Q49. - Algebra 2

Question

Megan exercises every morning for $40$ minutes. She does a combination of step aerobics, which burns about $11$ Calories per minute, and stretching, which burns about 4 Calories per minute. Her goal is to burn $335$ Calories during her routine.

How long should she participate in each activity in order to burn 335 Calories?

Step-by-Step Solution

Answer

She does $25$ minutes of step aerobics and $15$ minutes of stretching.

Step-1 – Apply the concept of system of equations

A set of two or more equations with the same variables is called a system of linear equations. The ordered pairs that satisfy all the equations in the system is called as solution of the equations.

Step-2 – Formation of the equations

Here, x represents the number of minutes Megan does step aerobics and y represents the number of minutes she does stretching. The equations so formed according to the question are:

$\begin{array}{c}x+y=40\\ 11x+4y=335\end{array}$

Step-3 – Apply the substitution method of solving equations

The algebraic method of substitution involves solving the one of the two equations for one variable in terms of other variable and then substituting the expression so formed for the variable in the second equation.

Step-4 – Solving one equation for x in terms of y

To solve the equation $x+y=40$ for x in terms of y, subtract y from both sides as shown below.

$\begin{array}{c}x+y-y=40-y\\ x=40-y\end{array}$

Step-5 – Substitute the expression

Now, substitute $x=40-y$ in the equation $11x+4y=335$ and solve.

$\begin{array}{c}11\left(40-y\right)+4y=335\\ 440-11y+4y=335\\ -7y=-105\\ y=15\end{array}$

Thus, the value of y is $15$.

Step-6 – Substitute the value of variable

To find the value of x, substitute $y=15$ in the equation $x+y=40$ and then solve as shown.

$\begin{array}{c}x+y=40\\ x+15=40\\ x=40-15\\ x=25\end{array}$

Thus, the value of x is $25$.

Hence, she does $25$ minutes of step aerobics and $15$ minutes of stretching.