consider the table of values for a system of two linear equations.\n\n| x | y₁ | y₂ |\n| --- | --- |…

consider the table of values for a system of two linear equations.\n\n| x | y₁ | y₂ |\n| --- | --- | --- |\n| 0 | 15 | 28 |\n| 1 | 18 | 25 |\n| 2 | 21 | 22 |\n| 3 | 24 | 19 |\n| 4 | 27 | 16 |\n\nwhere does the solution to the system of equations occur?\n\n- between ( x = 0 ) and ( x = 1 )\n- between ( x = 1 ) and ( x = 2 )\n- between ( x = 2 ) and ( x = 3 )\n- between ( x = 3 ) and ( x = 4 )

consider the table of values for a system of two linear equations.\n\n| x | y₁ | y₂ |\n| --- | --- | --- |\n| 0 | 15 | 28 |\n| 1 | 18 | 25 |\n| 2 | 21 | 22 |\n| 3 | 24 | 19 |\n| 4 | 27 | 16 |\n\nwhere does the solution to the system of equations occur?\n\n- between ( x = 0 ) and ( x = 1 )\n- between ( x = 1 ) and ( x = 2 )\n- between ( x = 2 ) and ( x = 3 )\n- between ( x = 3 ) and ( x = 4 )

Answer

Explanation:

Step1: Analyze ( y_1 ) and ( y_2 ) trends

For ( y_1 ), as ( x ) increases by 1 (from 0 to 1, 1 to 2, etc.), ( y_1 ) increases by 3 (e.g., ( 18 - 15 = 3 ), ( 21 - 18 = 3 )). So ( y_1 ) is increasing. For ( y_2 ), as ( x ) increases by 1, ( y_2 ) decreases by 3 (e.g., ( 25 - 28 = -3 ), ( 22 - 25 = -3 )). So ( y_2 ) is decreasing.

Step2: Check ( y_1 ) and ( y_2 ) at each interval

  • Between ( x = 0 ) and ( x = 1 ): ( y_1(0)=15 ), ( y_1(1)=18 ); ( y_2(0)=28 ), ( y_2(1)=25 ). Here, ( y_1 < y_2 ) at both ( x = 0 ) and ( x = 1 ) (15 < 28, 18 < 25). No crossing.

  • Between ( x = 1 ) and ( x = 2 ): ( y_1(1)=18 ), ( y_1(2)=21 ); ( y_2(1)=25 ), ( y_2(2)=22 ). At ( x = 1 ), ( y_1 = 18 < y_2 = 25 ); at ( x = 2 ), ( y_1 = 21 < y_2 = 22 )? Wait, 21 < 22? Yes, but wait, ( y_1 ) is increasing, ( y_2 ) is decreasing. Wait, at ( x = 2 ), ( y_1 = 21 ), ( y_2 = 22 ). Still ( y_1 < y_2 )? Wait, no, wait at ( x = 2 ), ( y_1 = 21 ), ( y_2 = 22 ) (21 < 22). At ( x = 3 ), ( y_1(3)=24 ), ( y_2(3)=19 ). Wait, wait, let's check again. Wait, between ( x = 2 ) and ( x = 3 ): ( y_1(2)=21 ), ( y_1(3)=24 ); ( y_2(2)=22 ), ( y_2(3)=19 ). At ( x = 2 ), ( y_1 = 21 ), ( y_2 = 22 ) ( ( y_1 < y_2 ) ). At ( x = 3 ), ( y_1 = 24 ), ( y_2 = 19 ) ( ( y_1 > y_2 ) ). So between ( x = 2 ) and ( x = 3 ), ( y_1 ) crosses ( y_2 ) (since ( y_1 ) goes from less than ( y_2 ) to more than ( y_2 )). Wait, no, wait between ( x = 1 ) and ( x = 2 ): Wait, ( y_1(1)=18 ), ( y_2(1)=25 ); ( y_1(2)=21 ), ( y_2(2)=22 ). So at ( x = 1 ), ( y_1 = 18 < y_2 = 25 ); at ( x = 2 ), ( y_1 = 21 < y_2 = 22 )? Wait, 21 < 22, so still ( y_1 < y_2 ). Then between ( x = 2 ) and ( x = 3 ): ( y_1(2)=21 ), ( y_2(2)=22 ) ( ( y_1 < y_2 ) ); ( y_1(3)=24 ), ( y_2(3)=19 ) ( ( y_1 > y_2 ) ). So the crossing happens here. Wait, but wait, let's check the options again. Wait, maybe I made a mistake. Wait, between ( x = 2 ) and ( x = 3 ): ( y_1 ) goes from 21 to 24, ( y_2 ) goes from 22 to 19. So at ( x = 2 ), ( y_1 = 21 ), ( y_2 = 22 ) ( ( y_1 < y_2 ) ); at ( x = 3 ), ( y_1 = 24 ), ( y_2 = 19 ) ( ( y_1 > y_2 ) ). So the solution (where ( y_1 = y_2 )) is between ( x = 2 ) and ( x = 3 )? Wait, no, wait between ( x = 1 ) and ( x = 2 ): ( y_1(1)=18 ), ( y_2(1)=25 ); ( y_1(2)=21 ), ( y_2(2)=22 ). So ( y_1 ) is catching up. Then between ( x = 2 ) and ( x = 3 ), ( y_1 ) overtakes ( y_2 ). So the solution is between ( x = 2 ) and ( x = 3 )? Wait, but let's check the options. Wait, the options are between 0-1, 1-2, 2-3, 3-4. Wait, maybe I miscalculated. Wait, at ( x = 2 ), ( y_1 = 21 ), ( y_2 = 22 ) ( ( y_1 < y_2 ) ). At ( x = 3 ), ( y_1 = 24 ), ( y_2 = 19 ) ( ( y_1 > y_2 ) ). So the two lines cross between ( x = 2 ) and ( x = 3 ), because ( y_1 ) goes from below ( y_2 ) to above ( y_2 ) in that interval. Wait, but let's check the other intervals again. Between ( x = 1 ) and ( x = 2 ): ( y_1(1)=18 ), ( y_2(1)=25 ); ( y_1(2)=21 ), ( y_2(2)=22 ). So ( y_1 ) is still below ( y_2 ). Between ( x = 2 ) and ( x = 3 ), ( y_1 ) becomes above ( y_2 ). So the solution (where ( y_1 = y_2 )) is between ( x = 2 ) and ( x = 3 ). Wait, but let's check the options. The options are: between 0-1, 1-2, 2-3, 3-4. So the correct interval is between ( x = 2 ) and ( x = 3 ). Wait, but wait, maybe I made a mistake. Let's recalculate ( y_1 ) and ( y_2 ) at each x:

  • ( x = 0 ): ( y_1 = 15 ), ( y_2 = 28 ) ( ( y_1 < y_2 ) )

  • ( x = 1 ): ( y_1 = 18 ), ( y_2 = 25 ) ( ( y_1 < y_2 ) )

  • ( x = 2 ): ( y_1 = 21 ), ( y_2 = 22 ) ( ( y_1 < y_2 ) )

  • ( x = 3 ): ( y_1 = 24 ), ( y_2 = 19 ) ( ( y_1 > y_2 ) )

  • ( x = 4 ): ( y_1 = 27 ), ( y_2 = 16 ) ( ( y_1 > y_2 ) )

So the point where ( y_1 = y_2 ) must be between ( x = 2 ) and ( x = 3 ), because at ( x = 2 ), ( y_1 < y_2 ), and at ( x = 3 ), ( y_1 > y_2 ). So the solution occurs between ( x = 2 ) and ( x = 3 ).

Answer:

between ( x = 2 ) and ( x = 3 ) (the option corresponding to "between ( x = 2 ) and ( x = 3 )")