SOLUTION: how do i solve
5x+2y=4
3x+4y+2z=6
7x+3y+4z=29
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Question 53806This question is from textbook Algebra 2
: how do i solve
5x+2y=4
3x+4y+2z=6
7x+3y+4z=29
This question is from textbook Algebra 2
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
5x+2y=4
3x+4y+2z=6
7x+3y+4z=29
:
Notice that the 1st equation only has two unknowns x & y, so we can use the
elimination method on the 2nd & 3rd equations to eliminate z. Then we can
pair the results of that, with the 1st equation, and work with two equations
and two unknowns, which is pretty managable.
:
Mult the 2nd equation by 2 and subtract it from the 3rd equation:
:
7x + 3y + 4z = 29
6x + 8y + 4z = 12
------------------subtract
x - 5y + 0z = 17
or just:
x - 5y = 17
:
Do the following to eliminate x from x - 5y = 17 and the 1st equation
Mult it by 5 and subtract it from the 1st equation:
:
5x + 2y = 4
5x -25y = 85
----------------- subtract
0x + 27y = -81
y = -81/27
y = -3
:
Find x: substitute -3 for y in the 1st equation
5x + 2(-3) = 4
5x - 6 = 4
5x = 4 +6
x = 10/5
x = +2
:
Substitute for x & y in the 2nd equation and find z
3(2) + 4(-3) + 2z = 6
+6 - 12 + 2z = 6
-6 + 2z = 6
2z = 6 + 6
z = 12/2
z = +6
:
Finally, check all three solutions in the 3rd equation:
7(2) + 3(-3) + 4(6) =
14 - 9 + 24 = 29
:
A lot of steps, but not that complicated, hope it makes sense to you.
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