If #2,-2 and then (a) x = , y = Ž (b) x = 5, y =Ż (0) x = Ž, y = { (a) x = 1, y = 2
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Answer:
-3
Step-by-step explanation:
△ =
∣
∣
∣
∣
∣
∣
∣
∣
x
2x
3x
x+y
5x+2y
7x+3y
x+y+z
7x+5y+2z
9x+7y+3z
∣
∣
∣
∣
∣
∣
∣
∣
= −16
∣
∣
∣
∣
∣
∣
∣
∣
x
0
0
x+y
3x
4x
x+y+z
5x+3y
6x+4y
∣
∣
∣
∣
∣
∣
∣
∣
= −16 (R
2
→R
2
−2R
1
) & (R
3
→R
3
−3R
1
)
x[(3x)(6x+4y)−4x(5x+3y)] = −16
x[18x
2
+12xy−20x
2
−12xy] = −16
−2x
3
=−16
x
3
=8
x=2
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