How many solutions does the system of equations below have?−2x−2y−z=3x−3y+z=173x+2y−z=−13Choices:(A)no solution(B)one solution(C)infinitely many solutions
Q. How many solutions does the system of equations below have?−2x−2y−z=3x−3y+z=173x+2y−z=−13Choices:(A)no solution(B)one solution(C)infinitely many solutions
Eliminate z: First, let's add the first and second equations to eliminate z.(−2x−2y−z)+(x−3y+z)=3+17−2x−2y−z+x−3y+z=20−2x+x−2y−3y=20−x−5y=20
Eliminate z again: Now, let's multiply the second equation by 2 and add it to the third equation to also eliminate z.2(x−3y+z)+(3x+2y−z)=2(17)+(−13)2x−6y+2z+3x+2y−z=34−132x+3x−6y+2y+2z−z=215x−4y+z=21
Eliminate y: We can now add the modified first equation to the modified third equation to eliminate y.(−x−5y)+(5x−4y)=20+21−x+5x−5y−4y=414x−9y=41
Solve for x: Let's solve for x using the equation 4x−9y=41. Since we don't have a value for y, we can't solve for x directly. We need to find another equation that relates x and y to solve the system.
Find x and y: We can use the modified first equation −x−5y=20 and the modified third equation 5x−4y+z=21 to solve for x and y. Let's multiply the first equation by 5 to match the coefficient of x in the third equation. 5(−x−5y)=5(20)−5x−25y=100
Eliminate x: Now, let's add this new equation to the third equation to eliminate x.(−5x−25y)+(5x−4y+z)=100+21−5x+5x−25y−4y+z=121−29y+z=121
Solve for y: We now have two equations with two variables:4x−9y=41−29y+z=121We can't solve for a specific value of y or z without another equation relating them. We need to go back and check our steps to see if we can find another relationship.
Find x: Looking back at the original equations, we can multiply the second equation by 2 and add it to the first to eliminate z.2(x−3y+z)+(−2x−2y−z)=2(17)+32x−6y+2z−2x−2y−z=34+3−6y−2y+2z−z=37−8y+z=37
Find z: We now have three equations with two variables:4x−9y=41−29y+z=121−8y+z=37We can use the last two equations to solve for y and z.
Find z: We now have three equations with two variables:4x−9y=41−29y+z=121−8y+z=37We can use the last two equations to solve for y and z.Subtract the third equation from the second to eliminate z.(−29y+z)−(−8y+z)=121−37−29y+z+8y−z=84−21y=84y=−4
Find z: We now have three equations with two variables:4x−9y=41−29y+z=121−8y+z=37We can use the last two equations to solve for y and z.Subtract the third equation from the second to eliminate z.(−29y+z)−(−8y+z)=121−37−29y+z+8y−z=84−21y=84y=−4Now that we have y, we can substitute it back into the equation 4x−9y=41 to find x.4x−9(−4)=414x+36=414x=41−36−29y+z=1210−29y+z=1211
Find z: We now have three equations with two variables:4x−9y=41−29y+z=121−8y+z=37We can use the last two equations to solve for y and z.Subtract the third equation from the second to eliminate z.(−29y+z)−(−8y+z)=121−37−29y+z+8y−z=84−21y=84y=−4Now that we have y, we can substitute it back into the equation 4x−9y=41 to find x.4x−9(−4)=414x+36=41−29y+z=1210−29y+z=1211−29y+z=1212Finally, we can substitute y=−4 into the equation −8y+z=37 to find z.−29y+z=1215−29y+z=1216−29y+z=1217−29y+z=1218
More problems from Determine the number of solutions to a system of equations in three variables