5y - z = -31
z = 5y + 31
2x - 2y + z = 3
( Now put z = 5y + 31 in equation )
2x - 2y + 5y + 31 = 3
2x + 3y = 3 - 31
2x + 3y = - 28
2x = - 28 - 3y Divide both sides with 2
x = - 14 - ( 3 / 2 ) y
x + 3y + 2z = -21
( Now put x = - 14 - ( 3 / 2 ) y and z = 5y + 31 in equation )
- 14 - ( 3 / 2 ) y + 3y + 2 ( 5y + 31 ) = - 21
- 14 - ( 3 / 2 ) y + 3y + 10y + 62 = - 21
- ( 3 / 2 ) y + 13y = - 21 + 14 - 62
- ( 3 / 2 ) y + ( 26 / 2 ) y = - 69
( 23 / 2 ) y = - 69 Multiply both sides with 2
23y = - 138 Divide both sides with 23
y = - 6
z = 5y + 31
z = 5 ( -6 ) + 31
z = - 30 + 31
z = 1
x = - 14 - ( 3 / 2 ) y
x = - 14 - ( 3 / 2 ) -6
x = - 14 + 18 / 2
x = - 14 + 9
x = - 5
So solutions are :
x = - 5
y = - 6
z = 1
Solve the system using any algebraic method.
2x - 2y + z = 3
5y - z = -31
x + 3y + 2z = -21
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The answer I got for this is ( -26, 6,67) but I know its wrong and I don't know why. Help please >< my work is at the bottom
I used substitution method changed 5y-z=-31 to z=5y+31 then I:
Plugged it in the first equation=
2x-2y+5y+31=3 -> 2x+3y+31=3 -> 2x+3y=-28 is my new 1st equation...
Plugged in z=5y+31 to equation 3 =
x+3y+2(5y+31)=-21 -> x+3y+10y+62=-21 -> x+13y+62=-21 -> x+13y=-83 is my new 2nd equation.
2(=13y=82)+3y=-28 -> -26-166+3y=-28 -> -23y=138 - > y=6
2x+3(6)=-28 -> 2x+18=-28 -> 2x=-46 -> x=-23
2(-26)-2*6)+z=3 -> -52-12+z=3 -> -64+z=3 -> z=67
1 answer