Solve the system of equations. show your work.
x^2-3y^2=13 x-2y=1
from the 2nd: x = 2y + 1
sub that into the 1st
(2y+1)^2 - 3y^2 = 13
4y^2 + 4y + 1 - 3y^2 - 13 = 0
y^2 + 4y - 12 = 0
it factors very nicely, and you will two values of y
sub each into x = 2y+1 to get their corresponding x values
Thank you!! So much!
To solve the system of equations:
1. Rearrange the second equation to express x in terms of y.
x - 2y = 1
x = 2y + 1
2. Substitute the value of x from the rearranged equation into the first equation.
(2y + 1)^2 - 3y^2 = 13
3. Expand and simplify the equation.
4y^2 + 4y + 1 - 3y^2 = 13
y^2 + 4y - 12 = 0
4. Factor the quadratic equation.
(y + 6)(y - 2) = 0
5. Set each factor equal to zero and solve for y values.
y + 6 = 0 -> y = -6
y - 2 = 0 -> y = 2
6. Substitute the y values into the rearranged equation for x and solve for x values.
For y = -6:
x = 2y + 1 -> x = 2(-6) + 1 -> x = -11
For y = 2:
x = 2y + 1 -> x = 2(2) + 1 -> x = 5
Thus, the solution to the system of equations is x = -11, y = -6 and x = 5, y = 2.