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Algebra help please?!?
hey, i have an algebra problem i can't solve.. here it is
4x+7y-z=42
-2x+2y+3z=-26
2x-3y+5z=10
I have to solve it using elimination.. and I keep getting weird numbers so I don't think they're correct.
2 Answers
- Jeff AaronLv 71 decade agoFavourite answer
From equation 1:
-z = 42 - 4x - 7y
z = 4x + 7y - 42
Substitute z into equation 2:
-2x + 2y + 3(4x + 7y - 42) = -26
-2x + 2y + 12x + 21y - 126 = -26
10x + 23y = -26 + 126
10x + 23y = 100
23y = 100 - 10x
y = (100 - 10x) / 23
y = (100/23) - (10/23)x
Substitute y into z:
z = 4x + 7((100/23) - (10/23)x) - 42
z = 4x + (700/23) - (70/23)x - 42
z = (22/23)x - (10/23)x + (700/23) - (966/23)
z = (22/23)x - (266/23)
Substitute y and z into equation 3:
2x - 3((100/23) - (10/23)x) + 5((22/23)x - (266/23)) = 10
2x - (300/23) + (30/23)x + (110/23)x - (1330/23) - 10 = 0
Multiply through by 23:
46x - 300 + 30x + 110x - 1330 - 230 = 0
186x - 1860 = 0
186x = 1860
x = 1860/186
x = 10
Substitute x into y:
y = (100/23) - (10/23)(10)
y = (100/23) - (100/23)
y = 0
Substitute x into z:
z = (22/23)x - (266/23)
z = (22/23)(10) - (266/23)
z = (220/23) - (266/23)
z = (220 - 266) / 23
z = -46/23
z = -2
- 1 decade ago
4x+7y-z=42-----------------(1)
-2x+2y+3z=-26------------(2)
2x-3y+5z=10---------------(3)
(1)z=4x+4y-42------------(4)
(2)z=(2x-2y-26)/3--------(5)
(3)z=(10-2x+3y)/5-------(6)
(4)=(5)
4x+4y-42=(2x-2y-26)/3
=>12x+12y-126=2x-2y-26
=>10x+14y=100
=>5x+7y=50-----------(7)
(5)=(6)
(2x-2y-26)/3=(10-2x+3y)/5
=>10x-10y-130=30-6x+9y
=>16x-19y=160--------(8)
Solving (7) and (8),
x=10,y=0..
substituting these values in (4),
z=4x+4y-42
=>z=4(10)+4(0)-42
=>z=40+0-42
=>z=-2
Hence values are x=10,y=0 and z=-2...
