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Emma is planning her summer and would like to work enough to travel and buy a new laptop.
She can earn $90 each day, after deductions, and she can work a maximum of 40 days in July and
August, combined. She expects each day of travel will cost her $150 and the laptop she hopes to
buy costs $700.

a)Write a linear equation that represents the number of days Emma can work and travel and
still earn enough for her laptop.

b)Sketch the graph of this relation.

c)State the domain and range of this relation.

d)If Emma plans to travel for 6 days, how many days does she need to work?

Verified Answer :

MissPhiladelphia

For the answer to the question above, f x is the number of days she works, she'll earn $90x 
After buying the laptop, she'll have $90x - $700 left over, which will pay for ($90x - $700) / $150 days of travel. So we have y = ($90x - $700) / $150 = (9x - 70) / 15 = 0.6x - (14/3) 
Note that y can't be negative. Also, if y = 0, then Emma doesn't get to travel at all, so we should avoid that. So we have: 
0.6x - (14/3) > 0 
0.6x > 14/3 
x > (14/3) / 0.6 
x > 70/9 
The question says that x can be up to 40, so the domain is 70/9 < x <= 40 
That's approximately 7.777... < x <= 40 
Multiply those numbers by 0.6 and then subtract 700 to get the range: 
0 < y <= 58/3 
That's approximately 0 < y <= 19.333

Answer:

(a) The required equation is \(90x-150y=700\).

(b) The graph of given relation is shown below.

(c) \(Domain=\{x|x\in Z,0\leq x\leq 40\}\) and \(Range=\{y|y\in Z,0\leq x\leq \frac{58}{3}\}\).

(d) She needs to work for 18 days.

Step-by-step explanation:

Let x be the number of days she can work and y is the number of days she can travel.

She can earn $90 each day. She expects each day of travel will cost her $150 and the laptop she hopes to  buy costs $700.

\(90x-150y=700\)

Therefore the required equation is 90x-150y=700.

(b)

The given relation is

\(90x-150y=700\)             .... (1)

Rewrite the above equation in slope intercept form.

\(150y=90x-700\)

Divide both sides by 150.

\(y=\frac{90x-700}{150}\)

\(y=\frac{3}{5}x-\frac{14}{3}\)

It is a straight line with slope 3/5 and y-intercept (14)/3. The graph of given relation is shown below.

(c)

The relation is

\(y=\frac{3}{5}x-\frac{14}{3}\)

Here, x is the number of days she can work and she can work a maximum of 40 days. So the domain of the function is

\(Domain=\{x|x\in Z,0\leq x\leq 40\}\)

At x=0,

\(y=\frac{3}{5}(0)-\frac{14}{3}=-\frac{14}{3}\)

At x=40,

\(y=\frac{3}{5}(40)-\frac{14}{3}=\frac{58}{3}\)

y is the number of days she can travel, so y cannot be negative.

The range of the relation is

\(Range=\{y|y\in Z,0\leq x\leq \frac{58}{3}\}\)

(d)

We need to find the number of days she need to work if she plans to travel for 6 days.

Substitute y=6 in equation (1) to find the value of x.

\(90x-150(6)=700\)

\(90x-900=700\)

\(90x=700+900\)

\(90x=1600\)

\(90x=\frac{1600}{90}\)

\(x=17.77\approx 18\)

Therefore she needs to work for 18 days.

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