2x + 2y + 3z =13
-3x + 4y-z = 5
5x -3y + z =2
2007-03-13 15:13:38 · 4 answers · asked by noggle4 2 in Education & Reference ➔ Homework Help
Start with low numbers because answers are low. Assume z is 1(because it has least multiples),From equation 1 X and Y are 2 and 3. Assume x is 2, y is 3
1. 4 + 6 +3 =13
2. -6 +12 -1 = 5
3. 10 - 9 +1 =2.
Works, hope you understand why.
2007-03-13 15:21:13 · answer #1 · answered by Hosebeast-ess to be 4 · 1⤊ 0⤋
(1)...2x + 2y + 3z = 13
(2)...-3x + 4y – z = 5
(3)...5x – 3y + z = 2
Add eqn (2) to (3):
(4)...2x + y = 7
Add eqn (1) to 3x(2):
(5)...-7x + 14y = 28,
-x + 2y = 4
Solve simultaneous eqn (4) & (5) to find the values of x & y,
2x + y + 2(-x + 2y) = 7 + 2(4)
5y = 15
Therefore, y = 3
Substitue y=3 into eqn (4) to find x,
2x + 3 = 7
Substitue both values of x & y into eqn(3) and you will find the value of z.
2007-03-13 15:31:40 · answer #2 · answered by kanlim 3 · 0⤊ 0⤋
Do you mean subtraction method?
(3)2x+2y+3z=13
(2)-3x+4y-z=5
=6x+6y+9z=39
-6x+8y-2z=10
=14y+7z=49
(5)2x+2y+3z=13
(-2)5x-3y+z=2
=10x+10y+15z=65
-10x+ 6y - 2z=-4
=16y+13z=61
(16)14y+ 7z=49
(-14)16y+13z=61
=224y+112z=784
- 224y- 182z=-854
=-70z=-70
z=1
14y+7(1)=49
14y+7=49
14y=42
y=3
2x+2(3)+3(1)=13
2x+6+3=13
2x+9=13
2x=4
x=2
2007-03-13 15:33:57 · answer #3 · answered by jake s 4 · 0⤊ 0⤋
[a million] Eqn1 - 2eqn3 : -4x+5y=3 , eqn2 - 3eqn3 : -8x+12y=4 , multily the 1st eqn by using 2 & subtract the 2nd eqn : -2y=2 , y=-a million , -4x+5(-a million)=3 , x=-2 , 3(-2)-2(-a million)+z=0 , z=4 . [2] j+b+c=1480 , j=a hundred and twenty+b then a hundred and twenty+b+b+c=1480 , 2b+c=1360 & b+c=j+280 , j+j+280=1480 , j=six hundred , b+c=880 , b=480 , c=4 hundred
2016-12-18 13:04:46 · answer #4 · answered by ? 4 · 0⤊ 0⤋