Solve the system of equations using substitution to find the exact solution.
Approximate the solution to be x ≈ − 3.216 and y ≈ 2.514 .
Compare the given options to the calculated solution and determine the closest approximation.
The closest approximation is ( − 4 13 , 2 5 ) .
Explanation
Understanding the Problem We are given a system of two equations:
− 2 x + 5 y = 19 y = − 6 5 x − 6 1
Our goal is to approximate the solution (x, y) of this system using the graph. The solution is the point where the two lines intersect. We need to find the option that is closest to the intersection point.
Solving the System of Equations To find the exact solution, we can solve the system of equations. Substitute the second equation into the first:
− 2 x + 5 ( − 6 5 x − 6 1 ) = 19 − 2 x − 6 25 x − 6 5 = 19
Multiply by 6 to eliminate fractions:
− 12 x − 25 x − 5 = 114 − 37 x = 119 x = − 37 119 ≈ − 3.216
Now substitute x back into the second equation to find y:
y = − 6 5 ∗ ( − 37 119 ) − 6 1 y = 222 595 − 6 1 y = 222 595 − 222 37 y = 222 558 = 37 93 ≈ 2.514
So the exact solution is approximately (-3.216, 2.514).
Analyzing the Options Now, let's examine the given options:
A. ( 2 5 , − 4 13 ) = ( 2.5 , − 3.25 ) B. ( − 4 13 , 2 5 ) = ( − 3.25 , 2.5 ) C. ( 4 13 , − 2 5 ) = ( 3.25 , − 2.5 ) D. ( − 2 5 , 4 13 ) = ( − 2.5 , 3.25 )
We want to find the option that is closest to our calculated solution (-3.216, 2.514).
Finding the Closest Approximation Comparing the options, we see that option B (-3.25, 2.5) is the closest to our calculated solution (-3.216, 2.514).
To confirm, let's calculate the distances between our solution and each option:
Distance A: ( − 3.216 − 2.5 ) 2 + ( 2.514 − ( − 3.25 ) ) 2 ≈ 8.117 Distance B: ( − 3.216 − ( − 3.25 ) ) 2 + ( 2.514 − 2.5 ) 2 ≈ 0.036 Distance C: ( − 3.216 − 3.25 ) 2 + ( 2.514 − ( − 2.5 ) ) 2 ≈ 8.182 Distance D: ( − 3.216 − ( − 2.5 ) ) 2 + ( 2.514 − 3.25 ) 2 ≈ 1.027
The smallest distance is for option B.
Conclusion Therefore, the closest approximation to the solution of the system of equations is option B.
Examples
Systems of equations are used in various real-world applications, such as determining the break-even point for a business, modeling supply and demand in economics, and calculating the trajectory of a projectile in physics. In this case, the two equations could represent the cost and revenue functions of a company, and the solution would represent the point where the company starts making a profit. Understanding how to solve and approximate solutions to systems of equations is crucial for making informed decisions in these fields. For example, consider a scenario where you're comparing two phone plans. One plan has a higher monthly fee but lower per-minute charges, while the other has a lower monthly fee but higher per-minute charges. By setting up a system of equations, you can determine at what usage level the two plans cost the same, helping you choose the most cost-effective plan for your needs. This involves solving for the intersection point of the two cost functions, similar to the problem we solved here.
We solved the system of equations by substituting, simplifying, and finding the values for x and y . The approximate solution was found to be ( − 3.216 , 2.514 ) . Among the provided options, the one that is closest to this solution is B, ( − 4 13 , 2 5 ) .
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