Roxanne's solution (-3, -3) is incorrect because substituting x = − 3 into both equations yields y-values of -7 and 9, not -3. Therefore, she likely made a mistake in interpreting the intersection point from her graph. The correct answer is B: No. She switched the x and y values of the intersection point when writing the solution.
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Substitute x = − 3 into the first equation: y = 3 2 ( − 3 ) − 5 = − 7 .
Substitute x = − 3 into the second equation: y = − 2 ( − 3 ) + 3 = 9 .
Since the y values are not equal to − 3 , Roxanne's solution is incorrect.
The correct answer is that Roxanne's solution is incorrect.
Explanation
Analyze the problem We are given a system of two equations:
Equation 1: y = 3 2 x − 5 Equation 2: y = − 2 x + 3
Roxanne believes the solution to this system is ( − 3 , − 3 ) . We need to check if this is correct.
Plan of action To check if Roxanne's solution is correct, we will substitute x = − 3 into both equations and see if we get y = − 3 in both cases.
Check Equation 1 Let's substitute x = − 3 into Equation 1:
y = 3 2 ( − 3 ) − 5 = − 2 − 5 = − 7
So, when x = − 3 , y = − 7 for Equation 1.
Check Equation 2 Now, let's substitute x = − 3 into Equation 2:
y = − 2 ( − 3 ) + 3 = 6 + 3 = 9
So, when x = − 3 , y = 9 for Equation 2.
Conclusion Since the y values obtained from the two equations when x = − 3 are not equal to − 3 , Roxanne's solution ( − 3 , − 3 ) is incorrect. The y values are − 7 and 9 , respectively.
Reason for error Now, let's analyze why Roxanne's solution might be incorrect. Since we don't have information about her graphing process, we can't say for sure if she used the wrong y-intercepts or slopes. However, we know that the point ( − 3 , − 3 ) does not satisfy either equation. Therefore, the most likely reason is that she misread the intersection point from her graph.
Final Answer Therefore, Roxanne is incorrect.
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 $5000 and variable costs of 10 p er u ni t , t h e t o t a l cos t c anb ere p rese n t e d b y t h ee q u a t i o n y = 10x + 5000 , w h ere x$ is the number of units produced. If the company sells each unit for 25 , t h ere v e n u ec anb ere p rese n t e d b y t h ee q u a t i o n y = 25x$. By solving this system of equations, the company can find the number of units they need to sell to break even.