xy = 12
y = 2x + 2
xy = 12 ....(1)
y = 2x + 2 ....(2)
Substitute (2) into (1) to get:
x(2x+2) = 12
simplify and solve for the quadratic in x.
substitute the first equation to the second:
x(2x + 2) = 12
2x^2 + 2x - 12 = 0
x^2 + x - 6 = 0
(x+3)(x-2)=0
x = -3 and x = 2
substitute x values to either equations and solve for y:
first equation:
if x = -3,
(-3)y = 12
y = -4
if x = 2,
(2)y = 12
y = 6
so there,, =)
To solve the system of equations:
xy = 12 ..............(1)
y = 2x + 2 ...........(2)
We can substitute the value of y from equation (2) into equation (1).
Replacing y in equation (1) with 2x + 2, we get:
x(2x + 2) = 12
Expanding the brackets, we have:
2x^2 + 2x = 12
Rearranging the equation to standard quadratic form, we have:
2x^2 + 2x - 12 = 0
To solve the quadratic equation, we can either factorize it or use the quadratic formula.
Factoring:
In this case, we can divide both sides of the equation by 2 to simplify it:
x^2 + x - 6 = 0
The factors of -6 that add up to 1 are -2 and 3. Therefore, we can factorize the equation as:
(x - 2)(x + 3) = 0
Setting each factor to zero and solving for x, we get:
x - 2 = 0 or x + 3 = 0
Solving these two equations, we find:
x = 2 or x = -3
Now that we have the values of x, we can substitute them back into equation (2) to find the corresponding values of y.
For x = 2:
y = 2(2) + 2
y = 4 + 2
y = 6
Thus, one solution to the system of equations is x = 2 and y = 6.
For x = -3:
y = 2(-3) + 2
y = -6 + 2
y = -4
Thus, another solution to the system of equations is x = -3 and y = -4.
Therefore, the system of equations has two solutions: (2, 6) and (-3, -4).