Solve the simultaneous equation:
x^2 + y^2 = 13 and x + 3y = -3
x = -3 (y + 1)
x^2 = 9 (y^2 + 2y + 1)
substitute and solve for y
there will probably be two solutions
... the intersection(s) of a circle and a line
x^2 + y^2 = 13 and x + 3y = -3
To solve the simultaneous equations, we will use the substitution method. Let's solve for x in terms of y from the second equation and substitute it into the first equation.
Step 1: Solve the second equation for x:
x + 3y = -3
x = -3 - 3y
Step 2: Substitute x = -3 - 3y into the first equation:
(-3 - 3y)^2 + y^2 = 13
Step 3: Expand and simplify:
9 + 18y + 9y^2 + y^2 = 13
10y^2 + 18y + 9 - 13 = 0
10y^2 + 18y - 4 = 0
Step 4: Solve the quadratic equation for y using factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
For the equation 10y^2 + 18y - 4 = 0, we have a = 10, b = 18, and c = -4. Substituting these values into the formula, we get:
y = (-18 ± √(18^2 - 4 * 10 * -4)) / (2 * 10)
Simplifying further:
y = (-18 ± √(324 + 160)) / 20
y = (-18 ± √484) / 20
y = (-18 ± 22) / 20
Simplifying the formula gives us two possible values for y:
y1 = (22 - 18) / 20 = 4/20 = 1/5
y2 = (-22 - 18) / 20 = -40/20 = -2
Step 5: Substitute the values of y back into the equation x = -3 - 3y to find the corresponding x-values:
For y = 1/5:
x = -3 - 3(1/5) = -3 - 3/5 = -15/5 - 3/5 = -18/5
For y = -2:
x = -3 - 3(-2) = -3 + 6 = 3
Step 6: The solution to the simultaneous equations is the set of ordered pairs (x, y):
(x1, y1) = (-18/5, 1/5)
(x2, y2) = (3, -2)
Therefore, the simultaneous equations x^2 + y^2 = 13 and x + 3y = -3 have two solutions: (-18/5, 1/5) and (3, -2).
To solve the simultaneous equations:
1. Start by rearranging one of the equations to solve for one variable in terms of the other variable. Let's rearrange the second equation, x + 3y = -3, to solve for x:
x = -3 - 3y
2. Substitute the value of x from the rearranged equation into the first equation, x^2 + y^2 = 13, to eliminate the x variable:
(-3 - 3y)^2 + y^2 = 13
3. Simplify the equation by expanding the squared term and combining like terms:
9 + 18y + 9y^2 + y^2 = 13
10y^2 + 18y - 4 = 0
4. Rearrange the quadratic equation to the standard form:
10y^2 + 18y - 4 = 0
5. Solve the quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, it can be factored as:
(5y - 2)(2y + 2) = 0
Setting each factor equal to zero:
5y - 2 = 0 or 2y + 2 = 0
Solving each equation for y:
5y = 2 or 2y = -2
y = 2/5 or y = -1
6. Substitute the values of y back into the rearranged equation x = -3 - 3y to find the corresponding values of x:
For y = 2/5:
x = -3 - 3(2/5)
x = -3 - 6/5
x = (-15 - 6)/5
x = -21/5
For y = -1:
x = -3 - 3(-1)
x = -3 + 3
x = 0
7. The solution to the simultaneous equations is:
x = 0 and y = -1
x = -21/5 and y = 2/5