Question

Solve the system of linear equations below. X + y = 4 2x + 3y = 0 A. X = -6, y = 2 B. X = -1, y = 5 C. X = 11 5 , y = 9 5 D. X = 12, y = -8

Answer 1

Answer:

Correct Answer :

(27/2, 3/2)

Step-by-step explanation:

The centre of the circumscribing the quadrilateral whose sides are 3x+y=22, x-3y=14 and 3x+ y=62 is

A.  (3/2, 27/2)

B.  (27/2, 3/2)

C.  (27, 3)

D.  (1, 2/3)

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The roots of the equation a(b-c) x2+b(c-a)x +c(a-b) =0 are

The distance of (1, -2) from the common chord of x2 + y2 – 5x + 4y – 2 =0 and x2 + y2 – 2x + 8y + 3 =0

If [(3x+4)/(x2-3x+2)]=[A/(x-2)]+[B/(x-1)] then (A,B)=

The value of (λ > 0) so that the line 3x – 4y =λ may touch the circle x2 + y2 -4x -8y -5=0 is

1+[(1/2).(3/5)]+[(1.3/2.4)(3/5)2]+[(1.3.5/2.4.6)(3/5)3]+-------∞=

Answer 2

X+y=4, so 2x+2y=8, 2x+3y=0, (2x+3y)-(2x+2y)=y, 0-8=-8, so y=-8. plug y=-8 in, x=12