To solve: (x−2y=22x+3y=11y−4z=−7)

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Alara Mccarthy · Accepted Answer

Calculation: We have to solve x−2y=2…(1) 2x+3y=11…(2) y−4z=−7…(3) We have to choose equation 1 and equation 2 and eliminate the same variable x. Multiply both sides of first equation by -2 and we have to add equation 1 and 2, we get (1)×−2→−2x+4y=−4 (2)→2x+3y=11 7y=7 y=1 Now we have to choose a different pair of equations and eliminate the same variable. Substitude y=1 in equation 1, we get x−2y=2 x−2(1)=2 x−2=2 x=2+2 x=4 Substitute y=1 in equation 3, we get y−4z=−7 1−4z=−7 −4z=−7−1 −4z=−8 Divide both sides by -4 −4z−4=−8−4 z=2 The solution set for the system is {(4, 1, 2)} To check: Substitute x=4,y=1 in equation 1 x−2y=2 4−2(1)=2 4−2=2 2=2[True] Substitute x=4,y=1 in equation 2 2x+3y=11 2(4)+3(1)=11 8+3=11 11=11 [True] Substitute y=1,z=2 in equation 3 y−4z=−7 1−4(2)=−7 1−8=−7 −7=−7 [True] The solution set ={(4,1,2)} Conclusion: The solution set for the system is {(4, 1, 2)}

To solve: ((x-2y=2),(2x+3y=11),(y-4z=-7))

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