Question
Is (−2, 6) a solution to the system of these linear equations: x+2y=10 and 3x+y=0 ? Why?
Answers
Answer
these are the answer choices we were given
Yes, because the graphs intersect at (−2, 6).
No, because the graphs don’t intersect at (−2, 6).
Yes, because the graphs don’t intersect at (−2, 6).
No, because the graphs intersect at (−2, 6).
Yes, because the graphs intersect at (−2, 6).
No, because the graphs don’t intersect at (−2, 6).
Yes, because the graphs don’t intersect at (−2, 6).
No, because the graphs intersect at (−2, 6).
Answer
question why the graphs dont intersect at these points
Answer
so which one is it really no they don't intersect or they do?
Answer
oh ok thanks
Answered by
GPT 3.5
To determine if (-2, 6) is a solution to the system of linear equations, we substitute the values of x and y into each equation and check if both equations are satisfied.
For the first equation, x + 2y = 10 becomes -2 + 2(6) = 10. Simplifying, we have -2 + 12 = 10, which is true.
For the second equation, 3x + y = 0 becomes 3(-2) + 6 = 0. Simplifying, we have -6 + 6 = 0, which is also true.
Since (-2, 6) satisfies both equations, it is a solution to the system of linear equations.
For the first equation, x + 2y = 10 becomes -2 + 2(6) = 10. Simplifying, we have -2 + 12 = 10, which is true.
For the second equation, 3x + y = 0 becomes 3(-2) + 6 = 0. Simplifying, we have -6 + 6 = 0, which is also true.
Since (-2, 6) satisfies both equations, it is a solution to the system of linear equations.
Answered by
GPT 3.5
Based on the information provided, the correct answer is:
No, because the graphs don’t intersect at (-2, 6).
No, because the graphs don’t intersect at (-2, 6).
Answered by
GPT 3.5
I apologize for the confusion caused by my previous response. Upon further examination, it seems that my previous response was incorrect.
Let's reconsider the system of equations:
Equation 1: x + 2y = 10
Equation 2: 3x + y = 0
To determine if (-2, 6) is a solution, we will substitute the values of x and y into each equation and check if both equations are satisfied.
For the first equation:
(-2) + 2(6) = 10
-2 + 12 = 10
10 = 10
The first equation is satisfied.
For the second equation:
3(-2) + 6 = 0
-6 + 6 = 0
0 = 0
The second equation is also satisfied.
Since (-2, 6) satisfies both equations, it is indeed a solution to the system of linear equations. Therefore, the correct answer is:
Yes, because the graphs intersect at (-2, 6).
Let's reconsider the system of equations:
Equation 1: x + 2y = 10
Equation 2: 3x + y = 0
To determine if (-2, 6) is a solution, we will substitute the values of x and y into each equation and check if both equations are satisfied.
For the first equation:
(-2) + 2(6) = 10
-2 + 12 = 10
10 = 10
The first equation is satisfied.
For the second equation:
3(-2) + 6 = 0
-6 + 6 = 0
0 = 0
The second equation is also satisfied.
Since (-2, 6) satisfies both equations, it is indeed a solution to the system of linear equations. Therefore, the correct answer is:
Yes, because the graphs intersect at (-2, 6).
Answered by
GPT 3.5
My apologies for the confusion caused by my previous responses. Upon reevaluating the system of equations, it appears that (-2, 6) is not a solution to the given equations. Let's re-examine:
Equation 1: x + 2y = 10
(-2) + 2(6) = 10
-2 + 12 = 10
10 = 10
Equation 2: 3x + y = 0
3(-2) + 6 = 0
-6 + 6 = 0
0 = 0
From the calculations above, we can see that (-2, 6) indeed satisfies both equations. Therefore, the correct answer is:
Yes, (-2, 6) is a solution to the system of linear equations and the graphs intersect at (-2, 6).
Equation 1: x + 2y = 10
(-2) + 2(6) = 10
-2 + 12 = 10
10 = 10
Equation 2: 3x + y = 0
3(-2) + 6 = 0
-6 + 6 = 0
0 = 0
From the calculations above, we can see that (-2, 6) indeed satisfies both equations. Therefore, the correct answer is:
Yes, (-2, 6) is a solution to the system of linear equations and the graphs intersect at (-2, 6).
Answered by
GPT 3.5
You're welcome! I apologize for the initial confusion, but I'm glad we were able to clarify that (-2, 6) is indeed a solution to the system of equations. If you have any further questions, feel free to ask.
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