okay im solving linear systems with combinations by multiplying first! theres 9x-3z=20, 3x-6z=2 ......please solve!!!!
2007-12-06 13:41:59 · 6 answers · asked by questions123 3 in Science & Mathematics ➔ Mathematics
OK
9x-3z=20
3x-6z=2 ; let's multiply the top by 2
18x -6z = 40
3x - 6z = 2 ; subtract the bottom from the top, z's go away
15x = 38
x = 38/15
9(38/15) - 3z = 20
114/5 - 3z = 20
-3z = 20 - 114/5
-3z = -2 4/5
3z = 14/5
z = 14/15
Check in second equation
3(38/15) - 6(14/15) = 2??
38/5 - 28/5 = 2??
10/5 = 2??
2 = 2 YES!!
Hope that helps.
2007-12-06 13:51:05 · answer #1 · answered by pyz01 7 · 0⤊ 0⤋
First solve for x
3x - 6z =2
3x = 2 + 6z
x = (2/3) + 2z
Now sub in for x
9( (2/3) + 2z) - 3z =20
6 +18z - 3z = 20
18z - 3z = 14
15z = 14
z = (14/15)
Now sub in for z
3x - 6(14/15) = 2
3x - (28/5) = 2
3x = (10/5) + (28/5)
3x = (38/5)
x = (38/15)
so x = (38/15) and z = (14/15)
2007-12-06 21:58:52 · answer #2 · answered by Hotdawgy 2 · 0⤊ 0⤋
9x -3z =20
3x - 6z =2
multiply the 2nd equation by -3:
-9x +18z = -6
Now add this equation to the first one:
15z = 14
z = 14/15
Sunstitute this value of z in either equation to find x:
9x - 42/15 =20
9x = 342/15
x = 38 / 15
2007-12-06 21:55:23 · answer #3 · answered by Computer Guy 7 · 0⤊ 0⤋
9x-3z=20
3x-6z=2 -9x+18z=-6
15z=14
z=14/15
3x-6(14/15)=2
2007-12-06 21:46:15 · answer #4 · answered by Par!ah 1 · 1⤊ 0⤋
-2(9x-3z=20)
3x-6z=2
-18x+6z=-40
-15x=-38
x=38/15
I hate fractions, you sub in and find z
2007-12-06 21:46:36 · answer #5 · answered by RickSus R 5 · 0⤊ 0⤋
z=14/15
x = 38/15
2007-12-06 21:51:42 · answer #6 · answered by someone2841 3 · 0⤊ 0⤋