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Express the situation as a system of two equations in two variables. Be sure to state clearly the meaning of your x- and y-variables. Solve the system by row-reducing the corresponding augmented matrix. State your final answer in terms of the original question.

For the final days before the election, the campaign manager has a total of $37,000 to spend on TV and radio campaign advertisements. Each TV ad costs $3000 and is seen by 10,000 voters, while each radio ad costs $500 and is heard by 2000 voters. Ignoring repeated exposures to the same voter, how many TV and radio ads will contact 130,000voters using the allocated funds?

x = TV ads
y = radio ads

1 Answer

1 vote

Answer:

9 TV ads and 20 radio ads

Explanation:

He has $37,000 to spend. He has to sum that amount of money between the TV and radio ads. Each TV ad costs $3000 while each radio ad costs $500, so the equation that represents that is 37000 = 3000x + 500y

The same happens with the amount of voters he needs to reach, the equation is 130000 = 10000x + 2000y

The system that represents this is


\left \{ {{3000x+500y=37000} \atop {10000x+2000y=130000}} \right.

And the augmented matrix is


\left[\begin{array}cc3000&500&37000\\10000&2000&130000\end{array}\right]

First we divide the first row by 3000 and the second by 10000:


\left[\begin{array}c1&1/6&37/3\\1&1/5&13\end{array}\right]

Then we multiply the second row by (-1) and we add the first row:


\left[\begin{array}cc1&1/6&37/3\\0&-1/30&-2/3\end{array}\right]

Now we multiply the second row by -30:


\left[\begin{array}c1&1/6&37/3\\0&1&20\end{array}\right]

Finally, to the first row we add the second one multiply by (-1/6):


\left[\begin{array}cc1&0&9\\0&1&20\end{array}\right]

So, x = 9 and y = 20

That means 9 TV and 20 radio ads will contact 130,000 voters using the allocated funds

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