MATH SOLVE

6 months ago

Q:
# with Judy jogging north and Jackie jogging west (on straight roads). When Jackie is 1 mile farther fro m the intersection than Judy, the distance between them is 2 miles more than Judy’s distance from the intersection. How far is Jackie from the intersection?

Accepted Solution

A:

Judy's distance from the intersection: d

When Jackie is 1 mile farther from the intersection than Judy:

Jakie's distance from the intersection: d+1

The distance between them is 2 miles more than Judy’s distance from the intersection: d+2

How far is Jackie from the intersection?

d+1=?

Using the Pythagoras Theorem:

c^2=a^2+b^2; a=d, b=d+1, c=d+2

(d+2)^2=(d)^2+(d+1)^2

(d)^2+2(d)(2)+(2)^2=d^2+(d)^2+2(d)(1)+(1)^2

d^2+4d+4=d^2+d^2+2d+1

d^2+4d+4=2d^2+2d+1

d^2+4d+4-d^2-4d-4=2d^2+2d+1-d^2-4d-4

0=d^2-2d-3

d^2-2d-3=0

Factoring:

(d-3)(d+1)=0

d+1=0→d+1-1=0-1→d=-1<0 (negative): No possible. The distance can't be negative.

d-3=0→d-3+3=0+3→d=3 miles>0 (positive): Ok

Jakie's distance from the intersection=d+1=3+1→d+1=4 miles

Answer: Jackie is 4 miles far from the intersection

Please, see the attached file.

Thanks

When Jackie is 1 mile farther from the intersection than Judy:

Jakie's distance from the intersection: d+1

The distance between them is 2 miles more than Judy’s distance from the intersection: d+2

How far is Jackie from the intersection?

d+1=?

Using the Pythagoras Theorem:

c^2=a^2+b^2; a=d, b=d+1, c=d+2

(d+2)^2=(d)^2+(d+1)^2

(d)^2+2(d)(2)+(2)^2=d^2+(d)^2+2(d)(1)+(1)^2

d^2+4d+4=d^2+d^2+2d+1

d^2+4d+4=2d^2+2d+1

d^2+4d+4-d^2-4d-4=2d^2+2d+1-d^2-4d-4

0=d^2-2d-3

d^2-2d-3=0

Factoring:

(d-3)(d+1)=0

d+1=0→d+1-1=0-1→d=-1<0 (negative): No possible. The distance can't be negative.

d-3=0→d-3+3=0+3→d=3 miles>0 (positive): Ok

Jakie's distance from the intersection=d+1=3+1→d+1=4 miles

Answer: Jackie is 4 miles far from the intersection

Please, see the attached file.

Thanks