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What are systems of equations?

A system of equations is a set of one or more equations involving a number of variables..

The solutions to systems of equations are the variable mappings such that all component equations are satisfied—in other words, the locations at which all of these equations intersect. To solve a system is to find all such common solutions or points of intersection.

Systems of linear equations are a common and applicable subset of systems of equations. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. The system is said to be inconsistent otherwise, having no solutions. Systems of linear equations involving more than two variables work similarly, having either one solution, no solutions or infinite solutions (the latter in the case that all component equations are equivalent).

More general systems involving nonlinear functions are possible as well. These possess more complicated solution sets involving one, zero, infinite or any number of solutions, but work similarly to linear systems in that their solutions are the points satisfying all equations involved. Going further, more general systems of constraints are possible, such as ones that involve inequalities or have requirements that certain variables be integers.

Solving systems of equations is a very general and important idea, and one that is fundamental in many areas of mathematics, engineering and science.

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Solving Systems of Equations Real World Problems

Wow! You have learned many different strategies for solving systems of equations! First we started with Graphing Systems of Equations . Then we moved onto solving systems using the Substitution Method . In our last lesson we used the Linear Combinations or Addition Method to solve systems of equations.

Now we are ready to apply these strategies to solve real world problems! Are you ready? First let's look at some guidelines for solving real world problems and then we'll look at a few examples.

Steps For Solving Real World Problems

• Highlight the important information in the problem that will help write two equations.
• Define your variables
• Write two equations
• Use one of the methods for solving systems of equations to solve.
• Check your answers by substituting your ordered pair into the original equations.
• Answer the questions in the real world problems. Always write your answer in complete sentences!

Ok... let's look at a few examples. Follow along with me. (Having a calculator will make it easier for you to follow along.)

Example 1: Systems Word Problems

You are running a concession stand at a basketball game. You are selling hot dogs and sodas. Each hot dog costs $1.50 and each soda costs $0.50. At the end of the night you made a total of $78.50. You sold a total of 87 hot dogs and sodas combined. You must report the number of hot dogs sold and the number of sodas sold. How many hot dogs were sold and how many sodas were sold?

1.  Let's start by identifying the important information:

• hot dogs cost $1.50
• Sodas cost $0.50
• Made a total of $78.50
• Sold 87 hot dogs and sodas combined

2.  Define your variables.

• Ask yourself, "What am I trying to solve for? What don't I know?

In this problem, I don't know how many hot dogs or sodas were sold. So this is what each variable will stand for. (Usually the question at the end will give you this information).

Let x = the number of hot dogs sold

Let y = the number of sodas sold

3. Write two equations.

One equation will be related to the price and one equation will be related to the quantity (or number) of hot dogs and sodas sold.

1.50x + 0.50y = 78.50    (Equation related to cost)

 x + y = 87   (Equation related to the number sold)

4.  Solve! 

We can choose any method that we like to solve the system of equations. I am going to choose the substitution method since I can easily solve the 2nd equation for y.

5. Think about what this solution means.

x is the number of hot dogs and x = 35. That means that 35 hot dogs were sold.

y is the number of sodas and y = 52. That means that 52 sodas were sold.

6.  Write your answer in a complete sentence.

35 hot dogs were sold and 52 sodas were sold.

7.  Check your work by substituting.

1.50x + 0.50y = 78.50

1.50(35) + 0.50(52) = 78.50

52.50 + 26 = 78.50

35 + 52 = 87

Since both equations check properly, we know that our answers are correct!

That wasn't too bad, was it? The hardest part is writing the equations. From there you already know the strategies for solving. Think carefully about what's happening in the problem when trying to write the two equations.

Example 2: Another Word Problem

You and a friend go to Tacos Galore for lunch. You order three soft tacos and three burritos and your total bill is $11.25. Your friend's bill is $10.00 for four soft tacos and two burritos. How much do soft tacos cost? How much do burritos cost?

• 3 soft tacos + 3 burritos cost $11.25
• 4 soft tacos + 2 burritos cost $10.00

In this problem, I don't know the price of the soft tacos or the price of the burritos.

Let x = the price of 1 soft taco

Let y = the price of 1 burrito

One equation will be related your lunch and one equation will be related to your friend's lunch.

3x + 3y = 11.25  (Equation representing your lunch)

4x + 2y = 10   (Equation representing your friend's lunch)

We can choose any method that we like to solve the system of equations. I am going to choose the combinations method.

5. Think about what the solution means in context of the problem.

x = the price of 1 soft taco and x = 1.25.

That means that 1 soft tacos costs $1.25.

y = the price of 1 burrito and y = 2.5.

That means that 1 burrito costs $2.50.

Yes, I know that word problems can be intimidating, but this is the whole reason why we are learning these skills. You must be able to apply your knowledge!

If you have difficulty with real world problems, you can find more examples and practice problems in the Algebra Class E-course.

Take a look at the questions that other students have submitted:

Problem about the WNBA

Systems problem about ages

Problem about milk consumption in the U.S.

Vans and Buses? How many rode in each?

Telephone Plans problem

Systems problem about hats and scarves

Apples and guavas please!

How much did Alice spend on shoes?

All about stamps

Going to the movies

Small pitchers and large pitchers - how much will they hold?

Chickens and dogs in the farm yard

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Systems of Equations

Take your solving skills to new levels with systems of equations.

Describe and unlock situations with more than one relationship.

Substitution

Put one equation inside another.

Elimination

Find solutions by combining equations.

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Use graphing to solve equations with variables on both sides.

Systems With Quadratics

Find intersections that involve parabolas.

Problem Solving With Systems

Find solutions to non-linear systems of equations.

End of Unit 2

Speed, Distance, and Time

Use proportional reasoning to solve rate problems.

Direct and Joint Variation

Learn how multiplying variables together affects the results.

Inverse Variation

When one value gets bigger, the other gets smaller.

Mixing Problems

Not too little, not too much — use algebra to get mixtures just right.

Investigate how scaling works across different dimensions.

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Course description

Learn different methods for solving systems of equations, and use visual models and interactive graphing to understand why they work.

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5.3: Solve Systems of Equations by Elimination

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Learning Objectives

By the end of this section, you will be able to:

• Solve a system of equations by elimination
• Solve applications of systems of equations by elimination
• Choose the most convenient method to solve a system of linear equations

Before you get started, take this readiness quiz.

• Simplify −5(6−3a). If you missed this problem, review Example 1.10.43 .
• Solve the equation \(\frac{1}{3}x+\frac{5}{8}=\frac{31}{24}\). If you missed this problem, review Example 2.5.1 .

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

Solve a System of Equations by Elimination

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a , b , c , and d ,

\[\begin{array}{lc} \text{ if } & a=b \\ \text { and } & c=d \\ \text { then } &a+c =b+d \end{array}\]

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

\[\begin{array}{l} 3x+y=5 \\ \underline{2x-y=0} \\ 5x\quad\quad=5\end{array}\]

The y ’s add to zero and we have one equation with one variable.

Let’s try another one:

\[\left\{\begin{array}{l}{x+4 y=2} \\ {2 x+5 y=-2}\end{array}\right.\]

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2, we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Add the equations yourself—the result should be −3 y = −6. And that looks easy to solve, doesn’t it? Here is what it would look like.

We’ll do one more:

\[\left\{\begin{array}{l}{4 x-3 y=10} \\ {3 x+5 y=-7}\end{array}\right.\]

It doesn’t appear that we can get the coefficients of one variable to be opposites by multiplying one of the equations by a constant, unless we use fractions. So instead, we’ll have to multiply both equations by a constant.

We can make the coefficients of x be opposites if we multiply the first equation by 3 and the second by −4, so we get 12 x and −12 x .

This gives us these two new equations:

\[\left\{\begin{aligned} 12 x-9 y &=30 \\-12 x-20 y &=28 \end{aligned}\right.\]

When we add these equations,

\[ \left\{\begin{array}{r}{12 x-9 y=30} \\ {\underline{-12 x-20 y=28}} \\\end{array}\right.\\\quad\qquad {-29 y=58}\]

the x ’s are eliminated and we just have −29 y = 58.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

Example \(\PageIndex{1}\): How to Solve a System of Equations by Elimination

Solve the system by elimination. \(\left\{\begin{array}{l}{2 x+y=7} \\ {x-2 y=6}\end{array}\right.\)

Try It \(\PageIndex{2}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{3 x+y=5} \\ {2 x-3 y=7}\end{array}\right.\)

(2,−1)

Try It \(\PageIndex{3}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{4 x+y=-5} \\ {-2 x-2 y=-2}\end{array}\right.\)

(−2,3)

The steps are listed below for easy reference.

HOW TO SOLVE A SYSTEM OF EQUATIONS BY ELIMINATION.

• Write both equations in standard form. If any coefficients are fractions, clear them.
• Decide which variable you will eliminate.
• Multiply one or both equations so that the coefficients of that variable are opposites.
• Add the equations resulting from Step 2 to eliminate one variable.
• Solve for the remaining variable.
• Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
• Write the solution as an ordered pair.
• Check that the ordered pair is a solution to both original equations.

First we’ll do an example where we can eliminate one variable right away.

Example \(\PageIndex{4}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{x+y=10} \\ {x-y=12}\end{array}\right.\)

Try It \(\PageIndex{5}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{2 x+y=5} \\ {x-y=4}\end{array}\right.\)

(3,−1)

Try It \(\PageIndex{6}\)

Solve the system by elimination.\(\left\{\begin{array}{l}{x+y=3} \\ {-2 x-y=-1}\end{array}\right.\)

(−2,5)

In Example \(\PageIndex{7}\), we will be able to make the coefficients of one variable opposites by multiplying one equation by a constant.

Example \(\PageIndex{7}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{3 x-2 y=-2} \\ {5 x-6 y=10}\end{array}\right.\)

Try It \(\PageIndex{8}\)

Solve the system by elimination.\(\left\{\begin{array}{l}{4 x-3 y=1} \\ {5 x-9 y=-4}\end{array}\right.\)

Try It \(\PageIndex{9}\)

Solve the system by elimination.\(\left\{\begin{array}{l}{3 x+2 y=2} \\ {6 x+5 y=8}\end{array}\right.\)

(−2,4)

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Example \(\PageIndex{10}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{4 x-3 y=9} \\ {7 x+2 y=-6}\end{array}\right.\)

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by a constant to get the opposites.

Try It \(\PageIndex{11}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{3 x-4 y=-9} \\ {5 x+3 y=14}\end{array}\right.\)

Try It \(\PageIndex{12}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{7 x+8 y=4} \\ {3 x-5 y=27}\end{array}\right.\)

(4,−3)

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by its LCD.

Example \(\PageIndex{13}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{x+\frac{1}{2} y=6} \\ {\frac{3}{2} x+\frac{2}{3} y=\frac{17}{2}}\end{array}\right.\)

In this example, both equations have fractions. Our first step will be to multiply each equation by its LCD to clear the fractions.

Try It \(\PageIndex{14}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{\frac{1}{3} x-\frac{1}{2} y=1} \\ {\frac{3}{4} x-y=\frac{5}{2}}\end{array}\right.\)

Try It \(\PageIndex{15}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{x+\frac{3}{5} y=-\frac{1}{5}} \\ {-\frac{1}{2} x-\frac{2}{3} y=\frac{5}{6}}\end{array}\right.\)

(1,−2)

In the Solving Systems of Equations by Graphing we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

Example \(\PageIndex{16}\)

Solve the system by elimination.\(\left\{\begin{array}{l}{3 x+4 y=12} \\ {y=3-\frac{3}{4} x}\end{array}\right.\)

\(\begin{array} {ll} & \left\{\begin{aligned} 3 x+4 y &=12 \\ y &=3-\frac{3}{4} x \end{aligned}\right. \\\\\text{Write the second equation in standard form.} & \left\{\begin{array}{l}{3 x+4 y=12} \\ {\frac{3}{4} x+y=3}\end{array}\right.\\ \\ \text{Clear the fractions by multiplying thesecond equation by 4.} & \left\{\begin{aligned} 3 x+4 y &=12 \\ 4\left(\frac{3}{4} x+y\right) &=4(3) \end{aligned}\right. \\\\ \text{Simplify.} & \left\{\begin{array}{l}{3 x+4 y=12} \\ {3 x+4 y=12}\end{array}\right.\\\\ \text{To eliminate a variable, we multiply thesecond equation by −1.} & \left\{\begin{array}{c}{3 x+4 y=12} \\ \underline{-3 x-4 y=-12} \end{array}\right.\\ &\qquad\qquad\quad 0=0 \\ \text{Simplify and add.} \end{array}\)

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Try It \(\PageIndex{17}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{5 x-3 y=15} \\ {y=-5+\frac{5}{3} x}\end{array}\right.\)

infinitely many solutions

Try It \(\PageIndex{18}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{x+2 y=6} \\ {y=-\frac{1}{2} x+3}\end{array}\right.\)

Example \(\PageIndex{19}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{-6 x+15 y=10} \\ {2 x-5 y=-5}\end{array}\right.\)

\(\begin{array} {ll} \text{The equations are in standard form.}& \left\{\begin{aligned}-6 x+15 y &=10 \\ 2 x-5 y &=-5 \end{aligned}\right. \\\\ \text{Multiply the second equation by 3 to eliminate a variable.} & \left\{\begin{array}{l}{-6 x+15 y=10} \\ {3(2 x-5 y)=3(-5)}\end{array}\right. \\\\ \text{Simplify and add.} & \left\{\begin{aligned}{-6 x+15 y =10} \\ \underline{6 x-15 y =-15} \end{aligned}\right. \\ & \qquad \qquad \quad0\neq 5 \end{array}\)

This statement is false. The equations are inconsistent and so their graphs would be parallel lines.

The system does not have a solution.

Try It \(\PageIndex{20}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{-3 x+2 y=8} \\ {9 x-6 y=13}\end{array}\right.\)

no solution

Try It \(\PageIndex{21}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{7 x-3 y=-2} \\ {-14 x+6 y=8}\end{array}\right.\)

Solve Applications of Systems of Equations by Elimination

Some applications problems translate directly into equations in standard form, so we will use the elimination method to solve them. As before, we use our Problem Solving Strategy to help us stay focused and organized.

Example \(\PageIndex{22}\)

The sum of two numbers is 39. Their difference is 9. Find the numbers.

\(\begin{array} {ll} \textbf{Step 1. Read}\text{ the problem}& \\ \textbf{Step 2. Identify} \text{ what we are looking for.} & \text{We are looking for two numbers.} \\\textbf{Step 3. Name} \text{ what we are looking for.} & \text{Let n = the first number.} \\ & \text{ m = the second number} \\\textbf{Step 4. Translate} \text{ into a system of equations.}& \\ & \text{The sum of two numbers is 39.} \\ & n+m=39\\ & \text{Their difference is 9.} \\ & n−m=9 \\ \\ \text{The system is:} & \left\{\begin{array}{l}{n+m=39} \\ {n-m=9}\end{array}\right. \\\\ \textbf{Step 5. Solve} \text{ the system of equations. } & \\ \text{To solve the system of equations, use} \\ \text{elimination. The equations are in standard} \\ \text{form and the coefficients of m are} & \\ \text{opposites. Add.} & \left\{\begin{array}{l}{n+m=39} \\ \underline{n-m=9}\end{array}\right. \\ &\quad 2n\qquad=48 \\ \\\text{Solve for n.} & n=24 \\ \\ \text{Substitute n=24 into one of the original} &n+m=39 \\ \text{equations and solve form.} & 24+m=39 \\ & m=15 \\ \textbf{Step 6. Check}\text{ the answer.} & \text{Since 24+15=39 and 24−15=9, the answers check.}\\ \textbf{Step 7. Answer} \text{ the question.} & \text{The numbers are 24 and 15.} \end{array}\)

Try It \(\PageIndex{23}\)

The sum of two numbers is 42. Their difference is 8. Find the numbers.

The numbers are 25 and 17.

Try It \(\PageIndex{24}\)

The sum of two numbers is −15. Their difference is −35. Find the numbers.

The numbers are −25 and 10.

Example \(\PageIndex{25}\)

Joe stops at a burger restaurant every day on his way to work. Monday he had one order of medium fries and two small sodas, which had a total of 620 calories. Tuesday he had two orders of medium fries and one small soda, for a total of 820 calories. How many calories are there in one order of medium fries? How many calories in one small soda?

Try It \(\PageIndex{26}\)

Malik stops at the grocery store to buy a bag of diapers and 2 cans of formula. He spends a total of $37. The next week he stops and buys 2 bags of diapers and 5 cans of formula for a total of $87. How much does a bag of diapers cost? How much is one can of formula?

The bag of diapers costs $11 and the can of formula costs $13.

Try It \(\PageIndex{27}\)

To get her daily intake of fruit for the day, Sasha eats a banana and 8 strawberries on Wednesday for a calorie count of 145. On the following Wednesday, she eats two bananas and 5 strawberries for a total of 235 calories for the fruit. How many calories are there in a banana? How many calories are in a strawberry?

There are 105 calories in a banana and 5 calories in a strawberry.

Choose the Most Convenient Method to Solve a System of Linear Equations

When you will have to solve a system of linear equations in a later math class, you will usually not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

Example \(\PageIndex{28}\)

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

• \(\left\{\begin{array}{l}{3 x+8 y=40} \\ {7 x-4 y=-32}\end{array}\right.\)
• \(\left\{\begin{array}{l}{5 x+6 y=12} \\ {y=\frac{2}{3} x-1}\end{array}\right.\)

1. \(\left\{\begin{array}{l}{3 x+8 y=40} \\ {7 x-4 y=-32}\end{array}\right.\)

• Since both equations are in standard form, using elimination will be most convenient.

2. \(\left\{\begin{array}{l}{5 x+6 y=12} \\ {y=\frac{2}{3} x-1}\end{array}\right.\)

Since one equation is already solved for y , using substitution will be most convenient.

Try It \(\PageIndex{29}\)

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

• \(\left\{\begin{array}{l}{4 x-5 y=-32} \\ {3 x+2 y=-1}\end{array}\right.\)
• \(\left\{\begin{array}{l}{x=2 y-1} \\ {3 x-5 y=-7}\end{array}\right.\)
• Since one equation is already solved for xx, using substitution will be most convenient.

Try It \(\PageIndex{30}\)

• \(\left\{\begin{array}{l}{y=2 x-1} \\ {3 x-4 y=-6}\end{array}\right.\)
• \(\left\{\begin{array}{l}{6 x-2 y=12} \\ {3 x+7 y=-13}\end{array}\right.\)
• Since one equation is already solved for \(y\), using substitution will be most convenient;

Access these online resources for additional instruction and practice with solving systems of linear equations by elimination.

• Instructional Video-Solving Systems of Equations by Elimination
• Instructional Video-Solving by Elimination
• Instructional Video-Solving Systems by Elimination

Key Concepts

An adaptive finite element PML method for Helmholtz equations in periodic heterogeneous media

• Published: 16 May 2024
• Volume 43 , article number  242 , ( 2024 )

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• Xue Jiang   ORCID: orcid.org/0000-0002-3160-0282 1 ,
• Zhongjiang Sun 1 ,
• Lijuan Sun 1 &
• Qiang Ma 2  

The paper concerns the numerical solution for the wave propagation problem in periodic heterogeneous media. The homogenization method is utilized for the solution in the bounded periodic structure with highly oscillating coefficients. The perfectly matched layer (PML) technique is adopted to truncate the unbounded physical domain into a bounded computational domain, and the exponential convergence of Cartesian PML is generalized to the Helmholtz transmission problem in periodic heterogeneous media. An efficient adaptive finite element algorithm based on reliable a posteriori error estimate is extended to solve the homogenized PML problem, and the reliability of the estimator is established. Numerical experiments are included to demonstrate the competitive behavior of the proposed method.

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The datasets generated and/or analyzed during the current study are available from the corresponding author on reasonable request.

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Xue Jiang, Zhongjiang Sun & Lijuan Sun

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Jiang, X., Sun, Z., Sun, L. et al. An adaptive finite element PML method for Helmholtz equations in periodic heterogeneous media. Comp. Appl. Math. 43 , 242 (2024). https://doi.org/10.1007/s40314-024-02770-y

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Wavefunction matching for solving quantum many-body problems

Strongly interacting systems play an important role in quantum physics and quantum chemistry. Stochastic methods such as Monte Carlo simulations are a proven method for investigating such systems. However, these methods reach their limits when so-called sign oscillations occur. This problem has now been solved by an international team of researchers from Germany, Turkey, the USA, China, South Korea and France using the new method of wavefunction matching. As an example, the masses and radii of all nuclei up to mass number 50 were calculated using this method. The results agree with the measurements, the researchers now report in the journal " Nature ."

All matter on Earth consists of tiny particles known as atoms. Each atom contains even smaller particles: protons, neutrons and electrons. Each of these particles follows the rules of quantum mechanics. Quantum mechanics forms the basis of quantum many-body theory, which describes systems with many particles, such as atomic nuclei.

One class of methods used by nuclear physicists to study atomic nuclei is the ab initio approach. It describes complex systems by starting from a description of their elementary components and their interactions. In the case of nuclear physics, the elementary components are protons and neutrons. Some key questions that ab initio calculations can help answer are the binding energies and properties of atomic nuclei and the link between nuclear structure and the underlying interactions between protons and neutrons.

However, these ab initio methods have difficulties in performing reliable calculations for systems with complex interactions. One of these methods is quantum Monte Carlo simulations. Here, quantities are calculated using random or stochastic processes. Although quantum Monte Carlo simulations can be efficient and powerful, they have a significant weakness: the sign problem. It arises in processes with positive and negative weights, which cancel each other. This cancellation leads to inaccurate final predictions.

A new approach, known as wavefunction matching, is intended to help solve such calculation problems for ab initio methods. "This problem is solved by the new method of wavefunction matching by mapping the complicated problem in a first approximation to a simple model system that does not have such sign oscillations and then treating the differences in perturbation theory," says Prof. Ulf-G. Meißner from the Helmholtz Institute for Radiation and Nuclear Physics at the University of Bonn and from the Institute of Nuclear Physics and the Center for Advanced Simulation and Analytics at Forschungszentrum Jülich. "As an example, the masses and radii of all nuclei up to mass number 50 were calculated -- and the results agree with the measurements," reports Meißner, who is also a member of the Transdisciplinary Research Areas "Modeling" and "Matter" at the University of Bonn.

"In quantum many-body theory, we are often faced with the situation that we can perform calculations using a simple approximate interaction, but realistic high-fidelity interactions cause severe computational problems," says Dean Lee, Professor of Physics from the Facility for Rare Istope Beams and Department of Physics and Astronomy (FRIB) at Michigan State University and head of the Department of Theoretical Nuclear Sciences.

Wavefunction matching solves this problem by removing the short-distance part of the high-fidelity interaction and replacing it with the short-distance part of an easily calculable interaction. This transformation is done in a way that preserves all the important properties of the original realistic interaction. Since the new wavefunctions are similar to those of the easily computable interaction, the researchers can now perform calculations with the easily computable interaction and apply a standard procedure for handling small corrections -- called perturbation theory.

The research team applied this new method to lattice quantum Monte Carlo simulations for light nuclei, medium-mass nuclei, neutron matter and nuclear matter. Using precise ab initio calculations, the results closely matched real-world data on nuclear properties such as size, structure and binding energy. Calculations that were once impossible due to the sign problem can now be performed with wavefunction matching.

While the research team focused exclusively on quantum Monte Carlo simulations, wavefunction matching should be useful for many different ab initio approaches. "This method can be used in both classical computing and quantum computing, for example to better predict the properties of so-called topological materials, which are important for quantum computing," says Meißner.

The first author is Prof. Dr. Serdar Elhatisari, who worked for two years as a Fellow in Prof. Meißner's ERC Advanced Grant EXOTIC. According to Meißner, a large part of the work was carried out during this time. Part of the computing time on supercomputers at Forschungszentrum Jülich was provided by the IAS-4 institute, which Meißner heads.

• Quantum Computers
• Computers and Internet
• Computer Modeling
• Spintronics Research
• Mathematics
• Quantum mechanics
• Quantum entanglement
• Introduction to quantum mechanics
• Computer simulation
• Quantum computer
• Quantum dot
• Quantum tunnelling
• Security engineering

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• Serdar Elhatisari, Lukas Bovermann, Yuan-Zhuo Ma, Evgeny Epelbaum, Dillon Frame, Fabian Hildenbrand, Myungkuk Kim, Youngman Kim, Hermann Krebs, Timo A. Lähde, Dean Lee, Ning Li, Bing-Nan Lu, Ulf-G. Meißner, Gautam Rupak, Shihang Shen, Young-Ho Song, Gianluca Stellin. Wavefunction matching for solving quantum many-body problems . Nature , 2024; DOI: 10.1038/s41586-024-07422-z

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Reliable analysis for obtaining exact soliton solutions of (2+1)-dimensional Chaffee-Infante equation

• Naveed Iqbal 1 , 
• Muhammad Bilal Riaz 2,3 ,  ,  , 
• Meshari Alesemi 4 , 
• Taher S. Hassan 1,5,6 , 
• Ali M. Mahnashi 7 , 
• Ahmad Shafee 8
• 1. Deparment of Mathematics, College of Science, University of Ha'il, Ha'il 2440, Saudi Arabia
• 2. IT4Innovations, VSB-Technical University of Ostrava, Ostrava, Czech Republic
• 3. Department of Computer Science and Mathematics, Lebanese American University, Byblos, Lebanon
• 4. Department of Mathematics, College of Science, University of Bisha, P.O. Box 511, Bisha 61922, Saudi Arabia
• 5. Department of Mathematics, Faculty of Science, Mansoura University, Mansoura, 35516, Egypt
• 6. Section of Mathematics, International Telematic University Uninettuno, Corso Vittorio Emanuele Ⅱ, 39, 00186 Roma, Italy
• 7. Department of Mathematics, Faculty of Science, Jazan University, P.O. Box 2097, Jazan 45142, Kingdom of Saudi Arabia
• 8. PAAET, College of Technological Studies, Laboratory Technology Department, Shuwaikh 70654, Kuwait
• Received: 08 March 2024 Revised: 23 April 2024 Accepted: 30 April 2024 Published: 14 May 2024

MSC : 34G20, 35A20, 35A22, 35R11

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The (2+1)-dimensional Chaffee-Infante equation (CIE) is a significant model of the ion-acoustic waves in plasma. The primary objective of this paper was to establish and examine closed-form soliton solutions to the CIE using the modified extended direct algebraic method (m-EDAM), a mathematical technique. By using a variable transformation to convert CIE into a nonlinear ordinary differential equation (NODE), which was then reduced to a system of nonlinear algebraic equations with the assumption of a closed-form solution, the strategic m-EDAM was implemented. When the resulting problem was solved using the Maple tool, many soliton solutions in the shapes of rational, exponential, trigonometric, and hyperbolic functions were produced. By using illustrated 3D and density plots to evaluate several soliton solutions for the provided definite values of the parameters, it was possible to determine if the soliton solutions produced for CIE are cuspon or kink solitons. Additionally, it has been shown that the m-EDAM is a robust, useful, and user-friendly instrument that provides extra generic wave solutions for nonlinear models in mathematical physics and engineering.

• nonlinear partial differential equations ,
• (2+1)-dimensional Chaffee-Infante equation ,
• modified extended direct algebraic method ,
• kink soliton ,

Citation: Naveed Iqbal, Muhammad Bilal Riaz, Meshari Alesemi, Taher S. Hassan, Ali M. Mahnashi, Ahmad Shafee. Reliable analysis for obtaining exact soliton solutions of (2+1)-dimensional Chaffee-Infante equation[J]. AIMS Mathematics, 2024, 9(6): 16666-16686. doi: 10.3934/math.2024808

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• Figure 1. The 3D and contour plots of the kink soliton solution $ z_{1, 6} $ stated in (3.12) are plotted for $ P = 1, Q = 3, R = 2, \sigma = 10, t = 0 $
• Figure 2. The 3D and contour plots of the bright kink soliton solution $ z_{1, 9} $ stated in (3.15) are plotted for $ P = 8, Q = 10, R = 2, \sigma = 50, t = 1 $
• Figure 3. The 3D and contour plots of the bright kink soliton solution $ z_{1, 18} $ stated in (3.24) are plotted for $ P = 9, Q = 0, R = -4, \sigma = 20, t = 10 $
• Figure 4. The 3D and contour plots of the imaginary part of cuspon soliton solution $ z_{1, 21} $ stated in (3.27) are plotted for $ P = 3, Q = 0, R = 3, \sigma = 30, t = 100 $
• Figure 5. The 3D and contour plots of the imaginary part of cuspon soliton solution $ z_{1, 24} $ stated in (3.30) are plotted for $ P = 6, Q = 0, R = 6, \sigma = 40, t = 50 $
• Figure 6. The 3D and contour plots of the kink soliton solution $ z_{1, 32} $ stated in (3.38) are plotted for $ P = 0, Q = 10, R = 5, \sigma = 100, t = 0 $
• Figure 7. The 3D and contour plots of the dark kink soliton solution $ z_{2, 10} $ stated in (3.49) are plotted for $ P = 2, Q = 10, R = 8, \sigma = 1, t = 5 $
• Figure 8. The 3D and contour plots of the kink soliton solution $ z_{2, 19} $ stated in (3.58) are plotted for $ P = 5, Q = 0, R = -5, \sigma = 4, t = 15 $
• Figure 9. The 3D and contour plots of the bright-dark soliton solution $ z_{2, 23} $ stated in (3.62) are plotted for $ P = 6, Q = 0, R = 6, \sigma = 100, t = 0 $
• Figure 10. The 3D and contour plots of the kink soliton solution $ z_{2, 31} $ stated in (3.70) are plotted for $ P = 10, \mu = 5, p = 2, Q = 5, R = 0, \sigma = 0, t = 2 $

5.3 Solve Systems of Equations by Elimination

Learning objectives.

By the end of this section, you will be able to:

• Solve a system of equations by elimination
• Solve applications of systems of equations by elimination
• Choose the most convenient method to solve a system of linear equations

Be Prepared 5.8

Before you get started, take this readiness quiz.

Simplify −5 ( 6 − 3 a ) −5 ( 6 − 3 a ) . If you missed this problem, review Example 1.136 .

Be Prepared 5.9

Solve the equation 1 3 x + 5 8 = 31 24 1 3 x + 5 8 = 31 24 . If you missed this problem, review Example 2.48 .

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

Solve a System of Equations by Elimination

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a , b , c , and d ,

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

The y ’s add to zero and we have one equation with one variable.

Let’s try another one:

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2, we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Add the equations yourself—the result should be −3 y = −6. And that looks easy to solve, doesn’t it? Here is what it would look like.

We’ll do one more:

It doesn’t appear that we can get the coefficients of one variable to be opposites by multiplying one of the equations by a constant, unless we use fractions. So instead, we’ll have to multiply both equations by a constant.

We can make the coefficients of x be opposites if we multiply the first equation by 3 and the second by −4, so we get 12 x and −12 x .

This gives us these two new equations:

When we add these equations,

the x ’s are eliminated and we just have −29 y = 58.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

Example 5.25

How to solve a system of equations by elimination.

Solve the system by elimination. { 2 x + y = 7 x − 2 y = 6 { 2 x + y = 7 x − 2 y = 6

Try It 5.49

Solve the system by elimination. { 3 x + y = 5 2 x − 3 y = 7 { 3 x + y = 5 2 x − 3 y = 7

Try It 5.50

Solve the system by elimination. { 4 x + y = −5 −2 x − 2 y = −2 { 4 x + y = −5 −2 x − 2 y = −2

The steps are listed below for easy reference.

How to solve a system of equations by elimination.

• Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
• Decide which variable you will eliminate.
• Multiply one or both equations so that the coefficients of that variable are opposites.
• Step 3. Add the equations resulting from Step 2 to eliminate one variable.
• Step 4. Solve for the remaining variable.
• Step 5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
• Step 6. Write the solution as an ordered pair.
• Step 7. Check that the ordered pair is a solution to both original equations.

First we’ll do an example where we can eliminate one variable right away.

Example 5.26

Solve the system by elimination. { x + y = 10 x − y = 12 { x + y = 10 x − y = 12

Try It 5.51

Solve the system by elimination. { 2 x + y = 5 x − y = 4 { 2 x + y = 5 x − y = 4

Try It 5.52

Solve the system by elimination. { x + y = 3 −2 x − y = −1 { x + y = 3 −2 x − y = −1

In Example 5.27 , we will be able to make the coefficients of one variable opposites by multiplying one equation by a constant.

Example 5.27

Solve the system by elimination. { 3 x − 2 y = −2 5 x − 6 y = 10 { 3 x − 2 y = −2 5 x − 6 y = 10

Try It 5.53

Solve the system by elimination. { 4 x − 3 y = 1 5 x − 9 y = −4 { 4 x − 3 y = 1 5 x − 9 y = −4

Try It 5.54

Solve the system by elimination. { 3 x + 2 y = 2 6 x + 5 y = 8 { 3 x + 2 y = 2 6 x + 5 y = 8

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Example 5.28

Solve the system by elimination. { 4 x − 3 y = 9 7 x + 2 y = −6 { 4 x − 3 y = 9 7 x + 2 y = −6

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by a constant to get the opposites.

What other constants could we have chosen to eliminate one of the variables? Would the solution be the same?

Try It 5.55

Solve the system by elimination. { 3 x − 4 y = −9 5 x + 3 y = 14 { 3 x − 4 y = −9 5 x + 3 y = 14

Try It 5.56

Solve the system by elimination. { 7 x + 8 y = 4 3 x − 5 y = 27 { 7 x + 8 y = 4 3 x − 5 y = 27

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by its LCD.

Example 5.29

Solve the system by elimination. { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2

In this example, both equations have fractions. Our first step will be to multiply each equation by its LCD to clear the fractions.

Try It 5.57

Solve the system by elimination. { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2

Try It 5.58

Solve the system by elimination. { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6

In the Solving Systems of Equations by Graphing we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

Example 5.30

Solve the system by elimination. { 3 x + 4 y = 12 y = 3 − 3 4 x { 3 x + 4 y = 12 y = 3 − 3 4 x

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Try It 5.59

Solve the system by elimination. { 5 x − 3 y = 15 y = −5 + 5 3 x { 5 x − 3 y = 15 y = −5 + 5 3 x

Try It 5.60

Solve the system by elimination. { x + 2 y = 6 y = − 1 2 x + 3 { x + 2 y = 6 y = − 1 2 x + 3

Example 5.31

Solve the system by elimination. { −6 x + 15 y = 10 2 x − 5 y = −5 { −6 x + 15 y = 10 2 x − 5 y = −5

This statement is false. The equations are inconsistent and so their graphs would be parallel lines.

The system does not have a solution.

Try It 5.61

Solve the system by elimination. { −3 x + 2 y = 8 9 x − 6 y = 13 { −3 x + 2 y = 8 9 x − 6 y = 13

Try It 5.62

Solve the system by elimination. { 7 x − 3 y = − 2 −14 x + 6 y = 8 { 7 x − 3 y = − 2 −14 x + 6 y = 8

Solve Applications of Systems of Equations by Elimination

Some applications problems translate directly into equations in standard form, so we will use the elimination method to solve them. As before, we use our Problem Solving Strategy to help us stay focused and organized.

Example 5.32

The sum of two numbers is 39. Their difference is 9. Find the numbers.

Try It 5.63

The sum of two numbers is 42. Their difference is 8. Find the numbers.

Try It 5.64

The sum of two numbers is −15. Their difference is −35. Find the numbers.

Example 5.33

Joe stops at a burger restaurant every day on his way to work. Monday he had one order of medium fries and two small sodas, which had a total of 620 calories. Tuesday he had two orders of medium fries and one small soda, for a total of 820 calories. How many calories are there in one order of medium fries? How many calories in one small soda?

Try It 5.65

Malik stops at the grocery store to buy a bag of diapers and 2 cans of formula. He spends a total of $37. The next week he stops and buys 2 bags of diapers and 5 cans of formula for a total of $87. How much does a bag of diapers cost? How much is one can of formula?

Try It 5.66

To get her daily intake of fruit for the day, Sasha eats a banana and 8 strawberries on Wednesday for a calorie count of 145. On the following Wednesday, she eats two bananas and 5 strawberries for a total of 235 calories for the fruit. How many calories are there in a banana? How many calories are in a strawberry?

Choose the Most Convenient Method to Solve a System of Linear Equations

When you will have to solve a system of linear equations in a later math class, you will usually not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

Example 5.34

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 3 x + 8 y = 40 7 x − 4 y = −32 { 3 x + 8 y = 40 7 x − 4 y = −32 ⓑ { 5 x + 6 y = 12 y = 2 3 x − 1 { 5 x + 6 y = 12 y = 2 3 x − 1

• ⓐ { 3 x + 8 y = 40 7 x − 4 y = −32 { 3 x + 8 y = 40 7 x − 4 y = −32 Since both equations are in standard form, using elimination will be most convenient.
• ⓑ { 5 x + 6 y = 12 y = 2 3 x − 1 { 5 x + 6 y = 12 y = 2 3 x − 1

>Since one equation is already solved for y , using substitution will be most convenient.

Try It 5.67

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 4 x − 5 y = −32 3 x + 2 y = −1 { 4 x − 5 y = −32 3 x + 2 y = −1 ⓑ { x = 2 y − 1 3 x − 5 y = − 7 { x = 2 y − 1 3 x − 5 y = − 7

Try It 5.68

ⓐ { y = 2 x − 1 3 x − 4 y = − 6 { y = 2 x − 1 3 x − 4 y = − 6 ⓑ { 6 x − 2 y = 12 3 x + 7 y = −13 { 6 x − 2 y = 12 3 x + 7 y = −13

Access these online resources for additional instruction and practice with solving systems of linear equations by elimination.

• Instructional Video-Solving Systems of Equations by Elimination
• Instructional Video-Solving by Elimination
• Instructional Video-Solving Systems by Elimination

Section 5.3 Exercises

Practice makes perfect.

In the following exercises, solve the systems of equations by elimination.

{ 5 x + 2 y = 2 −3 x − y = 0 { 5 x + 2 y = 2 −3 x − y = 0

{ −3 x + y = −9 x − 2 y = −12 { −3 x + y = −9 x − 2 y = −12

{ 6 x − 5 y = −1 2 x + y = 13 { 6 x − 5 y = −1 2 x + y = 13

{ 3 x − y = −7 4 x + 2 y = −6 { 3 x − y = −7 4 x + 2 y = −6

{ x + y = −1 x − y = −5 { x + y = −1 x − y = −5

{ x + y = −8 x − y = −6 { x + y = −8 x − y = −6

{ 3 x − 2 y = 1 − x + 2 y = 9 { 3 x − 2 y = 1 − x + 2 y = 9

{ −7 x + 6 y = −10 x − 6 y = 22 { −7 x + 6 y = −10 x − 6 y = 22

{ 3 x + 2 y = −3 − x − 2 y = −19 { 3 x + 2 y = −3 − x − 2 y = −19

{ 5 x + 2 y = 1 −5 x − 4 y = −7 { 5 x + 2 y = 1 −5 x − 4 y = −7

{ 6 x + 4 y = −4 −6 x − 5 y = 8 { 6 x + 4 y = −4 −6 x − 5 y = 8

{ 3 x − 4 y = −11 x − 2 y = −5 { 3 x − 4 y = −11 x − 2 y = −5

{ 5 x − 7 y = 29 x + 3 y = −3 { 5 x − 7 y = 29 x + 3 y = −3

{ 6 x − 5 y = −75 − x − 2 y = −13 { 6 x − 5 y = −75 − x − 2 y = −13

{ − x + 4 y = 8 3 x + 5 y = 10 { − x + 4 y = 8 3 x + 5 y = 10

{ 2 x − 5 y = 7 3 x − y = 17 { 2 x − 5 y = 7 3 x − y = 17

{ 5 x − 3 y = −1 2 x − y = 2 { 5 x − 3 y = −1 2 x − y = 2

{ 7 x + y = −4 13 x + 3 y = 4 { 7 x + y = −4 13 x + 3 y = 4

{ −3 x + 5 y = −13 2 x + y = −26 { −3 x + 5 y = −13 2 x + y = −26

{ 3 x − 5 y = −9 5 x + 2 y = 16 { 3 x − 5 y = −9 5 x + 2 y = 16

{ 4 x − 3 y = 3 2 x + 5 y = −31 { 4 x − 3 y = 3 2 x + 5 y = −31

{ 4 x + 7 y = 14 −2 x + 3 y = 32 { 4 x + 7 y = 14 −2 x + 3 y = 32

{ 5 x + 2 y = 21 7 x − 4 y = 9 { 5 x + 2 y = 21 7 x − 4 y = 9

{ 3 x + 8 y = −3 2 x + 5 y = −3 { 3 x + 8 y = −3 2 x + 5 y = −3

{ 11 x + 9 y = −5 7 x + 5 y = −1 { 11 x + 9 y = −5 7 x + 5 y = −1

{ 3 x + 8 y = 67 5 x + 3 y = 60 { 3 x + 8 y = 67 5 x + 3 y = 60

{ 2 x + 9 y = −4 3 x + 13 y = −7 { 2 x + 9 y = −4 3 x + 13 y = −7

{ 1 3 x − y = −3 x + 5 2 y = 2 { 1 3 x − y = −3 x + 5 2 y = 2

{ x + 1 2 y = 3 2 1 5 x − 1 5 y = 3 { x + 1 2 y = 3 2 1 5 x − 1 5 y = 3

{ x + 1 3 y = −1 1 2 x − 1 3 y = −2 { x + 1 3 y = −1 1 2 x − 1 3 y = −2

{ 1 3 x − y = −3 2 3 x + 5 2 y = 3 { 1 3 x − y = −3 2 3 x + 5 2 y = 3

{ 2 x + y = 3 6 x + 3 y = 9 { 2 x + y = 3 6 x + 3 y = 9

{ x − 4 y = −1 −3 x + 12 y = 3 { x − 4 y = −1 −3 x + 12 y = 3

{ −3 x − y = 8 6 x + 2 y = −16 { −3 x − y = 8 6 x + 2 y = −16

{ 4 x + 3 y = 2 20 x + 15 y = 10 { 4 x + 3 y = 2 20 x + 15 y = 10

{ 3 x + 2 y = 6 −6 x − 4 y = −12 { 3 x + 2 y = 6 −6 x − 4 y = −12

{ 5 x − 8 y = 12 10 x − 16 y = 20 { 5 x − 8 y = 12 10 x − 16 y = 20

{ −11 x + 12 y = 60 −22 x + 24 y = 90 { −11 x + 12 y = 60 −22 x + 24 y = 90

{ 7 x − 9 y = 16 −21 x + 27 y = −24 { 7 x − 9 y = 16 −21 x + 27 y = −24

{ 5 x − 3 y = 15 y = 5 3 x − 2 { 5 x − 3 y = 15 y = 5 3 x − 2

{ 2 x + 4 y = 7 y = − 1 2 x − 4 { 2 x + 4 y = 7 y = − 1 2 x − 4

In the following exercises, translate to a system of equations and solve.

The sum of two numbers is 65. Their difference is 25. Find the numbers.

The sum of two numbers is 37. Their difference is 9. Find the numbers.

The sum of two numbers is −27. Their difference is −59. Find the numbers.

The sum of two numbers is −45. Their difference is −89. Find the numbers.

Andrea is buying some new shirts and sweaters. She is able to buy 3 shirts and 2 sweaters for $114 or she is able to buy 2 shirts and 4 sweaters for $164. How much does a shirt cost? How much does a sweater cost?

Peter is buying office supplies. He is able to buy 3 packages of paper and 4 staplers for $40 or he is able to buy 5 packages of paper and 6 staplers for $62. How much does a package of paper cost? How much does a stapler cost?

The total amount of sodium in 2 hot dogs and 3 cups of cottage cheese is 4720 mg. The total amount of sodium in 5 hot dogs and 2 cups of cottage cheese is 6300 mg. How much sodium is in a hot dog? How much sodium is in a cup of cottage cheese?

The total number of calories in 2 hot dogs and 3 cups of cottage cheese is 960 calories. The total number of calories in 5 hot dogs and 2 cups of cottage cheese is 1190 calories. How many calories are in a hot dog? How many calories are in a cup of cottage cheese?

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

ⓐ { 8 x − 15 y = −32 6 x + 3 y = −5 { 8 x − 15 y = −32 6 x + 3 y = −5 ⓑ { x = 4 y − 3 4 x − 2 y = −6 { x = 4 y − 3 4 x − 2 y = −6

ⓐ { y = 7 x − 5 3 x − 2 y = 16 { y = 7 x − 5 3 x − 2 y = 16 ⓑ { 12 x − 5 y = −42 3 x + 7 y = −15 { 12 x − 5 y = −42 3 x + 7 y = −15

ⓐ { y = 4 x + 9 5 x − 2 y = −21 { y = 4 x + 9 5 x − 2 y = −21 ⓑ { 9 x − 4 y = 24 3 x + 5 y = −14 { 9 x − 4 y = 24 3 x + 5 y = −14

ⓐ { 14 x − 15 y = −30 7 x + 2 y = 10 { 14 x − 15 y = −30 7 x + 2 y = 10 ⓑ { x = 9 y − 11 2 x − 7 y = −27 { x = 9 y − 11 2 x − 7 y = −27

Everyday Math

Norris can row 3 miles upstream against the current in 1 hour, the same amount of time it takes him to row 5 miles downstream, with the current. Solve the system. { r − c = 3 r + c = 5 { r − c = 3 r + c = 5

• ⓐ for r r , his rowing speed in still water.
• ⓑ Then solve for c c , the speed of the river current.

Josie wants to make 10 pounds of trail mix using nuts and raisins, and she wants the total cost of the trail mix to be $54. Nuts cost $6 per pound and raisins cost $3 per pound. Solve the system { n + r = 10 6 n + 3 r = 54 { n + r = 10 6 n + 3 r = 54 to find n n , the number of pounds of nuts, and r r , the number of pounds of raisins she should use.

Writing Exercises

Solve the system { x + y = 10 5 x + 8 y = 56 { x + y = 10 5 x + 8 y = 56

ⓐ by substitution ⓑ by graphing ⓒ Which method do you prefer? Why?

Solve the system { x + y = −12 y = 4 − 1 2 x { x + y = −12 y = 4 − 1 2 x

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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• Authors: Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis
• Publisher/website: OpenStax
• Book title: Elementary Algebra 2e
• Publication date: Apr 22, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/elementary-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/elementary-algebra-2e/pages/5-3-solve-systems-of-equations-by-elimination

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