To determine how much more Paula can shovel in one minute compared to Melinda, we will solve the given system of linear equations to find their respective shoveling rates.
The equations are:
1. [tex]\(30x + 30y = 450\)[/tex]
2. [tex]\(20x + 25y = 345\)[/tex]
Here, [tex]\(x\)[/tex] represents the rate at which Melinda shovels (in square feet per minute), and [tex]\(y\)[/tex] represents the rate at which Paula shovels (in square feet per minute).
First, let's simplify the equations:
1. [tex]\(30(x + y) = 450\)[/tex]
[tex]\[
x + y = \frac{450}{30}
\][/tex]
[tex]\[
x + y = 15
\][/tex]
2. Leave the second equation as it is for substitution later:
[tex]\[
20x + 25y = 345
\][/tex]
Now, let's solve for one variable in terms of the other using the simplified first equation. For simplicity, solve for [tex]\(x\)[/tex]:
[tex]\[
x = 15 - y
\][/tex]
Substitute [tex]\(x\)[/tex] in the second equation:
[tex]\[
20(15 - y) + 25y = 345
\][/tex]
[tex]\[
300 - 20y + 25y = 345
\][/tex]
[tex]\[
300 + 5y = 345
\][/tex]
[tex]\[
5y = 45
\][/tex]
[tex]\[
y = 9
\][/tex]
So, Paula's rate (y) is 9 square feet per minute. Now, substitute [tex]\(y = 9\)[/tex] back into the equation [tex]\(x + y = 15\)[/tex] to find [tex]\(x\)[/tex]:
[tex]\[
x + 9 = 15
\][/tex]
[tex]\[
x = 6
\][/tex]
Thus, Melinda's rate (x) is 6 square feet per minute.
The difference in their shoveling rates is:
[tex]\[
y - x = 9 - 6 = 3
\][/tex]
Therefore, Paula can shovel 3 square feet per minute more than Melinda.
So, the correct answer is:
3 square feet per minute.
