Multiply equations to prepare for elimination.
Eliminate x by subtracting the equations.
Solve for y : y = − 5 .
Solve for x : x = 2 . The solution is ( 2 , − 5 ) .
Explanation
Analyze the problem We are given a system of two linear equations:
3 x − 4 y = 26
2 x + 8 y = − 36
Our goal is to find the values of x and y that satisfy both equations. We can use the elimination method to solve this system.
Multiply equations Multiply the first equation by 2 and the second equation by 1 to make the coefficients of x in both equations equal:
6x - 8y = 52"> 2 ∗ ( 3 x − 4 y ) = 2 ∗ 26 => 6 x − 8 y = 52 2x + 8y = -36"> 1 ∗ ( 2 x + 8 y ) = 1 ∗ ( − 36 ) => 2 x + 8 y = − 36
Now, multiply the second equation by 3 to make the coefficients of x equal: 6x + 24y = -108"> 3 ∗ ( 2 x + 8 y ) = 3 ∗ ( − 36 ) => 6 x + 24 y = − 108
Eliminate x Subtract the first transformed equation from the second transformed equation to eliminate x :
( 6 x + 24 y ) − ( 6 x − 8 y ) = − 108 − 52 6 x + 24 y − 6 x + 8 y = − 160 32 y = − 160
Solve for y Solve for y :
y = 32 − 160 = − 5
Solve for x Substitute the value of y back into the first original equation to solve for x :
3 x − 4 ( − 5 ) = 26 3 x + 20 = 26 3 x = 26 − 20 3 x = 6 x = 3 6 = 2
Final Answer The solution to the system of equations is x = 2 and y = − 5 . Therefore, the ordered pair is ( 2 , − 5 ) .
Examples
Systems of equations are used in various real-world applications, such as determining the break-even point for a business. For example, a company might want to know how many units they need to sell to cover their costs. This involves setting up equations for cost and revenue and solving for the quantity at which they are equal. Understanding how to solve systems of equations is crucial for making informed business decisions and optimizing resource allocation.
We solved the system of equations by using the elimination method, leading us to find that x = 2 and y = − 5 . Therefore, the solution is the ordered pair ( 2 , − 5 ) . The correct answer is option A.
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