6.2
Systems of Linear Equations in Two Variables
An airplane trip between two countries presents a classic puzzle for solving a system of equations.
The distance is 1260 kilometers. Flying against the wind takes 3 hours, while flying with the wind takes only 2 hours.
Although both trips cover the same distance, the time differs due to the wind’s speed, which remains constant throughout the journey.
Let x be the airplane’s speed in still air, and y be the wind’s speed.
Since distance equals speed multiplied by time, the effective speed against the wind is the airplane’s speed minus the wind’s speed. Multiplying this by 3 hours gives 1260 kilometers.
For the journey with the wind, the effective speed is the airplane’s speed plus the wind’s speed. Multiplying this by 2 hours also gives 1260 kilometers.
These two relationships form a pair of linear equations. Using the elimination method, adding the two equations removes the wind’s speed, leaving a simpler equation in x.
This reveals the airplane’s true speed.
Substituting this value into either equation gives the wind’s speed.
Solving a system of linear equations is a fundamental concept in algebra. A system of equations consists of two or more linear equations involving the same set of variables. One of the most efficient algebraic methods for solving such systems is the substitution method. This technique involves expressing one variable in terms of the other from one equation and substituting it into the second equation. This method is particularly useful when one of the equations is easily rearranged.
Consider the system of equations:
First, we isolate one variable in one of the equations. From the second equation, solve for y:
Next, substitute the value of y into the first equation:
Expand and simplify to find the value of x:
Substitute x into the expression for y:
Since both equations are satisfied when substituting the calculated values of x and y, the solution is confirmed to be correct. Therefore, the solution to the system of equations is x = 27/14 and y = 19/7. This means the pair represents the intersection point of the two lines described by the given equations. The substitution method has thus successfully provided a unique solution to the system.
An airplane trip between two countries presents a classic puzzle for solving a system of equations.
The distance is 1260 kilometers. Flying against the wind takes 3 hours, while flying with the wind takes only 2 hours.
Although both trips cover the same distance, the time differs due to the wind’s speed, which remains constant throughout the journey.
Let x be the airplane’s speed in still air, and y be the wind’s speed.
Since distance equals speed multiplied by time, the effective speed against the wind is the airplane’s speed minus the wind’s speed. Multiplying this by 3 hours gives 1260 kilometers.
For the journey with the wind, the effective speed is the airplane’s speed plus the wind’s speed. Multiplying this by 2 hours also gives 1260 kilometers.
These two relationships form a pair of linear equations. Using the elimination method, adding the two equations removes the wind’s speed, leaving a simpler equation in x.
This reveals the airplane’s true speed.
Substituting this value into either equation gives the wind’s speed.
From Chapter 6:
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