Question 660956

Please help me solve this system of equations:

Directions: Solve using elimination and using the matrix method
x+y+z=6
2x-3y+5z=-11
x+3y-4z=19

Work I have done so far:

x+y+z=6
-x-3y+4z=-19              This equals -2y+5z=-13     equation #1

-2x-6y+8z=-38
2x-3y+5z=-11              This equals -9y+13z=-49    equation #2

Multiply equation #1 by -9 and multiply equation #2 by 2

18y+45z=117
-18y-26z=-98             z= -19/71

At this point I got stuck because my teacher said that all answers would be nice even numbers.  When I solved the equations using the matrix method (I know how to do this part) I got (3,4,-1) which I checked and is the correct answer.


It would seem as though you multiplied - 9y + 13z = - 49 by 2, which results in: - 18y + 26z = - 98, NOT - 18y - 26z = - 98. This is where you went wrong. Look into this and you should be fine. My elimination-solution is below.


{{{system (x + y + z = 6_______eq (i),
2x - 3y + 5z = - 11______eq (ii),
x + 3y - 4z = 19______eq (iii))}}}
- 2x – 2y – 2z = - 12 ------- Multiplying eq (i) by – 2 ------ eq (iv)


– 5y + 3z = - 23 ------- Adding eqs (iv) and (ii) ----- eq (v)
- 2y + 5z = - 13 ------- Subtract eq (iii) from eq (i) ----- eq (vi)


10y – 6z = 46 ------- Multiplying eq (v) by – 2 ------ eq (vii)
- 10y + 25z = - 65 ------- Multiplying eq (vi) by 5 ------ eq (viii)
19z = - 19 -------- Adding eqs (viii) & (vii)
z = {{{(- 19)/19}}}, or {{{highlight_green(z = - 1)}}}


– 5y + 3(- 1) = - 23 ------- Substituting – 1 for z in eq (v) 
- 5y – 3 = - 23 
– 5y = - 23 + 3 
– 5y = - 20
y = {{{(- 20)/- 5}}}, or {{{highlight_green(y = 4)}}}


x + 4 - 1 = 6 ----- Substituting 4 for y and – 1 for z in eq (i)
x + 3 = 6
x = 6 - 3, or {{{highlight_green(x = 3)}}}


Send comments and “thank-yous” to “D” at MathMadEzy@aol.com