I am really having a hard time with these 2 problems: Please help me!! :)
--Solve any way you can:
a) 2x - y + z = 8
x + 2y + z = -3
x - 2y - z = 7
Add the 1st and 3rd equations and the z's
will cancel out
2x - y + z = 8
x - 2y - z = 7
------------------
3x - 3y = 15
And what's more, that can be divided through
by 3
x - y = 5
Next add the 2nd and 3rd equations and the z's
will also cancel out
x + 2y + z = -3
x - 2y - z = 7
------------------
2x = 4
x = 2
Substitute in
x - y = 5
2 - y = 5
-y = 3
y = -3
Substitute in this original equation:
2x - y + z = 8
2(2) - (-3) + z = 8
4 + 3 + z = 8
7 + z = 8
z = 1
(x,y,z) = (2,-3,1)
-----------------------------------------
b) 3x - y - 2z = 11
x - 2y + 3z = 12
x + y - 2z = 5
Add the 1st and 3rd equations and the y's
will cancel out
3x - y - 2z = 11
x + y - 2z = 5
-----------------
4x - 4z = 16
And what's more, that can be divided through
by 4 (that rhymes)
x - z = 4
To make the y's cancel in the 2nd and 3rd
original equations which are:
x - 2y + 3z = 12
x + y - 2z = 5
we will have to multiply the bottom one through by 2
and add them:
x - 2y + 3z = 12
2x + 2y - 4z = 10
-------------------
3x - z = 22
Now we have a system of two equations in two unknowns:
x - z = 4
3x - z = 22
Multiply the 1st one through by -1 and the z's will cancel
when you add them:
-x + z = -4
3x - z = 22
: -----------
2x = 18
x = 9
Substitute in
x - z = 4
9 - z = 4
-z = 4-9
-z = -5
z = 5
Substitute in the original equation:
x + y - 2z = 5
(9) + y - 2(5) = 5
9 + y - 10 = 5
y - 1 = 5
y = 6
(x,y,z) = (9,6,5)
Edwin
