# 2 ships leave port one sailing east and the other south

2 ships leave port one sailing east and the other south. some time later they are 17 mi apart, with the eastbound ship 7 mi farther from the port than the southbound ship. how far is each from port?
I thank you very much for your assistance. I constantly struggle with all of my word problems.
Once again your time and assistance is appreciated.
Sincerely,
Steph

2 ships leave port one sailing east and the other south. some time later they are 17 mi apart, with the eastbound ship 7 mi farther from the port than the southbound ship. how far is each from port? In this type of questions, we have to use the Pythagorean Theorem. It states that in a right triangle, the square of the first leg plus the square of the sqcond lerg is equal to the square of the hypotenuse. In this case, let us assume that x is the distance of the southbound ship from the port, then it follows that x + 7 is the distance of the eastbound ship from the port. They are 17 miles apart. Equating we have: {{{x^2 + (x+7)^2 = 17^2}}} Solving: {{{x^2 + x^2 + 14x + 49 = 289}}} {{{2x^2 + 14x - 240 = 0}}} Applying the quadratic formula {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}, we have: {{{x = (-14 +- sqrt( 14^2-4*2*-240 ))/(2*2) }}} Solving: {{{x = (-14 +- sqrt( 196-(-1920)))/(4) }}} {{{x = (-14 +- sqrt( 2116))/(4) }}} {{{x = (-14 + 46 )/(4) }}} or {{{x = (-14 - 46 )/(4) }}} {{{x = (32)/4}}} or {{{x = -60/4}}} {{{x=8}}} or {{{x=-15}}} Since we are solving real problems, it cannot be a negative integer. So we have to use the positive root which is {{{x=8}}} Then the distance of the southbound ship from the port is 8 miles, while the distance of the eastbound ship from the port is 15 miles. Thank you. ~kmcruz09~
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