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Use the given information to build a system of linear equations.
30 min on the elliptical trainer and 10 min on the stationary bike.
We are told that we have a total of 40 minutes to exercise. If we let t be the time we exercise on the elliptical trainer and b be the time we exercise on the stationary bike, we can write an equation to represent the situation.
40=t+b
We can rearrange this equation, so it will be easier to plot it later.
b=- t+40
| Verbal expression | Algebraic expression |
|---|---|
| Calories burnt on the elliptical trainer per minute | 8 |
| Time spent on the elliptical trainer | t |
| Total calories burnt on the elliptical trainer | 8t |
| Calories burnt on the stationary bike per minute | 6 |
| Time spent on the stationary bike | b |
| Total calories burnt on the stationary bike | 6b |
| Total calories burnt on both machines | 8t+6b |
Let's build an equation that compares the total calories burnt with the given goal of 300 calories. 300=8t+6b Isolating b will help us graph this equation.
LHS-8t=RHS-8t
.LHS /6.=.RHS /6.
Put minus sign in front of fraction
a* b/c=a/c* b
.a /2./.b /2.=a/b
Calculate quotient
Rearrange equation
Now that we have both equations in the slope-intercept form, we can plot them and find the point of intersection.
From the graph, we can see that the solution, or point of intersection, to this system of linear equations is the point (30,10). However, we should check the point by substituting it into the equations. Let's start with the first one.
Now let's check the second one.
t= 30, b= 10
a/c* b = a* b/c
Calculate quotient
Add terms
Since both equations hold true, the point (30,10) is a solution. This means that we should exercise 30 minutes on the elliptical trainer and 10 minutes on the stationary bike.