Solve the system by the addition method
-10x^2 + 10y^2-320=0
30x^2 + 6y^2 - 336 = 0
multiply the first by 3
-30x^2 + 30y^2 = 960
30x^2 + 6y^2 = 336, add them
36y^2 = 1296
y^2 = 36
y =±6
plug y = 6 into one of the equations and find the x
do the same with y = -6
To solve the system by the addition method, we need to add or subtract the two equations in a way that eliminates one of the variables.
Let's start by multiplying the first equation by 3 and the second equation by -1, to make the coefficients of x^2 the same.
Equation 1: -30x^2 + 30y^2 - 960 = 0
Equation 2: -30x^2 - 6y^2 + 336 = 0
Now, we can add the two equations together to eliminate the x^2 variable:
(-30x^2 + 30y^2 - 960) + (-30x^2 - 6y^2 + 336) = 0
Combining like terms, we get:
-60x^2 + 24y^2 - 624 = 0
Next, let's divide the equation by -12 to simplify it:
5x^2 - 2y^2 + 52 = 0
Now, we have eliminated the x^2 variable.
To continue solving the system, we need to use the substitution method or solve for one variable in terms of the other.
Let's solve the new equation for x^2 in terms of y^2:
5x^2 = 2y^2 - 52
Dividing both sides by 5:
x^2 = 0.4y^2 - 10.4
Now, we can substitute this expression for x^2 into one of the original equations. Let's use the first equation:
-10(0.4y^2 - 10.4) + 10y^2 - 320 = 0
Simplifying this equation:
-4y^2 + 104 + 10y^2 - 320 = 0
Combining like terms:
6y^2 - 216 = 0
Dividing the equation by 6:
y^2 - 36 = 0
Now, let's solve for y:
y^2 = 36
Taking the square root of both sides:
y = ± 6
Now that we have the values of y, we can substitute them back into the equation we found for x^2 to solve for x.
Let's start with y = 6:
x^2 = 0.4(6)^2 - 10.4
x^2 = 14.4 - 10.4
x^2 = 4
x = ± 2
So, for y = 6, we have two solutions: (2, 6) and (-2, 6).
Now, let's try y = -6:
x^2 = 0.4(-6)^2 - 10.4
x^2 = 14.4 - 10.4
x^2 = 4
x = ± 2
So, for y = -6, we also have two solutions: (2, -6) and (-2, -6).
Therefore, the system has a total of four solutions: (2, 6), (-2, 6), (2, -6), and (-2, -6).