Question

y=x^2 2y+6=2(x+3) If (x,y) is a solution of the system of equations above and x>0, what is the value of xy

Answer 1

To solve the system of equations, we simplified and solved to find that x = 1 and y = 1. The required value of xy is therefore 1.

Step-by-step explanation:

The student provided two equations which form a system:

y = x^2

2y + 6 = 2(x + 3)

To find the solution to this system, we first simplify the second equation:

2y + 6 = 2x + 6

By simplifying, we get:

y = x

Since we have y = x and also y = x^2 from the first equation, we can equate them as they both equal to y:

x = x^2

Now solve for x, knowing that x > 0:

x^2 - x = 0

x(x - 1) = 0

This gives us two solutions for x: x = 0 or x = 1, but since x > 0, the valid solution here is x = 1.

Substituting x = 1 into y = x^2 gives us y = 1^2 = 1.

Therefore, we have solved for both x and y, and the point is (1,1).

Finally, to find the value of xy, we multiply x and y together:

xy = 1 * 1 = 1

Answer 2

a.) 1

Step-by-step explanation:

y = x2 -------- eqn(1)

2y +6 = 2(x +3) ------- eqn(2)

substitute eqn(1) into eqn(2)

2(x2) + 6 = 2x + 6

2x2 - 2x + 6 - 6 = 0

2x2 - 2x = 0

divide through by 2

x2 - x = 0

x2 = x

x = 1

Substitute x = 1 into eqn(1)

y = (1)2

y = 1

xy = 1 x 1

= 1

Ans A