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When using the Substitution Method to solve a system of equations, it is necessary to isolate a variable.
(2,1,- 1)
(II): LHS-(y+4z)=RHS-(y+4z)
(I), (III): x= - 1-y-4z
(I): Distribute 2
(III): Distribute 3
(I), (III): Add and subtract terms
(I): LHS+2=RHS+2
(I): LHS+11z=RHS+11z
(III): LHS+3=RHS+3
(III): LHS+14z=RHS+14z
(I): - 3y= 11 + 14z
(III): z= - 1
(III): Multiplication Property of -1
(III): Subtract term
(III): .LHS /(- 3).=.RHS /(- 3).
(II): y= 1, z= - 1
(II): Multiplication Property of -1
(II): Add and subtract terms
(I), (II), (III): Substitute values
(I), (II), (III): Multiplication Property of -1
(I), (II), (III): Multiply
(I), (II), (III): Add and subtract terms