Select the correct answer. Solve the system of equations given below: [tex]
\begin{array}{l}
-5 x=y-5 \
-2 y=-x-21
\end{array}
[/tex]
A. (-1,-23)
B. (10,16)
C. (10,-45)
D. (-1,10)
Answer (2)
Rewrite the given equations in standard form.
Solve for x in terms of y using the second equation: x = 2 y − 21 .
Substitute the expression for x into the first equation and solve for y : y = 10 .
Substitute the value of y back into the expression for x and find x = − 1 . The solution is ( − 1 , 10 ) .
Explanation
Analyze the problem We are given a system of two linear equations with two variables, x and y . Our goal is to find the values of x and y that satisfy both equations. The given equations are:
Equation 1: − 5 x = y − 5
Equation 2: − 2 y = − x − 21
We can solve this system using either the substitution or elimination method. Here, we will use the substitution method.
Rewrite the equations First, let's rewrite the equations in a more standard form:
Equation 1: − 5 x − y = − 5 Equation 2: − 2 y = − x − 21 can be rewritten as x − 2 y = − 21
Now, we can solve for x in terms of y from Equation 2: x = 2 y − 21
Solve for y Next, substitute the expression for x into Equation 1: − 5 ( 2 y − 21 ) − y = − 5
Now, simplify and solve for y :
− 10 y + 105 − y = − 5 − 11 y = − 110 y = 10
Solve for x Now that we have the value of y , we can substitute it back into the expression for x :
x = 2 ( 10 ) − 21 x = 20 − 21 x = − 1
Verify the solution So, the solution is x = − 1 and y = 10 . We can write this as an ordered pair ( − 1 , 10 ) .
Now, let's verify the solution by substituting x = − 1 and y = 10 into both original equations:
Equation 1: − 5 ( − 1 ) = 10 − 5 => 5 = 5 (Correct) Equation 2: − 2 ( 10 ) = − ( − 1 ) − 21 => − 20 = 1 − 21 => − 20 = − 20 (Correct)
Since the solution satisfies both equations, it is the correct solution.
Final Answer The solution to the system of equations is x = − 1 and y = 10 , which corresponds to option D.
Examples
Systems of equations are used in many real-world applications, 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 equations imply a potential solution set nearby ( − 1 , 10 ) , yet complexity in deductions shows neither alignment effectively with original formulations prompts exploration deeper focus across alternate observation criteria. Approximations reflect considerations within options while utilizing likely integer outcomes leads thorough transformations engage within coherence challenge recalibrated equations suitably. Such nuance highlights intricacies underpinning linear constraint alongside practical evaluations prompting substructures engage in intricacy assessment timely.
;
