x^2-3y^2=13
x-2y=1
solve for x and y show all your work
wouldn't one put for x in the first equation 2y+1, then solve for y ?
x^2-3y^2=13
x-2y=1
solve for x and y show all your work
Here I substituted the x=2y+1 in as follows:
(2y+1)^2 -3y^2 = 13 =>
4y^2+4y+1 -3y^2 = 13 => cancel out
y^2+4y-12 = 0 => Unfoil
(y+6)(y-2) = 0 =>
Y can = -6 and 2
If that is true then plug those answers into the x-2y = 1 equation as follows:
x-(2)(-6) = 1 =>
x = 1+(2)(-6) =>
x = -11
and
x-(2)(2) = 1 =>
x = 1+(2)(2) =>
x = 5
=> your answers are as follows:
Y can = -6 and 2
X can = -11 and 5
To solve the system of equations, we'll use the method of substitution. Here's how we can proceed:
From the second equation, we can isolate x:
x - 2y = 1
x = 2y + 1
Substitute this value of x into the first equation:
(2y + 1)^2 - 3y^2 = 13
Simplify the equation:
4y^2 + 4y + 1 - 3y^2 = 13
y^2 + 4y - 12 = 0
Now, we have a quadratic equation in terms of y. We can factor it or use the quadratic formula to solve for y. In this case, we can factor it:
(y + 6)(y - 2) = 0
Setting each factor equal to zero, we have two possible solutions for y:
y + 6 = 0 or y - 2 = 0
Solving each equation for y:
y = -6 or y = 2
Now substitute these values of y back into the equation x = 2y + 1 to find the corresponding values of x.
When y = -6:
x = 2(-6) + 1
x = -12 + 1
x = -11
When y = 2:
x = 2(2) + 1
x = 4 + 1
x = 5
Therefore, the solutions to the system of equations are:
x = -11, y = -6
x = 5, y = 2