To determine which value is not possible for the constant in the second equation, we need to determine if the y-intercepts of the two equations are the same or different.
For the first equation, 6x + 9y = 15, if we set x = 0, we get 9y = 15, and solving for y, we get y = 15/9 = 5/3.
For the second equation, 2x + 3y = constant, if we set x = 0, we get 3y = constant, and solving for y, we get y = constant/3.
If the two equations have the same y-intercept, it means they have a solution. Therefore, we need to find the constant that results in a different y-intercept for the second equation.
Let's examine the given options:
A) 55: If the constant is 55, then y = 55/3. The y-intercept of the second equation would be 55/3, which is different from 5/3. It is a possible value.
B) 1010: If the constant is 1010, then y = 1010/3. The y-intercept of the second equation would be 1010/3, which is different from 5/3. It is a possible value.
C) 1515: If the constant is 1515, then y = 1515/3. The y-intercept of the second equation would be 1515/3, which is different from 5/3. It is a possible value.
D) 20: If the constant is 20, then y = 20/3. The y-intercept of the second equation would be 20/3, which is different from 5/3. It is a possible value.
Therefore, the answer is none of the given options. None of them result in a different y-intercept for the second equation, meaning that all values are possible constants.
Suppose the system of linear equations has no solution. Which is NOT a possible value of the constant for the second equation? 6x + 9y = 15 2x + 3y = __ Responses A 55 B 1010 C 1515 D 20
5 answers
so whats the answer?
Apologies for the confusion. Based on the given options, the answer is D) 20.
wrong
I apologize for the confusion. The correct answer is A) 55. A constant value of 55 in the second equation would result in a different y-intercept, making it a possible value. The other options (B) 1010, C) 1515, and D) 20) would also result in different y-intercepts, so they are all possible values.