Question

practice 2.03

1. the car and scooter are moving at constant speeds. the equation 21t = 19t + 7 represents the time, t, when they will be in the same position. when will the car and scooter be in the same position?

problems 2 - 3: the equation 10t = 2.5t represents the time, t, when two vehicles will be in the same position.

1. when will these two vehicles be in the same position? circle one. once never always
2. explain how you know.
3. here is kiandras work to solve 16x = 10x. she says there is no solution. is this correct? show or explain your thinking.
4. group the equations based on their number of solutions. a. 5t = 3t b. 2t = 10 - 2t c. 15 - 3(t + 5) = -3t d. 4t + 7 = 4(t + 2) e. 6t + 2 = -3 + 6t

one solution no solution infinitely many solutions

Explanation:

Step1: Solve problem 2

Given equation is \(10t = 2.5t\). Subtract \(2.5t\) from both sides: \(10t-2.5t=2.5t - 2.5t\), which simplifies to \(7.5t = 0\). Then \(t = 0\). So they are in the same - position once.

Step2: Solve problem 3

The equation \(10t = 2.5t\) can be thought of as two lines representing the positions of the vehicles. The left - hand side \(y = 10t\) and the right - hand side \(y = 2.5t\). They intersect at \(t = 0\), so they are in the same position once.

Step3: Solve problem 4

Kiandra's work is incorrect. When solving \(16x=10x\), we should not divide by \(x\) directly (because \(x\) could be \(0\)). Instead, subtract \(10x\) from both sides: \(16x−10x = 10x−10x\), getting \(6x = 0\), so \(x = 0\) is the solution.

Step4: Solve problem 5 for equation A

For \(5t = 3t\), subtract \(3t\) from both sides: \(5t−3t=3t - 3t\), \(2t = 0\), \(t = 0\) (one solution).

Step5: Solve problem 5 for equation B

For \(2t=10 - 2t\), add \(2t\) to both sides: \(2t + 2t=10-2t + 2t\), \(4t = 10\), \(t=\frac{10}{4}=2.5\) (one solution).

Step6: Solve problem 5 for equation C

Expand \(15−3(t + 5)=-3t\): \(15-3t-15=-3t\), \(-3t=-3t\). This is an identity, so there are infinitely many solutions.

Step7: Solve problem 5 for equation D

Expand \(4t + 7=4(t + 2)\): \(4t+7 = 4t+8\). Subtract \(4t\) from both sides: \(4t-4t + 7=4t-4t + 8\), \(7 = 8\), which is a contradiction, so no solution.

Step8: Solve problem 5 for equation E

For \(6t + 2=-3 + 6t\), subtract \(6t\) from both sides: \(6t-6t + 2=6t-6t-3\), \(2=-3\), which is a contradiction, so no solution.

Answer:

1. Once
2. The equation \(10t = 2.5t\) has a solution \(t = 0\), so the vehicles are in the same position once.
3. No. Subtracting \(10x\) from \(16x = 10x\) gives \(6x = 0\), so \(x = 0\) is the solution.
4. One Solution: A, B

No Solution: D, E
Infinitely Many Solutions: C