Answer - A
Explanation
The given equations are:
8x+5y = 9 ...(a)
3x+2y = 4 ...(b)
On multiplying (a) by 2, (b) by 5 and subtracting, we get: x= -2
Putting x = -2 in (b), we get:
-6 + 2y = 4 => 2y = 10 ∴ y = 5
Answer - C
Explanation
The given equation is:
5/x +6y = 13 ...(a)
3/x+4y =7 ...(b)
On multiplying (a) by 3, (b) by 5 and subtracting, we get:
-2y = 4 ∴ y = -2
Answer - D
Explanation
Taking first two parts, we get:
(x+y-8)/2 = (x+2y-14)/3
=> 3 (x+y-8) = 2(x+ 2y-14)
=> 3x+3y-24 = 2x+4y -28
=> x- y= -4 ...(1)
Taking last two parts, we get:
(x+2y-14)/3 = (3x+y-12)/11
=> 11 (x+2y-14) = 3(3x+y-12)
=> 11x+ 22y - 154 = 9x+3y -36
=> 2x+19y- 118 ...(2)
Multiplying (1) by 2 and subtracting from (2) we get,
21 y = 126
=> y = 6
Putting y = 6 in (1), we get: x= 2
=> x= 2, y= 6
Answer - B
Explanation
217x +131y= 913 ...(a)
131 x+ 217 y= 827 ...(b)
It is a special case in which coefficients of x and y in (a) are interchanged in (b)
Adding (a) and (b) , we get : 348(x+y)= 1740 => x+y = 5 ...(a)
Subtracting (b) from (a), we get: 86(x-y) = 86 => x-y =1 ...(b)
Adding (a) and (b), we get: x= 3, y= 2
Q 5 - For what value of h, the system of equations,hx-y-2=0 and 6x-2y-3=0 has a unique solution?
A - 2
B - 3
C - 4
D - 5
Answer - B
Explanation
For, a unique solution, we must have a1/a ≠ b1/b2
h/6 ≠ -1/-2 => h/6 ≠ 1/2 => h = 3
Q 6 - For what value of h, the system of equations, x+2y+7 = 0 and 2x+hy+14= 0 have an infinite number of solutions?
A - 3
B - 4
C - 5
D - 6
Answer - B
Explanation
For infinite solutions, we have a1/a2 = b1/b2= c1/c2;
h1/2 = 2/h = 7/14 => h=4.
Q 7 - For what value of h, the system of equations, hx-10y-3= 0 and 3x-5y-7=0 has no solutions?
A - 6
B - 5
C - 4
D - 3
Answer - A
Explanation
For no solution, we have a1/a2 = b1/b2 ≠ c1/c1
∴ h/3 = -10/-5 ≠-3/-7 => h = 6
