# SOLUTION: How do I solve by using a system of two equations in two variables? A duck can fly 2400 m in 10 min with the wind. Against the wind, it can fly only two thirds of this distance

Algebra ->  Algebra  -> Coordinate Systems and Linear Equations -> SOLUTION: How do I solve by using a system of two equations in two variables? A duck can fly 2400 m in 10 min with the wind. Against the wind, it can fly only two thirds of this distance      Log On

 Ad: You enter your algebra equation or inequality - Algebrator solves it step-by-step while providing clear explanations. Free on-line demo . Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Linear Solvers Practice Answers archive Word Problems Lessons In depth

 Click here to see ALL problems on Linear-systems Question 141089This question is from textbook Prentice hall algebra 1 : How do I solve by using a system of two equations in two variables? A duck can fly 2400 m in 10 min with the wind. Against the wind, it can fly only two thirds of this distance in 10 min. How fast could the duck fly in still air? What is the rate of the wind?This question is from textbook Prentice hall algebra 1 Answer by stanbon(57967)   (Show Source): You can put this solution on YOUR website!How do I solve by using a system of two equations in two variables? A duck can fly 2400 m in 10 min with the wind. Against the wind, it can fly only two thirds of this distance in 10 min. How fast could the duck fly in still air? What is the rate of the wind? ------------------------------------------------ Let still wind rate be "s"; Let wind rate to "w". ----------------------------- EQUATIONS: s + w = 2400 meter/10 min = 240 meters / minute s - w = (2/3)2400 meter/10min = 160 meters/ minute -------------------------- 2s = 400 meters/min s = 200 meters/min (duck in still air speed) -------------------- Substitute to solve for "w": 200 + w = 240 w = 40 meters/ min (wind speed) ============= Cheers, Stan H.