Math / Algebra

QuestionUnit 4: Rational Expressions and Equations Assignment Booklet 4
10. Consider each scenario: a. A high-speed passenger train in Europe completes an 800 km trip in 3 hours. The train travels the first 600 km at an average speed that was $100 \mathrm{~km} / \mathrm{h}$ faster than the last 200 km of the trip. If $x$ represents the average speed of the train on the second part of the trip, then an equation to represent this situation is: $\frac{600}{x+100}+\frac{200}{x}=3$
Identify all restrictions on the variable $x$ in this context. (2 marks)

Studdy Solution

STEP 1

1. $x$ represents the average speed of the train on the second part of the trip.
2. The speed of the train cannot be zero or negative.
3. The equation given is $\frac{600}{x+100}+\frac{200}{x}=3$ .

STEP 2

1. Identify restrictions based on the context of speed.
2. Identify restrictions based on the mathematical properties of the equation.

STEP 3

The speed of the train must be positive, so $x > 0$ .

STEP 4

The denominators in the equation $\frac{600}{x+100}+\frac{200}{x}=3$ cannot be zero, so we must consider the values that make each denominator zero.

STEP 5

For the term $\frac{600}{x+100}$ , the denominator $x + 100 \neq 0$ , which implies $x \neq -100$ .

STEP 6

For the term $\frac{200}{x}$ , the denominator $x \neq 0$ .
The restrictions on $x$ are:
1. $x > 0$ (since speed must be positive)
2. $x \neq -100$ (since $x + 100 \neq 0$ )
Thus, the restrictions are $x > 0$ .

Was this helpful?

Studdy Solution

STEP 1

1. $x$ represents the average speed of the train on the second part of the trip.
2. The speed of the train cannot be zero or negative.
3. The equation given is $\frac{600}{x+100}+\frac{200}{x}=3$ .

STEP 2

1. Identify restrictions based on the context of speed.
2. Identify restrictions based on the mathematical properties of the equation.

STEP 3

The speed of the train must be positive, so $x > 0$ .

STEP 4

The denominators in the equation $\frac{600}{x+100}+\frac{200}{x}=3$ cannot be zero, so we must consider the values that make each denominator zero.

STEP 5

For the term $\frac{600}{x+100}$ , the denominator $x + 100 \neq 0$ , which implies $x \neq -100$ .

STEP 6

For the term $\frac{200}{x}$ , the denominator $x \neq 0$ .
The restrictions on $x$ are:
1. $x > 0$ (since speed must be positive)
2. $x \neq -100$ (since $x + 100 \neq 0$ )
Thus, the restrictions are $x > 0$ .

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

1. x x x represents the average speed of the train on the second part of the trip.2. The speed of the train cannot be zero or negative.3. The equation given is 600x+100+200x=3 \frac{600}{x+100}+\frac{200}{x}=3 x+100600​+x200​=3.

STEP 2

1. Identify restrictions based on the context of speed.2. Identify restrictions based on the mathematical properties of the equation.

STEP 3

The speed of the train must be positive, so x>0 x > 0 x>0.

STEP 4

The denominators in the equation 600x+100+200x=3 \frac{600}{x+100}+\frac{200}{x}=3 x+100600​+x200​=3 cannot be zero, so we must consider the values that make each denominator zero.

STEP 5

For the term 600x+100 \frac{600}{x+100} x+100600​, the denominator x+100≠0 x + 100 \neq 0 x+100=0, which implies x≠−100 x \neq -100 x=−100.

STEP 6

For the term 200x \frac{200}{x} x200​, the denominator x≠0 x \neq 0 x=0. The restrictions on x x x are:1. x>0 x > 0 x>0 (since speed must be positive)2. x≠−100 x \neq -100 x=−100 (since x+100≠0 x + 100 \neq 0 x+100=0)Thus, the restrictions are x>0 x > 0 x>0.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. $x$ represents the average speed of the train on the second part of the trip.
2. The speed of the train cannot be zero or negative.
3. The equation given is $\frac{600}{x+100}+\frac{200}{x}=3$ .

1. Identify restrictions based on the context of speed.
2. Identify restrictions based on the mathematical properties of the equation.

The speed of the train must be positive, so $x > 0$ .

The denominators in the equation $\frac{600}{x+100}+\frac{200}{x}=3$ cannot be zero, so we must consider the values that make each denominator zero.

For the term $\frac{600}{x+100}$ , the denominator $x + 100 \neq 0$ , which implies $x \neq -100$ .

For the term $\frac{200}{x}$ , the denominator $x \neq 0$ .
The restrictions on $x$ are:
1. $x > 0$ (since speed must be positive)
2. $x \neq -100$ (since $x + 100 \neq 0$ )
Thus, the restrictions are $x > 0$ .