MATH SOLVE

4 months ago

Q:
# What is the solution to the following system? 4x+3y-z=-6 6x-y+3z=12 8x+2y+4z=6A. x = 1, y = –3, z = –1B. x = 1, y = –3, z = 1C. x = 1, y = 3, z = 19D. x = 1, y = 3, z = –2

Accepted Solution

A:

Answer:Option B is correct.Step-by-step explanation:4x+3y-z= -6 eq(1)6x-y+3z= 12 eq(2)8x+2y+4z=6 eq(3)We need to solve these equations and find the value of x, y and z.Multiply equation 2 with 3 and then add equation 1 and 218x -3y +9z = 364x +3y -z = -6_____________22x + 8z = 30 eq(3)Multiply equation 2 with 2 and add with equation 312x -2y + 6z = 248x +2y +4z = 6____________20x + 10 z = 30 eq(4)Multiply equation 3 with 10 and equation 4 with 8 and then subtract220x + 80z = 300160x + 80z = 240- - -_________________60x = 60x= 60/60x= 1Putting value of x in equation 322x + 8z = 3022(1) + 8z = 308z = 30 - 228z = 8z = 8/8z=1Putting value of x and z in equation 14(1)+3y-(1)=-64 + 3y -1 = -63 + 3y = -63y = -6 -33y = -9y = -3so, Option B x=1, y=3 and z=1 is correct