 times 2 is 14. now let's rewrite the third equation
• 02:00 - 02:30 right beneath the modified first equation and let's add them 2y and negative 2y cancels 4x plus x is 5x 2z plus z is 3z so here's the second equation that we have in terms of x and z 5x plus 3z is equal to 14.
• 02:30 - 03:00 now to solve those two equations we just need to multiply one of them by negative one let's multiply the first equation by minus one so it's going to be negative four x minus 3z which is equal to negative 13. and let's rewrite the other equation and right beneath it
• 03:00 - 03:30 so let's add the two equations so we can see that z will cancel negative 4x plus 5x is 1x negative 13 plus 14 is 1. so therefore x is equal to 1. now let's plug that value into the first equation we're not the first equation but the second equation in terms of x and z so it's going to be five times one plus three z is equal to fourteen if we subtract both sides by five
• 03:30 - 04:00 fourteen minus five is nine and then we'll divide both sides by three nine divided by three is three so z is equal to three now we need to find the value of y let's use the first equation in its unmodified state the original first equation so x is one we don't know the value of y z is three so let's combine like terms two plus three is five so y plus five is seven
• 04:00 - 04:30 now let's subtract both sides by five seven minus five is two so y is equal to two so we can write the answer like this one comma two comma three x y z so that's it for that problem now let's work on some word problems two investments totaling thirteen thousand were placed in separate accounts earning
• 04:30 - 05:00 fifteen percent and fourteen percent annually if the total interest received was nineteen hundred during the first year how much money was invested in the account paying 15 interest well let's write an equation let's say that x is the amount of money paid or placed in the first account and y is for the other count x plus y has to add up to the total investment of thirteen thousand
• 05:00 - 05:30 now the total interest is nineteen hundred so the first account which x amount of money was placed in let's say that account paid fifteen percent interest fifteen percent as the decimal is point one five to convert a percentage into a decimal divide by a hundred fourteen divided by a hundred is point one four so this should equal 1900 the total interest received from both accounts
• 05:30 - 06:00 so now that we have a system of equations we can solve it whenever you have two variables you need two equations to solve those two variables and since we have decimals let's multiply the second equation by a hundred point fifteen times a hundred is fifteen and point fourteen times a hundred is fourteen nineteen hundred times a hundred is a hundred ninety thousand just add two zeros
• 06:00 - 06:30 now let's solve this by elimination let's cancel the y variable so let's multiply the first equation by negative 14. so we're going to have negative 14x minus 14y and then 13000 times negative 14 is negative 182 000.
• 06:30 - 07:00 now let's add the two equations 15 minus 14 is x the y's will cancel and 190 000 minus 182 000 is 8 000. so that's how much money was invested in the account paying 15 interest because x is associated with 15 now the total investment is 13 000 which means that 5 000 was placed in the other account so that's the account that was paying 14
• 07:00 - 07:30 interest and so that's it for this problem now let's make sense of that problem so 8 000 was placed in the first account which paid 15 percent interest fifteen percent of eight thousand if you multiply these two numbers is twelve hundred so the first account earned twelve hundred in interest alone now the second account received five thousand dollars and was paying 14 interest
• 07:30 - 08:00 so 5 000 times 0.14 is 700. so that account received 700 in interest for that year the total interest is 1200 plus seven hundred which is nineteen hundred so as you can see the numbers make sense if five apples and eight bananas cost six dollars and fifty five cents and if nine apples and seven bananas cost nine dollars and twenty cents what is the cost of seven apples
• 08:00 - 08:30 and ten bananas we have two variables apples and bananas we'll use a and b to represent those variables so we need to write two equations if we could find the value of one apple and one banana then we could find a value of seven apples and ten bananas which is the goal of the problem so the first part of the problem states that 5 apples or 5a plus 8 bananas
• 08:30 - 09:00 has a cost of 6.55 cents so 5a plus 8b equals six point fifty five nine apples and seven bananas cost nine dollars and twenty cents let's use these two equations to find the value of each apple and each banana so what do you think we should do
• 09:00 - 09:30 now if you want to we can get rid of the decimal we can multiply both sides by 100 but we don't have to let's multiply the first equation by a negative seven to get rid of b and let's multiply the second equation by positive eight we want to get 56b 5a times negative 7
• 09:30 - 10:00 is going to be negative 35a 8b times negative 7. is negative 56b and 6.55 times negative seven that's going to be negative 45.85 9a times 8 is 72a 7b times 8 is positive 56b 9.20 times 8
• 10:00 - 10:30 is 73.60 now let's add the two equations negative 35 plus 72 that's going to be positive 37 or 37a negative 45.85 plus 73.60 that's 27.75
• 10:30 - 11:00 now a is going to be 27.75 divided by 37. so the cost of each apple is 75 cents now let's use the first equation in its unmodified form to find b so 5 times 0.75 plus 8b is equal to 6.55 so 5 times 0.75
• 11:00 - 11:30 is 3.75 now let's subtract both sides by 3.75 so 6.55 minus 3.75 and that's equal to 2.8 and b is going to be 2.8 divided by 8 which is 35 cents so now we have the cost of every apple
• 11:30 - 12:00 and every banana so the cost of a single apple is 75 cents and the cost of a single banana is 35 cents so now we can find the value of seven apples and 10 bananas so it's going to be 7 times 75 cents plus 10 times 35 cents 7 times 0.75
• 12:00 - 12:30 is 5. and 25 cents 10 times 0.35 you just got to move the decimal that's three dollars and fifty cents so if we add this this should be eight seventy five so that's the cost of seven apples and ten bananas it's eight dollars and seventy five cents
• 12:30 - 13:00 you