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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

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Final answer:

The solution to the system of linear equations X + Y = 4 and 2X + 3Y = 0 is obtained using the elimination method, resulting in X = 12 and Y = -8.

Step-by-step explanation:

To solve the system of linear equations X + Y = 4 and 2X + 3Y = 0, we can use the substitution or elimination method. Let's use the elimination method for this solution.

  1. Rewrite the first equation as Y = 4 - X.
  2. Substitute the expression for Y into the second equation: 2X + 3(4 - X) = 0.
  3. Simplify and solve for X: 2X + 12 - 3X = 0 which simplifies to -X + 12 = 0, yielding X = 12.
  4. Substitute X back into the first equation: Y = 4 - 12, giving Y = -8.

Therefore, the solution to the system is X = 12 and Y = -8, which corresponds to option D.

User Pureth
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6 votes

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)

View Answer "

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]+-------∞=

User HRM
by
8.4k points

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