## System of Equations Word Problem

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### System of Equations

Generally, a system of linear equations is defined by two or more equations that contain the same variables. When we solve a system of two linear equations, we find the point of intersection of the two lines represented by the equations (if it exists). A system of linear equations can have a unique solution, infinitely many solutions, or no solutions. There are four primary methods to solve a system of linear equations:

- Elimination
- Matrices
- Substitution
- Graphing

In this article, we will learn about the elimination method. The elimination method involves adding, subtracting, or multiplying the equations in a linear system to eliminate one variable, making it easier to solve for the other variable(s).

### Solving a system of equations word problem

**Question**: A class of \(82\) students went on a field trip. A total of \(7\) vehicles were taken, out of which some were cars and some buses. If each **car **holds \(7\) students and each **bus **hold \(18\) students, how many buses did they take?

Let’s consider the number of **buses **to be \(x\) and the number of **cars **to be \(y\). So, it is clearly given that the **total **number of vehicles is \(7\), so, \(x \ + \ y \ = \ 7\).

Now, it is given that **total **number of students is \(82\) and each car holds \(7\) students and each bus holds \(18\). So, \(18x \ + \ 7y \ = \ 82\).

Now, let’s solve both these equations.

- Firstly, we need to equate the
**coefficient**of any term (be it \(x\) or \(y\)) between the two equations. Here, we can see that if we multiply the first equation by \(18\) on both sides, we get the**same coefficient**of \(x\) as the second equation. So, let’s multiply the first equation by \(18\) to get \(18x \ + \ 18y \ = \ 126\). - Now, our aim is to add or subtract both equations so that one variable gets eliminated. So, if we
**subtract**both equations (equation \(1 \ –\) equation \(2\)) we get \(11y \ = \ 44\), hence we get the value of \(y \ = \ 4\). - Now, if we put the value of \(y\) in any equation (suppose equation \(1\)), we get \(x \ + \ 4 \ = \ 7\), hence we get the value of \(x \ = \ 3\).
- So, we can say that this system of equations has the solution \(x \ = \ 3\) and \(y \ = \ 4\). So, the number of buses is \(3\).

### Exercises for Systems of Equations Word Problems

**1)** The equations of the two lines are \(4x \ + \ 3y \ = \ 40\) and \(7x \ + \ 6y \ = \ 73\). Find the value of \(x\) and \(y\) in the solution for this system of equations.

\( \begin{align} 4x \ + \ 3y &= 40 \\ 7x \ + \ 6y &= 73 \\ \hline \end{align}\) \( \ \Rightarrow \ \)

**2) **A theater is selling tickets for a performance. Mr. Smith purchased \(16\) senior tickets and \(3\) child tickets for \($92\) for his friends and family. Mr. Jackson purchased \(1\) senior tickets and \(1\) child tickets for \($9\). What is the price of a senior ticket?

\( \begin{align} 16x \ + \ 3y &= 92 \\ x \ + \ y &= 9 \\ \hline \end{align}\) \( \ \Rightarrow \ \)

**3) **At a store, Eva bought \(1\) shirts and \(2\) hats for \($11\). Nicole bought \(1\) same shirts and \(1\) same hats for \($7\). What is the price of each shirt?

\( \begin{align} x \ + \ 2y &= 11 \\ x \ + \ y &= 7 \\ \hline \end{align}\) \( \ \Rightarrow \ \)

**4) **The equations of the two lines are \( 4x \ + \ 3y \ = \ 40\) and \(7x \ + \ 6y \ = \ 73\). Find the value of \(x\) and \(y\) in the solution for this system of equations.

\( \begin{align} 5x \ + \ 6y &= 50 \\ 3x \ + \ 5y &= 37 \\ \hline \end{align}\) \( \ \Rightarrow \ \)

**5) **At a store, Eva bought \(4\) shirts and \(2\) hats for \($30\). Nicole bought \(1\) same shirts and \(1\) same hats for \($10\). What is the price of each shirt?

\( \begin{align} 4x \ + \ 2y &= 30 \\ x \ + \ y &= 10 \\ \hline \end{align}\) \( \ \Rightarrow \ \)

**6) **A class of \(24\) students went on a field trip. They took \(9\) vehicles, some cars, and some buses. If each car holds \(2\) students and each bus hold \(4\) students, how many buses did they take?

\( \begin{align} 4x \ + \ 2y &= 24 \\ x \ + \ y &= 9 \\ \hline \end{align}\) \( \ \Rightarrow \ \)

**7) **Emma and Sepehr are selling Chocolate Chip cookies and Oreo cookies Emma sold \(3\) boxes of Chocolate Chip Cookies and \(7\) boxes of Oreo cookies for a total of \($46\). Sepehr sold \(5\) boxes of Chocolate Chip Cookies and \(6\) boxes of Oreo cookies for a total of \($54\). Find the cost of one box of Chocolate Chip cookies.

\( \begin{align} 3x \ + \ 7y &= 46 \\ 5x \ + \ 6y &= 54 \\ \hline \end{align}\) \( \ \Rightarrow \ \)

**8) **A theater is selling tickets for a performance. Mr. Smith purchased \(5\) senior tickets and \(7\) child tickets for \($58\) for his friends and family. Mr. Jackson purchased \(7\) senior tickets and \(6\) child tickets for \($66\). What is the price of a senior ticket?

\( \begin{align} 5x \ + \ 7y &= 58 \\ 7x \ + \ 6y &= 66 \\ \hline \end{align}\) \( \ \Rightarrow \ \)

**9) **The equations of the two lines are \(x \ - \ y \ = \ 5\) and \(x \ + \ y \ = \ 17\). Find the value of \(x\) and \(y\) in the solution for this system of equations.

\( \begin{align} x \ - \ y &= 5 \\ x \ + \ y &= 17 \\ \hline \end{align}\) \( \ \Rightarrow \ \)

**10) **Emma and Sepehr are selling Chocolate Chip cookies and Oreo cookies Emma sold \(7\) boxes of Chocolate Chip Cookies and \(5\) boxes of Oreo cookies for a total of \($58\). Sepehr sold \(7\) boxes of Chocolate Chip Cookies and \(2\) boxes of Oreo cookies for a total of \($40\). Find the cost of one box of Chocolate Chip cookies.

\( \begin{align} 7x \ + \ 5y &= 58 \\ 7x \ + \ 2y &= 40 \\ \hline \end{align}\) \( \ \Rightarrow \ \)

\( \begin{align} 4x \ + \ 3y &= 40 \\ 7x \ + \ 6y &= 73 \\ \hline \end{align}\) \( \ \Rightarrow \color{red}{ + \begin{align} -7(4x \ + \ 3y) &= -7(40) \\ 4(7x \ + \ 6y) &= 4(73) \\ \hline 3y &= 12 \end{align}}\) \( \ \Rightarrow \ \color{red}{ y = \frac{12 }{ 3} = 4} \)\( \ \Rightarrow \ \color{red}{ 4x \ + \ 3(4) = 40 }\) \( \ \Rightarrow \ \color{red}{ x = \frac{40 \ - \ (12)}{4} = 7}\)

\( \begin{align} 16x \ + \ 3y &= 92 \\ x \ + \ y &= 9 \\ \hline \end{align}\) \( \ \Rightarrow \color{red}{ + \begin{align} -1(16x \ + \ 3y) &= -1(92) \\ 16(x \ + \ y) &= 16(9) \\ \hline 13y &= 52 \end{align}}\) \( \ \Rightarrow \ \color{red}{ y = \frac{52 }{ 13} = 4} \)\( \ \Rightarrow \ \color{red}{ 16x \ + \ 3(4) = 92 }\) \( \ \Rightarrow \ \color{red}{ x = \frac{92 \ - \ (12)}{16} = 5}\)

\( \begin{align} x \ + \ 2y &= 11 \\ x \ + \ y &= 7 \\ \hline \end{align}\) \( \ \Rightarrow \color{red}{ + \begin{align} -1(x \ + \ 2y) &= -1(11) \\ x \ + \ y &= 7 \\ \hline -y &= -4 \end{align}}\) \( \ \Rightarrow \ \color{red}{ y = 4} \)\( \ \Rightarrow \ \color{red}{ x \ + \ 2(4) = 11 }\) \( \ \Rightarrow \ \color{red}{ x = 11 \ - \ 8 = 3}\)

\( \begin{align} 5x \ + \ 6y &= 50 \\ 3x \ + \ 5y &= 37 \\ \hline \end{align}\) \( \ \Rightarrow \color{red}{ + \begin{align} -3(5x \ + \ 6y) &= -3(50) \\ 5(3x \ + \ 5y) &= 5(37) \\ \hline 7y &= 35 \end{align}}\) \( \ \Rightarrow \ \color{red}{ y = \frac{35 }{ 7} = 5} \)\( \ \Rightarrow \ \color{red}{ 5x \ + \ 6(5) = 50 }\) \( \ \Rightarrow \ \color{red}{ x = \frac{50 \ - \ (30)}{5} = 4}\)

\( \begin{align} 4x \ + \ 2y &= 30 \\ x \ + \ y &= 10 \\ \hline \end{align}\) \( \ \Rightarrow \color{red}{ + \begin{align} -1(4x \ + \ 2y) &= -1(30) \\ 4(x \ + \ y) &= 4(10) \\ \hline 2y &= 10 \end{align}}\) \( \ \Rightarrow \ \color{red}{ y = \frac{10 }{ 2} = 5} \)\( \ \Rightarrow \ \color{red}{ 4x \ + \ 2(5) = 30 }\) \( \ \Rightarrow \ \color{red}{ x = \frac{30 \ - \ (10)}{4} = 5}\)

\( \begin{align} 4x \ + \ 2y &= 24 \\ x \ + \ y &= 9 \\ \hline \end{align}\) \( \ \Rightarrow \color{red}{ + \begin{align} -1(4x \ + \ 2y) &= -1(24) \\ 4(x \ + \ y) &= 4(9) \\ \hline 2y &= 12 \end{align}}\) \( \ \Rightarrow \ \color{red}{ y = \frac{12 }{ 2} = 6} \)\( \ \Rightarrow \ \color{red}{ 4x \ + \ 2(6) = 24 }\) \( \ \Rightarrow \ \color{red}{ x = \frac{24 \ - \ (12)}{4} = 3}\)

\( \begin{align} 3x \ + \ 7y &= 46 \\ 5x \ + \ 6y &= 54 \\ \hline \end{align}\) \( \ \Rightarrow \color{red}{ + \begin{align} -5(3x \ + \ 7y) &= -5(46) \\ 3(5x \ + \ 6y) &= 3(54) \\ \hline -17y &= -68 \end{align}}\) \( \ \Rightarrow \ \color{red}{ y = \frac{-68 }{ -17} = 4} \)\( \ \Rightarrow \ \color{red}{ 3x \ + \ 7(4) = 46 }\) \( \ \Rightarrow \ \color{red}{ x = \frac{46 \ - \ (28)}{3} = 6}\)

\( \begin{align} 5x \ + \ 7y &= 58 \\ 7x \ + \ 6y &= 66 \\ \hline \end{align}\) \( \ \Rightarrow \color{red}{ + \begin{align} -7(5x \ + \ 7y) &= -7(58) \\ 5(7x \ + \ 6y) &= 5(66) \\ \hline -19y &= -76 \end{align}}\) \( \ \Rightarrow \ \color{red}{ y = \frac{-76 }{ -19} = 4} \)\( \ \Rightarrow \ \color{red}{ 5x \ + \ 7(4) = 58 }\) \( \ \Rightarrow \ \color{red}{ x = \frac{58 \ - \ (28)}{5} = 6}\)

\( \begin{align} x \ - \ y &= 5 \\ x \ + \ y &= 17 \\ \hline \end{align}\) \( \ \Rightarrow \color{red}{ + \begin{align} -1(x \ - \ y) &= -1(5) \\ 1(x \ + \ y) &= 1(17) \\ \hline 2y &= 12 \end{align}}\) \( \ \Rightarrow \ \color{red}{ y = \frac{12 }{ 2} = 6} \)\( \ \Rightarrow \ \color{red}{ 1x \ + \ (-1)(6) = 5 }\) \( \ \Rightarrow \ \color{red}{ x = \frac{5 \ - \ (-6)}{1} = 11}\)

\( \begin{align} 7x \ + \ 5y &= 58 \\ 7x \ + \ 2y &= 40 \\ \hline \end{align}\) \( \ \Rightarrow \color{red}{ + \begin{align} -7(7x \ + \ 5y) &= -7(58) \\ 7(7x \ + \ 2y) &= 7(40) \\ \hline -21y &= -126 \end{align}}\) \( \ \Rightarrow \ \color{red}{ y = \frac{-126 }{ -21} = 6} \)\( \ \Rightarrow \ \color{red}{ 7x \ + \ 5(6) = 58 }\) \( \ \Rightarrow \ \color{red}{ x = \frac{58 \ - \ (30)}{7} = 4}\)