Chapter 0.1, Problem 39E

To determine

To determine the cost of a burger and the cost of an order of fries.

Answer to Problem 39E

The cost of a burger is $\$4.28$ and that of an order of fries is $\$2.34$ .

Explanation of Solution

Given:

Given that $5$ burgers and $4$ orders of fries cost $\$30.76$ and that $8$ burgers and $6$ orders of fries cost $\$48.28$ .

Concept Used:

Method of substitution in the case of linear equations involving two variables:

Given a system of linear equations in two variables:

  $\left\{\begin{array}{c}{a}_{1}x+b_{1}y=c_1\\ a_{2}x+b_{2}y=c_2\end{array}\right.$ .

The method of substitution employs solving one of the equations, say $a_{1}x+b_{1}y=c_1$ , to find an algebraic expression for one of its variables, say $x$ .

Now, this expression of $x$ is substituted in the other equation $a_{2}x+b_{2}y=c_2$ and then solved to find the value of $y$ .

The value of $y$ is now substituted in either $a_{1}x+b_{1}y=c_1$ or $a_{2}x+b_{2}y=c_2$ to find the value of $x$ .

Now, the value of $x$ and that of $y$ are recorded to be the solution of the system of equations.

Calculation:

Suppose that $x$ denotes the cost of a burger and that $y$ denotes the cost of an order of fries in dollars.

Since $5$ burgers and $4$ orders of fries cost $\$30.76$ it follows that:

  $5x+4y=30.76$ .

Again, since $8$ burgers and $6$ orders of fries cost $\$48.28$ , it follows that:

  $8x+6y=48.28$ .

Now, the problem is to solve the system of equations $\left\{\begin{array}{c}5x+4y=30.76\\ 8x+6y=48.28\end{array}\right.$ for $x$ and $y$ .

Solve the system of equations:

Consider the first equation $5x+4y=30.76$ .

Isolate $y$ on the left side of the equation $5x+4y=30.76$ :

  $\begin{array}{c}5x+4y=30.76\\ 4y=30.76-5x\\ y=\frac{30.76-5x}{4}\end{array}$ .

Substitute $y=\frac{30.76-5x}{4}$ in the second equation $8x+6y=48.28$ and solve for $x$ :

  $\begin{array}{c}8x+6y=48.28\\ 8x+6\left(\frac{30.76-5x}{4}\right)=48.28\\ 8x+6\cdot \frac{30.76}{4}-6\cdot \frac{5x}{4}=48.28\\ 8x-6\cdot \frac{5}{4}x=48.28-6\cdot \frac{30.76}{4}\\ \left(8-\frac{15}{2}\right)x=2.14\\ 0.5x=2.14\\ x=\frac{2.14}{0.5}\\ x=4.28\end{array}$ .

Thus, it is found that:

  $x=4.28$ .

Substitute $x=4.28$ in the second equation $8x+6y=48.28$ and solve for $y$ :

  $\begin{array}{c}8x+6y=48.28\\ 8\left(4.28\right)+6y=48.28\\ 34.24+6y=48.28\\ 6y=48.28-34.24\\ 6y=14.04\\ y=\frac{14.04}{6}\\ y=2.34\end{array}$ .

Thus, it is found that:

  $y=2.34$ .

Thus, the solution of the system of equations $\left\{\begin{array}{c}5x+4y=30.76\\ 8x+6y=48.28\end{array}\right.$ :

  $x=4.28$ and $y=2.34$ .

That is, the cost of a burger is $\$4.28$ and that of an order of fries is $\$2.34$ .

Conclusion:

The cost of a burger is $\$4.28$ and that of an order of fries is $\$2.34$ .