1.Algebra Booster

7.42 Algebra Booster
Thus,
D1
( d -b)( b -c)( c -d) ( d -b)( c -d)
x = = =
D ( a -b)( b -c)( c -a) ( a -b)(c-a)
D2
( a -d)( d -c)( c -a) ( a -d)( d -c)
y = = =
D ( a -b)( b -c)( c -a) ( a -b)( b -c)
D3
( a -b)( b -d)( d -a) ( b -d)( d -a)
z = = =
D ( a -b)(b -c)( c -a) (b -c)( c -a)
1 1 1
67. Let u = , v= , w=
x y z
The given system of equations reduces to
1
u + v - w=
,
4
9
2u - v + 3w=
4
and –u – 2v + 4w = 1
1 1 -1
Here, D = 2 -1 3
-1 -2 4
= 1(–4 + 6) – 1(8 + 3) – 1(–4 – 1)
= 2 – 11 + 5 = –4
1/4 1 -1
D1
= 9/4 -1 3
1 -2 4
1 Ê 9 ˆ
= ◊2 -(9-3)-Á- + 1
4 Ë
˜
2 ¯
1 7
= + - 6= 4- 6= -2
2 2
1 1/4 -1
Also, D2
= 2 9/4 3
-1 1 4
11 Ê 9ˆ
= (9 -3) - - Á2
+
4 Ë
˜
4¯
= 4 – 5 = –1
1 1 1/4
Again, D 3 = 2 -1 9/4
-1 -2 1
Ê 9ˆ Ê 9ˆ
5
= Á- 1+ - 1 2+ -
Ë
˜
2¯ Á
Ë
˜
4¯
4
7 14
= - 2– = -2
2 4
D1
-4 1
Now, u = = = 2 fi x=
D -2 2
D2
-1 1
v= = = fi y = 2
D -2 2
D3 -2
and w= = = 1fi z = 1
D -2
68. Given parabola is y = ax 2 + bx + c, which is passing
through (2, 4), (–1, 1) and (–2, 5), so,
4a + 2b + c = 4
a – b + c = 1
4a – 2b + c = 5
From Cramers rule,
D1 15 5
a = = =
D 12 4
where
D2
1
b = = and
D 12
D3
2 1
c = = =
D 12 6
4 2 1
D = 1 - 1 1 = 12
4 -2 1
4 2 1
D1
= 1 - 1 1 = 15
5 -2 1
4 4 1
D2
= 1 1 1 = 1
4 5 1
4 2 4
D3
= 1 –1 1 = 2
4 –2 5
Hence, the required equation of the parabola is
y = ax 2 + bx + c
5 2 x 1
fi y = x + +
4 12 6
69. Since the system of equations has a unique solution, so,
2 k
π
3 - 4
fi
8
k π-
3
8 .
3
Therefore, the value of k is k ŒR
-{-
}
70. Since the system of equations has infinitely many solutions,
so
3 4 5
l = 8 = 10
fi 3 1 l = 2
fi l = 6
Hence, the value of l is 6.