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# 5.4: Applications of 2 × 2 Systems of Linear Equations

Difficulty Level: At Grade Created by: CK-12

## Solving Problems Using 2 × 2 Systems of Linear Equations

Introduction

In this lesson you will learn to apply the different methods of solving systems of linear equations to real-world problem. You will learn to represent the real-world problems with a system of linear equations. The system of equations will then be solved, using one of the algebraic methods you’ve learned, to determine the solution to the problem. The method that you will use to solve the system will be determined by the makeup of the system of equations.

Objectives

The lesson objectives for Applications of $2 \times 2$ Systems of Linear Equations are:

• Identifying the unknown quantities
• Representing the unknown quantities with appropriate variables
• Writing a system of linear equations to represent the problem
• Solving the linear system to determine the solution of the real-world problem

Introduction

Many real-world problems can be modeled as a system of linear equations in two variables. If two variables are necessary to model the problem, there must be two quantities in the problem. When the quantities have been identified, appropriate variables must be used to represent them.

The following example is a very simple problem but it will give some insight into identifying and representing quantities.

Determine two numbers such that the sum of the numbers is 763 and the difference of the same two numbers is 179.

If the sum of two numbers is 763 and their difference is 179, the numbers are different numbers. If they were the same number, both the sum and the difference would be zero. Therefore, one quantity is a larger number and the other is a smaller number.

Let ‘$x$’ represent the larger number.

Let ‘$y$’ represent the smaller number.

Guidance

The above example showed the first step in solving a real-world problem by using a system of linear equations. Identifying the quantities and assigning a variable to each quantity was the beginning step to solving the problem. Now that the quantities have been identified and represented with variables, the equations can be written to model the problem. The system of equations can then be solved to determine the values for the variables.

Write an equation to model – the sum of two numbers is 763. $\rightarrow \boxed{x+y=763}$

Write an equation to model – the difference of two numbers is 179. $\rightarrow \boxed{x-y=179}$

The system of equations $\begin{Bmatrix}x+y=763\\x-y=179\end{Bmatrix}$ can be solved by a method of your choice. Choose one of the algebraic methods that you have learned to solve the system. Elimination would be a good choice since the variable ‘$y$’ has opposite numerical coefficients.

$& \begin{Bmatrix}x+y=763\\x-y=179\end{Bmatrix}\\\\& x + {\cancel{y}}=763\\& \underline{x -{\cancel{y}}=179} && \text{Add the equations to eliminate} \ y'.\\& \quad 2x=942\\\\& 2x=942 && \text{Solve the equation for the variable} \ x'.\\& \frac{2x}{2}=\frac{942}{2}\\& \frac{\cancel{2}x}{\cancel{2}}=\frac{\overset{{\color{red}471}}{\cancel{942}}}{\cancel{2}}\\& \boxed{x=471}\\\\& x+y=763 && \text{Substitute the value for} \ x' \ \text{into one of the original equations.}\\& ({\color{red}471})+y=763\\& 471 {\color{red}-471}+y=763 {\color{red}-471} && \text{Solve the equation.}\\& \boxed{y = 292}$

The larger number is 471 $(x)$ and the smaller number is 292 $(y)$.

Example A

The length of a rectangular plot of land is 255 yards longer than the width. If the perimeter is 1206 yards, find the dimensions of the rectangle.

The perimeter of a rectangle is found by the formula $P=2l+2w$ where $P$ is the perimeter, $l$ is the length and $w$ is the width. The quantities are the length and the width of the rectangle.

Let the length of the rectangle be represented by ‘$l$’.

Let the width of the rectangle be represented by ‘$w$’.

The equations would be:

The length is 255 yards longer than the width $\rightarrow \boxed{l=w+255}$

The perimeter is 1206 yards $\rightarrow \boxed{2l+2w=1206}$

The system of equations can now be solved to determine the dimensions of the rectangle. In the first equation the length is expressed in terms of the width. Substitution will be used to solve the system of equations.

$& \begin{Bmatrix}l=w+255\\2l+2w=1206\end{Bmatrix}\\& 2l+2w = 1206 && \text{Substitute} \ (w + 255) \ \text{for} \ l' \ \text{in the equation.}\\& 2({\color{red}w+255})+2w = 1206 && \text{Apply the distributive property.}\\& {\color{red}2w+510}+2w = 1206 && \text{Simplify}\\& {\color{red}4w}+510 =1206 && \text{Solve the equation.}\\& 4w+510 {\color{red}-510} = 1206 {\color{red}-510}\\& 4w = {\color{red}696}\\& \frac{4w}{{\color{red}4}} = \frac{696}{{\color{red}4}}\\& \frac{\cancel{4}w}{\cancel{4}} = \frac{\overset{{\color{red}174}}{\cancel{696}}}{\cancel{4}}\\& \boxed{w = 174}\\\\& l = w+255 && \text{Substitute the value for} \ w' \ \text{into the equation.}\\& l = {\color{red}174}+255 && \text{Solve the equation.}\\& \boxed{l = 429}$

The length of the rectangular plot of land is 429 yards and the width is 174 yards.

Example B

Maria had $12 100 to invest. She decided to invest her money in bonds and mutual funds. She invested a portion of the money in bonds paying 8 % interest per year and the remainder in a mutual fund paying 9 % per year. After one year the total income she had earned from the investments was$1043. How much had she invested at each rate?

The two quantities in this problem are the amount she had invested in bonds and the amount she had invested in mutual funds.

Let the amount invested in bonds be represented by ‘$b$’.

Let the amount invested in mutual funds be represented by ‘$m$’.

The equations would be:

The total amount of money she had to invest was $\12 \ 100 \rightarrow \boxed{b+m=12 \ 100}$

The amount of money she earned from the investments was $\1043 \rightarrow \boxed{.08b+.09m=1043}$

The system of equations will be solved using elimination.

$\begin{Bmatrix}b+m=12 \ 100\\.08b+.09m=1043\end{Bmatrix}$

NOTE: To avoid using decimals, multiply the second equation by 100 before you begin to solve the system of equations.

The first equation will be multiplied by (-.08)

$& {\color{red}-.08} (b+m=12 \ 100)\\& {\color{red}-.08b-.08m=-968}\\& -.08b-.08m=-968$

The equations now have opposite, numerical coefficients for the variable ‘$b$’. add the equations to eliminate the variable ‘$b$’.

$& -{\cancel{.08b}}-.08m=-968\\& \underline{\;\;\;\;\cancel{.08b}+.09m=1043\;} && \text{Solve the equation.}\\& {\color{red}.01m=75}\\& .01m = 75\\& \frac{.01m}{{\color{red}.01}} = \frac{75}{{\color{red}.01}}\\& \frac{\cancel{.01}m}{\cancel{.01}} = \frac{\overset{{\color{red}7500}}{\cancel{75}}}{\cancel{.01}}\\& \boxed{m = 7500}\\\\& b+m = 12 \ 100 && \text{Substitute the value for} \ m' \ \text{into the equation.}\\& b+ {\color{red}7500} = 12 \ 100 && \text{Solve the equation.}\\& b+7500 {\color{red}-7500} = 12 \ 100 {\color{red}-7500}\\& \boxed{b = 4600}$

Maria invested $4600 in bonds and$7500 in mutual funds.

Example C

Pedro was saving quarters and dimes to buy a new skateboard. After months of saving his coins in a bottle, he emptied its contents and counted the money. The 561 coins in the bottle had a total value of $107.85. How many of each coin were in the bottle? The two quantities in this problem are the number of quarters and the number of dimes. Let the number of quarters be represented by ‘$q$’. Let the number of dimes be represented by ‘$d$’. The equations would be: The total number of coins in the bottle was $561 \rightarrow \boxed{q+d=561}$ The amount of money in the bottle was $\ 107.85 \rightarrow \boxed{.25q+.10d=107.85}$ This system will be solved using the substitution method. Solve the first equation in terms of quarters. $& q+d = 561\\& q+d {\color{red}-d} = 561 {\color{red}-d}\\& \boxed{q = 561-d}\\\\& .25q+.10d = 107.85 && \text{Substitute the value for} \ q' \ \text{into the equation.}\\& .25({\color{red}561-d})+.10d = 107.85 && \text{Apply the distributive property.}\\& {\color{red}140.25-.25d}+.10d = 107.85 && \text{Solve the equation.}\\& 140.25 {\color{red}-.15d} = 107.85\\& 140.25 {\color{red}-140.25}-.15d = 107.85 {\color{red}-140.25}\\& -.15d = {\color{red}-32.40}\\& \frac{-.15d}{{\color{red}-.15}} = \frac{-32.40}{{\color{red}-.15}}\\& \frac{{\cancel{-.15}d}}{\cancel{-.15}} = \frac{\overset{{\color{red}216}}{\cancel{-32.40}}}{\cancel{-.15}}\\& \boxed{d = 216}\\\\& q+d = 561 && \text{Substitute the value for} \ d' \ \text{into the equation.}\\& q+ {\color{red}216} = 561 && \text{Solve the equation.}\\& q+216 {\color{red}-216} = 561 {\color{red}-216}\\& q = {\color{red}345}\\& \boxed{q = 345}$ The number of quarters that Pedro had saved in the bottle was 345 and the number of dimes was 216. Example D The Bayplex and Centre 200 rent their ice out to the community whenever possible. The Bayplex charges a flat rate of$20.00 plus $15.00 for every hour rented. Centre 200 charges$50.00 for a flat rate but only asks for $10.00 for every hour rented. a) Write an equation to model the cost of renting the ice surface for each arena. b) Determine the intersection point of the equations. What does this intersection point represent? c) Explain when it is best to use the Bayplex and when it is best to use Center 200. a) Begin by writing the equations to model the cost of renting the ice surface in each arena. The cost of renting the ice at the Bayplex $\rightarrow \boxed{c=15h+20}$ The cost of renting the ice at Centre $200 \rightarrow \boxed{c=10h+50}$ b) The intersection point of the costs for the arenas can be determined by using the comparison method. Both equations are equal to the variable ‘$c$’. $15h+20 &= 10h+50 && \text{Solve the equation.}\\15h+20 {\color{red}-20} &= 10h+50 {\color{red}-20}\\15h &= 10h {\color{red}+30}\\15h {\color{red}-10h} &= 10h {\color{red}-10h}+30\\{\color{red}5h} &= 30\\\frac{5h}{{\color{red}5}} &= \frac{30}{{\color{red}5}}\\\frac{\cancel{5}h}{\cancel{5}} &= \frac{\overset{{\color{red}6}}{\cancel{30}}}{\cancel{5}}\\h &= 6$ Substitute the value for ‘$h$’ into one of the original equations. $& c = 15h+20\\& c = 15 ({\color{red}6})+20 && \text{Evaluate the equation.}\\& c = {\color{red}90}+20\\& \boxed{c = \ 110}$ The intersection point of the system of linear equations is (6,110). This point is the time when the cost of renting the arena will be the same. Renting the Bayplex for six hours will cost$110 and renting Centre 200 for six hours will cost $110. c) The flat rate to rent the Bayplex is only$20.00. Therefore, it will cost less to rent the Bayplex for less than six hours but more to rent it after six hours. If you need to rent the ice surface for more than six hours, rent the Centre 200.

Guided Practice

1. The sum of Henry’s and his mother’s age is 67. Three times Henry’s age increased by 7 is his mother’s age. How old is Henry?
2. The Sydney Schooners played a total of 41 games of hockey. The number of games lost was ten less than one-half the number of games won. How many games did the Schooners win?
3. Tim invested $3000. A portion of the money was invested into a college fund that paid 8% interest per year and the remainder was invested into a retirement fund that paid 7 % interest per year. At the end of the first year, the interest from the college fund was$60 more than the interest from the retirement fund. How much money did Tim invest in each fund?

1. The two quantities in this problem are Henry’s age and mother’s age.

Let Henry’s age be represented by ‘$h$’.

Let mother’s age be represented by ‘$m$’.

The equations would be:

The sum of Henry’s age and his mother’s age is 67. $\rightarrow \boxed{h+m=67}$

Three times Henry‘s age increased by 7 is his mother’s age $\rightarrow 3h+7=m$

The system of equations will be solved by substitution.

$& \begin{Bmatrix}h+m=67\\3h+7=m\end{Bmatrix}\\& h+m = 67 && \text{Substitute} \ 3h+7 \ \text{into the equation for} \ m'.\\& h+({\color{red}3h+7}) = 67 && \text{Apply the distributive property.}\\& h {\color{red}+3h+7} = 67 && \text{Solve the equation.}\\& {\color{red}4h}+7 = 67 &&\\& 4h+7 {\color{red}-7} = 67 {\color{red}-7}\\& 4h = {\color{red}60}\\& \frac{4h}{{\color{red}4}} = \frac{60}{{\color{red}4}}\\& \frac{\cancel{4}h}{\cancel{h}} = \frac{\overset{{\color{red}15}}{\cancel{60}}}{\cancel{4}}\\& \boxed{h = 15} && \text{Henry is} \ 15 \ \text{years old. The problem asked for his mother's age.}\\\\& 3h+7 = m && \text{Substitute} \ 15 \ \text{into the equation for} \ h'.\\& 3({\color{red}15})+7 = m && \text{Solve the equation.}\\& {\color{red}45}+7 = m\\& \boxed{52 = m} && \text{Henry's mother is} \ 52 \ \text{years of age.}$

2. The two quantities in this problem are the games won and the games lost.

Let games won be represented by ‘$w$’.

Let games lost be represented by ‘$l$’.

The equations would be:

The total number of games played (losses and wins) is 41. $\rightarrow \boxed{l+w=41}$

The number of games lost was 10 less than one-half the number of games won. $\rightarrow \boxed{l=\frac{w}{2}-10}$

The system of equations will be solved by elimination.

$& l = \frac{w}{2}-10 && \text{Multiply the equation by} \ 2.\\& {\color{red}2}(l) = {\color{red}2} \left[\frac{w}{2}\right]-{\color{red}2}(10)\\& {\color{red}2}(l) = {\color{red}\cancel{2}} \left[\frac{w}{\cancel{2}}\right]-{\color{red}2}(10)\\& {\color{red}2l} = {\color{red}w}-{\color{red}20} && \text{Align the equations such that} \begin{Bmatrix} a_1 {\color{red}x}+b_1 {\color{blue}y}=c_1\\ a_2 {\color{red}x}+b_2 {\color{blue}y}=c_2\end{Bmatrix}\\& 2l {\color{red}-w} = w {\color{red}-w}-20\\& \boxed{2l-w = -20}$

The system of linear equations to solve by elimination is:

$& \begin{Bmatrix}l+w=41\\2l-w=-20\end{Bmatrix}\\\\& l + \cancel{w} = 41\\& \underline{2l -\cancel{w}=-20} && \text{Add the equations to eliminate} \ w'.\\& {\color{red}3l = 21}\\& 3l=21 && \text{Solve the equation.}\\& \frac{3l}{{\color{red}3}} = \frac{21}{{\color{red}3}}\\& \frac{\cancel{3}l}{\cancel{3}} = \frac{\overset{{\color{red}7}}{\cancel{21}}}{\cancel{3}}\\& \boxed{l=7} && \text{The number of games the Schooners lost was seven.}\\\\& l+w=41 && \text{Substitute} \ 7 \ \text{into the equation for the variable} \ l'.\\& {\color{red}7}+w=41 && \text{Solve the equation.}\\&7 {\color{red}-7}+w=41 {\color{red}-7}\\& w = {\color{red}34}\\& \boxed{w=34} && \text{The number of games the Schooners won was} \ 34.$

3. The two quantities in this problem are the amount of money invested in the college fund and the amount of money invested in the retirement fund.

Let the amount of money invested in the college fund be represented by ‘$c$’.

Let the amount of money invested in the retirement fund be represented by ‘$r$’.

The equations would be:

The total amount of money to be invested is $3000 $\rightarrow \boxed{c+r=3000}$ The amount of interest from the 8% college fund is$60 more than the amount of interest from the 7% retirement fund. $\rightarrow \boxed{.08c=.07r+60}$

The system of equations will be solved by substitution.

$& c+r = 3000 && \text{Solve the equation for} \ c'.\\& c+r {\color{red}-r} = 3000 {\color{red}-r}\\& \boxed{c = 3000-r}\\& .08c = .07r+60 && \text{Substitute} \ 3000-r \ \text{into the equation for the variable} \ c '.\\& .08({\color{red}3000-r}) = .07r+60 && \text{Apply the distributive property. Solve the equation.}\\& {\color{red}240-.08r} = .07r+60\\& 240 {\color{red}-240}-.08r = .07r+60 {\color{red}-240}\\& -.08r {\color{red}-.07r} = .07r {\color{red}-.07r}-180\\& {\color{red}-.15r} = -180\\& \frac{-.15r}{{\color{red}-.15}} = \frac{-180}{{\color{red}-.15}}\\& \frac{\cancel{-.15}r}{\cancel{-.15}} = \frac{\overset{{\color{red}1200}}{\cancel{-180}}}{\cancel{-.15}}\\& \boxed{r = 1200} && \text{The amount of money invested in the retirement fund was} \ \1200.\\\\& c = 3000-r && \text{Substitute} \ 1200 \ \text{into the equation for} \ `r'.\\& c = 3000 -{\color{red}1200} && \text{Solve the equation.}\\& c = {\color{red}1800}\\& \boxed{c = 1800} && \text{The amount of money invested in the college fund was} \ \1800.$

Summary

In this lesson you have learned that a system of linear equations can be used to represent a real-world problem. You have also learned that the system of equations can be solved by the substitution method or the elimination method. The solution to the system of equations is the answer to the real-world problem.

Problem Set

Write each of the following statements as a linear equation in two variables.

1. The sum of two numbers is 100.
2. When six times the larger of two numbers is added to 4 times the smaller, the result is 112.
3. The length of a rectangle is 8m more than 7 times its width.
4. Four times the number of nickels less three times the number of pennies is 56.
5. Jason has some $5 bills and some$1 bills which have a total of $91. Solve each of the following real-world problems using $2 \times 2$ systems of linear equations. The solution may be determined by using a method of your choice. 1. At a party, there were 72 people. The hostess counted the shoes and found that there were 32 more ladies’ shoes compared to the number of males’ shoes. How many males and females were at the party? 2. Two weight loss programs offer competitive services. Super Slim charges$33 to join and $1.50 per session whereas Think Thin charges$2.50 per session and $15 to join. Determine algebraically, under what circumstances you would choose each plan. 3. Today Sam is twice Jenny’s age. Three years ago the sum of their ages was 45. How old are Sam and Jenny today? 4. The parking lot at a local amusement park contained 123 vehicles (cars and buses). Each car is charged$3 to park for the day and each bus is charged $10. If the total revenue for the day was$481.00, how many cars were on the parking lot?
5. Seven times the larger of two numbers less three times the smaller is 351. Six times the larger less twice the smaller is 342. What are the numbers?

Write each of the following...

1. Let $x$ represent one number. Let $y$ represent the other number. $\boxed{x+y=100}$
1. Let $l$ represent the length. Let $w$ represent the width. $\boxed{l=8+7w}$
1. Let $x$ represent the number of $5 bills. Let $y$ represent the number of$1 bills. $\boxed{5x+1y=91}$

Solve each of the following...

1. The two quantities are the number of females and the number of males. Let $f$ represent the number of females. Let $m$ represent the number of males. If there are 32 more ladies’ shoes, then there must be 16 more females than males at the party. The equations are:

$\begin{Bmatrix}f+m=72\\f=m+16\end{Bmatrix}$

Solve by substitution:

$f+m &= 72\\(m+16)+m &= 72\\m+16+m &= 72\\2m+16 &= 72\\2m+16-16 &=72-16\\2m &= 56\\\frac{2m}{2} &= \frac{56}{2}\\\frac{\cancel{2}m}{\cancel{2}} &= \frac{\overset{28}{\cancel{56}}}{\cancel{2}}\\m &= 28\\\\f &= m+16\\f &= 28+16\\f &= 44$

There are 44 females and 28 males at the party.

1. The two quantities are Sam’s age and Jenny’s age. Let $s$ represent Sam’s age. Let $j$ represent Jenny’s age. The equations are:

$\begin{Bmatrix}s=2j\\(s-3)+(j-3)=45\end{Bmatrix}$

Solve by substitution:

$(s-3)+(j-3) &= 45\\s-3+j-3 &= 45\\s+j-6 &= 45\\s+j-6+6 &= 45+6\\s+j &= 51\\\\s+j &= 51\\2j+j &= 51\\3j &= 51\\\frac{3j}{3} &= \frac{51}{3}\\\frac{\cancel{3}j}{\cancel{3}} &= \frac{\overset{17}{\cancel{51}}}{\cancel{3}}\\j &= 17\\\\s &= 2j\\s &= 2(17)\\s &= 34$

Today Sam is 34 years of age and Jenny is 17 years of age.

1. The two quantities are a larger number and a smaller number. Let $x$ represent the larger number. Let $y$ represent the smaller number. The equations are

$\begin{Bmatrix}7x-3y=351\\6x-2y=342\end{Bmatrix}$

Solve by elimination:

$2(7x-3y=351) & \rightarrow 14x -{\cancel{6y}}=702\\-3(6x-2y=342) & \rightarrow \underline{-18x +{\cancel{6y}}=-1026}\\& \qquad \qquad \frac{\cancel{-4}x}{\cancel{-4}} = \frac{\overset{81}{\cancel{-324}}}{\cancel{-4}}\\& \qquad \qquad \quad \ x=81\\\\7x-3y &= 351\\7(81)-3y &= 351\\567-3y &= 351\\567-567-3y &= 351-567\\\frac{-3y}{-3} &= \frac{-216}{-3}\\\frac{\cancel{-3}y}{\cancel{-3}} &= \frac{\cancel{-216}}{\cancel{-3}}\\y &= 72$

The larger number is 81 and the smaller number is 72.

Jan 16, 2013

Jan 14, 2015