Melinda and paul shovel driveways and sidewalks in the winter as a way to earn extra money. together they shoveled 450 square feet of sidewalk in 30 minutes. then melinda shoveled for 20 minutes while paul shoveled for 25 minutes to complete 345 square feet of driveway.

30x + 30y = 450
20x + 25y = 345
how much more can paul shovel in 1 minute than melinda?

3 square feet per minute
6 square feet per minute
9 square feet per minute
15 square feet per minute

Hello mate!

we have the system of equations

30x + 30y = 450

20x + 25y = 345

where x is the number of square feet of snow that Melinda shovels in one minute, and y is the number of square feet of snow that Paul shovels in one minute.

Let's solve the system of equations, first, isolate one of the therms in the first equation, for example, x.

30x+30y = 450

x + y = 450/30 = 15

x = 15 - y

now we can replace it in the second equation and solve it for y.

20x + 25y = 345

20(15 - y) + 25y = 345

300 - 20y + 25y = 345

5y = 45

y = 45/5 = 9

Paul can shovel 9 square feets of snow in one minute, now replace it in the first equation and solve it for x.

x + y = 450/30 = 15

x + 9 = 15

x = 15 - 9 = 6

so melinda sholves 6 square feet of snow in one minute.

The question is:

How much more can Paul shovel in 1 minute than Melinda?

We need to see the difference y - x = 9 - 6 = 3

this means that paul shovels 3 more square feet per minute than Melinda.

answer:

c 60% on edg

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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