Express x in terms of y using the first equation: x = y + 2 .
Substitute the expression for x into the second equation and solve for y : y = 4 .
Substitute the value of y back into the expression for x and solve for x : x = 6 .
The solution to the system of equations is x = 6 , y = 4 .
Explanation
Analyze the problem We are given a system of two linear equations:
Equation 1: x − y = 2 Equation 2: 2 x + 4 y = 28
Our goal is to find the values of x and y that satisfy both equations.
Express x in terms of y From Equation 1, we can express x in terms of y :
x = y + 2
Solve for y Substitute the expression for x into Equation 2: 2 ( y + 2 ) + 4 y = 28 Simplify the equation: 2 y + 4 + 4 y = 28 Combine like terms: 6 y + 4 = 28 Subtract 4 from both sides: 6 y = 24 Divide by 6: y = 4
Solve for x Now that we have the value of y , we can substitute it back into the expression for x :
x = y + 2 x = 4 + 2 x = 6
State the solution Therefore, the solution to the system of equations is x = 6 and y = 4 .
Examples
Systems of linear equations can be used in various real-life scenarios, such as determining the break-even point for a business. For example, if a company has fixed costs of $10,000 and variable costs of $5 per unit, and they sell each unit for 15 , w ec an se t u p a sys t e m o f e q u a t i o n s t o f in d t h e n u mb ero f u ni t s t h ey n ee d t ose llt o b re ak e v e n . L e t x b e t h e n u mb ero f u ni t s an d y b e t h e t o t a l cos t / re v e n u e . T h ecos t e q u a t i o ni s y = 5x + 10000 an d t h ere v e n u ee q u a t i o ni s y = 15x$. Solving this system gives the break-even point.
The solution to the system of equations is x = 6 and y = 4 . To find these values, we first expressed x in terms of y , substituted it into the second equation, and solved for each variable. This method effectively allows us to find the intersecting values that satisfy both equations simultaneously.
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