Q9. - Algebra 1

Question

The cost of 5 notebooks and 3 pens is $\mathrm{\backslash (}9.75$. The cost of 4 notebooks and 6 pens is $\mathrm{\backslash )}10.50$. Which of the following systems can be used to find the cost of a notebook $n$ and a pen $p$?

a. Write a system of equations to model the situation.

b. Solve the system of equations. How much does each item cost?

Step-by-Step Solution

Answer

a. The system of equations to model the situation is:

$\begin{array}{c}5n+3p=9.75\\ 4n+6p=10.50\end{array}$

b. The cost of each notebook and pen are $\$1.5$ and $\$0.75$ respectively.

Part a. Step 1. Write the equations for the given problem.

Let the cost of 1 notebook be $n$ and the cost of 1 pen be $p$.

It is given that cost of 5 notebooks and 3 pens is $\$9.75$.

Therefore, the equation depicting the above information is: $5n+3p=9.75$

It is given that the cost of 4 notebooks and 6 pens is $\$10.50$.

Therefore, the equation depicting the above information is:

$4n+6p=10.50$

Part a. Step 2. Write a system of equations to model the situation .

 Therefore, the system of equations to model the situation are:

$\begin{array}{c}5n+3p=9.75\\ 4n+6p=10.50\end{array}$

Part b. Step 1. Number the equations in the system of equations.

The equations after numbering is:

$\begin{array}{c}5n+3p=9.75\left(1\right)\\ 4n+6p=10.50\left(2\right)\end{array}$

Multiply the both sides of the equation (1) by 2.

$\begin{array}{c}2\left(5n+3p\right)=2\left(9.75\right)\\ 10n+6p=19.50\left(3\right)\end{array}$

Subtract the equation (2) from the equation (3).

Therefore, it is obtained that:

$$\begin{gathered} 10n+6p-\left(4n+6p\right)=19.50-10.50 \\ 10n+6p-4n-6p=9 \\ 6n=9 \\ n=\frac{9}{6} \\ n=\frac{3}{2} \\ n=1.5 \end{gathered}$$

Therefore, the value of $n$ is 1.5.

Substitute the value of $n$ in equation (1).

$$\begin{gathered} 5n+3p=9.75 \\ 5\left(1.5\right)+3p=9.75 \\ 7.5+3p=9.75 \\ 3p=9.75-7.5 \\ 3p=2.25 \\ p=\frac{2.25}{3} \\ p=0.75 \end{gathered}$$

Therefore, the value of $p$ is 0.75.

Therefore, the cost of each notebook and pen $\$1.5$ are $\$0.75$ and respectively.