5 Ways to Solve Systems of Equations Word Problems

Solving systems of equations is a fundamental concept in mathematics, and word problems are an essential part of applying these skills to real-world situations. In this article, we’ll explore five ways to solve systems of equations word problems, including the substitution method, elimination method, graphing method, matrix method, and a combination of methods.

Understanding Systems of Equations Word Problems

Systems of equations word problems typically involve two or more variables and two or more equations. These problems often represent real-world situations, such as managing finances, optimizing resources, or modeling population growth. To solve these problems, you’ll need to use algebraic techniques to find the values of the variables that satisfy both equations simultaneously.

Method 1: Substitution Method

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This method is useful when one equation is easily solvable for one variable.

Example Problem:

Tom has been saving money for a new bike and has 120 in his savings account. He wants to buy a bike that costs 180. Tom’s parents have agreed to give him an allowance of 5 per week for doing chores. If Tom also earns 10 per week by walking his neighbor’s dog, how many weeks will it take him to have enough money to buy the bike?

Solution:

Let x be the number of weeks Tom needs to save. Since Tom earns 10 per week from dog walking and 5 per week from his allowance, his total weekly earnings are $15. The equation representing Tom’s savings is:

15x + 120 = 180

Subtract 120 from both sides:

15x = 60

Divide both sides by 15:

x = 4

It will take Tom 4 weeks to save enough money to buy the bike.

📝 Note: When using the substitution method, make sure to check your solution by plugging it back into both original equations.

Method 2: Elimination Method

The elimination method involves adding or subtracting the two equations to eliminate one variable. This method is useful when the coefficients of one variable are the same in both equations or are easily made the same by multiplying one equation by a constant.

Example Problem:

A bakery sells a total of 250 loaves of bread per day. They sell a combination of whole wheat and white bread. If the number of whole wheat loaves sold is 30 more than the number of white bread loaves sold, and the total number of white bread loaves sold is 120, how many whole wheat loaves are sold?

Solution:

Let x be the number of whole wheat loaves sold and y be the number of white bread loaves sold. The two equations are:

x + y = 250 x = y + 30

Subtract the second equation from the first equation:

y + 30 - y = 250 - y 30 = 250 - y

Subtract 250 from both sides:

-220 = -y

Divide both sides by -1:

y = 220

Since x = y + 30, we have:

x = 220 + 30 x = 250

However, this can’t be true since x + y = 250. We made an error in our calculations. Let’s re-examine the equations.

x + y = 250 x - y = 30

Add the two equations:

2x = 280

Divide both sides by 2:

x = 140

Now that we have x, we can find y:

140 + y = 250

Subtract 140 from both sides:

y = 110

The bakery sells 140 whole wheat loaves and 110 white bread loaves.

📝 Note: When using the elimination method, make sure to check your solution by plugging it back into both original equations.

Method 3: Graphing Method

The graphing method involves graphing both equations on the same coordinate plane and finding the point of intersection.

Example Problem:

A company produces two products, A and B. The profit from product A is 10 per unit, and the profit from product B is 20 per unit. The company has a limited production capacity of 100 units per day. If the company produces x units of product A and y units of product B, and the profit from product A is $200 more than the profit from product B, how many units of each product should the company produce?

Solution:

Let x be the number of units of product A produced and y be the number of units of product B produced. The two equations are:

10x - 20y = 200 x + y = 100

We can graph both equations on the same coordinate plane:

The point of intersection is (60, 40). Therefore, the company should produce 60 units of product A and 40 units of product B.

📝 Note: When using the graphing method, make sure to label the axes and include a scale.

Method 4: Matrix Method

The matrix method involves representing the system of equations as a matrix and using row operations to find the solution.

Example Problem:

A farmer has 100 acres of land to plant two crops, wheat and oats. Each acre of wheat requires 2 hours of labor, and each acre of oats requires 1 hour of labor. If the farmer has 240 hours of labor available and wants to plant 30 more acres of wheat than oats, how many acres of each crop should the farmer plant?

Solution:

Let x be the number of acres of wheat planted and y be the number of acres of oats planted. The two equations are:

2x + y = 240 x - y = 30

We can represent this system as a matrix:

| 2 1 | | x | | 240 | | 1 -1 | | y | | 30 |

We can use row operations to find the solution:

| 2 1 | | x | | 240 | | 0 -3 | | y | | -60 |

Divide the second row by -3:

| 2 1 | | x | | 240 | | 0 1 | | y | | 20 |

Subtract 2 times the second row from the first row:

| 0 -1 | | x | | 160 | | 0 1 | | y | | 20 |

Divide the first row by -1:

| 0 1 | | x | | -160 | | 0 1 | | y | | 20 |

Therefore, the farmer should plant 20 acres of oats and 50 acres of wheat.

📝 Note: When using the matrix method, make sure to check your solution by plugging it back into both original equations.

Method 5: Combination of Methods

Sometimes, it’s necessary to use a combination of methods to solve a system of equations.

Example Problem:

A company produces three products, A, B, and C. The profit from product A is 10 per unit, the profit from product B is 20 per unit, and the profit from product C is 30 per unit. The company has a limited production capacity of 100 units per day. If the company produces x units of product A, y units of product B, and z units of product C, and the profit from product A is 200 more than the profit from product B, and the profit from product C is $100 more than the profit from product B, how many units of each product should the company produce?

Solution:

Let x be the number of units of product A produced, y be the number of units of product B produced, and z be the number of units of product C produced. The three equations are:

10x - 20y = 200 30z - 20y = 100 x + y + z = 100

We can use the substitution method to solve the first two equations for x and z:

x = 20 + 2y z = 10 + 2y/3

Substitute these expressions into the third equation:

(20 + 2y) + y + (10 + 2y/3) = 100

Combine like terms:

30 + 11y/3 = 100

Subtract 30 from both sides:

11y/3 = 70

Multiply both sides by 3:

11y = 210

Divide both sides by 11:

y = 19

Now that we have y, we can find x and z:

x = 20 + 2(19) x = 58

z = 10 + 2(19)/3 z = 23

Therefore, the company should produce 58 units of product A, 19 units of product B, and 23 units of product C.

📝 Note: When using a combination of methods, make sure to check your solution by plugging it back into all original equations.

In conclusion, there are several ways to solve systems of equations word problems, including the substitution method, elimination method, graphing method, matrix method, and a combination of methods. Each method has its strengths and weaknesses, and the choice of method depends on the specific problem and the skills of the solver.