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Solve the following system of equations in terms of z : (1) x + 2y − 2z = 3(2) 3x + y + z = 8

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Given

1 ) x + 2y -2z =3

we will solve the above by letting Z be the subjet o our eqution as follows :

-2z = 3 -x -2y

Z = (-x /-2 ) -( 2y/-2) + (3 /-2) .{ divide by -2 both sides)

z = 1/2x -y -3/2

2)3x +y +z =8

z = -3x -y +8 ......{ set z as a subject of the formula )

since we have 2 equation we can equate , and cancel the variable that is

x + 2y -2z =3

3x + y + z =8

= 5x + 3y -z = 11 {I have added the two equation to get equation 3 )

lets multiply equation 3 by 2 , and add it to equation 1

2 *(5x + 3y -z )= 11

10 x +6y -2Z = 22 .....add to equation 1

x + 2y -2z = 3

11x +8y = 3

• we will now take the first equation and sutract x

x + 2y -2z =3

2y -2z +x = 3 -x ( then subtract x )

→ 2y -2z= 3-x (divide by 2 )

y -z = (3-x /2)

y = (3-x/2) + z ..... take this equation and substitute into equation 2

• 3x + y + z =8

3x + (3-x/2) + z +z = 8 .... we know that

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