What is one solution of the following system?
$\left\{\begin{array}{l}
2 y-2 x=12 \\
x^2+y^2=36
\end{array}\right.$
A. $(-6,0)$
B. $(-2,4)$
C. $(0,-6)$
D. $(4,-2)$
Answer (1)
To solve the system of equations:
Simplify the first equation 2 y − 2 x = 12 to y = x + 6 .
Substitute y = x + 6 into the second equation x 2 + y 2 = 36 , resulting in x 2 + ( x + 6 ) 2 = 36 .
Simplify and factor the equation to find x = 0 or x = − 6 .
Determine the corresponding y values, giving the solution ( − 6 , 0 ) .
Explanation
Analyze the problem We are given a system of two equations:
Equation 1 2 y − 2 x = 12
Equation 2 x 2 + y 2 = 36
Objective We need to find one solution (x, y) from the given options: (-6, 0), (-2, 4), (0, -6), (4, -2).
Simplify Equation 1 Let's simplify the first equation by dividing both sides by 2:
Simplified Equation 1 y − x = 6
Isolate y y = x + 6
Substitution Now, substitute y = x + 6 into the second equation:
Substitute into Equation 2 x 2 + ( x + 6 ) 2 = 36
Expand Expand and simplify the equation:
Expanded Equation x 2 + x 2 + 12 x + 36 = 36
Simplify 2 x 2 + 12 x = 0
Factor Factor the equation:
Factored Equation 2 x ( x + 6 ) = 0
Solve for x Solve for x:
Solutions for x x = 0 or x = − 6
Find y when x=0 If x = 0 , then y = x + 6 = 0 + 6 = 6 . So, one solution is (0, 6).
Find y when x=-6 If x = − 6 , then y = x + 6 = − 6 + 6 = 0 . So, another solution is (-6, 0).
Check options Check the given options to see if any of them are solutions. The option (-6, 0) is a solution.
Verification Let's check the given options in the original equations:
Verify (-6,0) For (-6, 0): 2 ( 0 ) − 2 ( − 6 ) = 12 , which is true. Also, ( − 6 ) 2 + ( 0 ) 2 = 36 , which is true. So, (-6, 0) is a solution.
Verify (-2,4) For (-2, 4): 2 ( 4 ) − 2 ( − 2 ) = 8 + 4 = 12 , which is true. Also, ( − 2 ) 2 + ( 4 ) 2 = 4 + 16 = 20 , which is not 36. So, (-2, 4) is not a solution.
Verify (0,-6) For (0, -6): 2 ( − 6 ) − 2 ( 0 ) = − 12 , which is not 12. So, (0, -6) is not a solution.
Verify (4,-2) For (4, -2): 2 ( − 2 ) − 2 ( 4 ) = − 4 − 8 = − 12 , which is not 12. So, (4, -2) is not a solution.
Final Answer Therefore, one solution of the system is (-6, 0).
Examples
Systems of equations are used in various real-world applications, such as determining the break-even point for a business. For example, if a company's cost equation is y = 2 x + 12 (where y is the total cost and x is the number of units produced) and the revenue equation is x 2 + y 2 = 36 (representing the total revenue), solving this system of equations will give the production level at which the company's cost equals its revenue. This helps in making informed business decisions.
