How many solutions does the system of equations below have?x+y+2z=19−x−y−2z=−192x−y+z=1Choices:(A)no solution(B)one solution(C)infinitely many solutions
Q. How many solutions does the system of equations below have?x+y+2z=19−x−y−2z=−192x−y+z=1Choices:(A)no solution(B)one solution(C)infinitely many solutions
Addition of Equations: First, let's add the first two equations to see if they simplify.(x+y+2z)+(−x−y−2z)=19+(−19)This simplifies to 0=0, which is always true.
Dependent Equations: Since 0=0 is true, the first two equations are dependent, meaning they represent the same plane in three-dimensional space.
Consistency Check: Now, let's check if the third equation, 2x−y+z=1, is consistent with the first two equations.We can substitute x, y, and z from the first equation into the third one to see if it holds true.But since the first two equations are the same, we can't actually find a unique value for x, y, and z to substitute.
Infinite Solutions: Since the first two equations are the same, and we don't have information that the third equation is different or contradictory, the system has infinitely many solutions. All points that satisfy the first equation will also satisfy the second, and potentially the third.
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