Question 110553
solve the system use the substitution method
:
Eq1: 2x - y - z =  1
Eq2: 4x +5y +2z = 10
Eq3: -x- 4y +3z =-20
:
Not sure why you would want to use the substitution method here. Elimination makes sense. But let's try to substitute and see what happens;
Using eq1:
2x - y - z =  1
-y = 1 - 2x + z
y = 2x - z - 1; make y positive multiply by -1
:
Substitute (2x-z-1) for y in  Eq2
4x + 5(2x-z-1) + 2z = 10
4x + 10x - 5z - 5 + 2z = 10
14x - 3z = 10 + 5
14x - 3z = 15
-3z = -14x + 15
3z = 14x - 15
z = {{{14/3}}}x - 5
:
Substitute (2x-z-1) for y in Eq3
-x - 4(2x-z-1) + 3z = -20
-x - 8x + 4z + 4 + 3z = -20
-9x + 7z = -20 - 4
-9x + 7z = -24
:
Substitute ({{{14/3}}}x-5) for z in the above equation
-9x + 7({{{14/3}}}x-5) = -24
-9x + {{{98/3}}}x - 35 = - 24
{{{-27/3}}}x + {{{98/3}}}x = - 24 + 35
{{{71/3}}}x = 11
x = 11*{{{3/71}}}
x = {{{33/71}}}
x = .46478; not even an integer, disgusting!
:
Using -9x + 7z = -24, substitute for x and find z:
-9(.46478) + 7z = -24
7z = -24 + 4.1831
z = -19.8169/7
z = -2.8309
:
Find y using y = 2x - z - 1
y = 2(.46478) - (-2.8309) -1
y + .92956 + 2.8309 - 1
y = 2.76056
:
All these solutions check out on the Ti83 Matrix function. Really a nasty problem, hope this helped you out.