How many solutions does the system of equations below have?3x+y−z=−5−2x−2y+z=03x+2y−3z=8Choices:(A)no solution(B)one solution(C)infinitely many solutions
Q. How many solutions does the system of equations below have?3x+y−z=−5−2x−2y+z=03x+2y−3z=8Choices:(A)no solution(B)one solution(C)infinitely many solutions
Solve using elimination or substitution: First, let's try to solve the system using elimination or substitution.
Multiply second equation: We can multiply the second equation by 1.5 to make the coefficients of z cancel out when we add it to the third equation.1.5(−2x−2y+z)=1.5(0)This gives us −3x−3y+1.5z=0
Add new equation to third: Now, add this new equation to the third equation:(3x+2y−3z)+(−3x−3y+1.5z)=8+0This simplifies to −y−1.5z=8
Correct previous mistake: Oops, I made a mistake in the previous step. I should have subtracted 1.5z, not added it. Let's correct that.(3x+2y−3z)+(−3x−3y+1.5z)=8+0This simplifies to −y−4.5z=8
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