solve each system of equation algebraically:
x2+y2=18
x+y=4
Do you mean the circle
x^2+y^2 = 18 ?
yes
Making that assumption, we are looking for where the straight line:
y = -1x + 4
which has a slope of -1 and goes through (0,4) and (4,0)
hits the circle with center at the origin and radius 3 sqrt 2 which is about 4.24
Sketch a graph so we will get an idea of how the solution should look.
So
substitute y = -x + 4 into the circle
x^2 + (-x+4)^2 = 18
x^2 + x^2 - 8 x + 16 = 18
2 x^2 -8 x -2 = 0
x^2 - 4 x -1 = 0
x = (1/2)[ 4 +/- sqrt (16 +4)]
x = (1/2) [ 4 +/- 2 sqrt 5]
x = 2 +/- sqrt 5
find the y for each of those x values
To solve this system of equations algebraically, we will use the method of substitution. First, we solve one equation for one variable, and then substitute that expression into the other equation.
Let's solve the second equation, x + y = 4, for x:
x + y = 4
x = 4 - y
Now, substitute this expression for x in the first equation, x^2 + y^2 = 18:
(4 - y)^2 + y^2 = 18
Expanding and simplifying the equation:
16 - 8y + y^2 + y^2 = 18
2y^2 - 8y - 2 = 0
Next, we solve this quadratic equation. To make it easier, we can divide the entire equation by 2:
y^2 - 4y - 1 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
y = (-b ± √(b^2 - 4ac))/2a
For this equation, the values are:
a = 1, b = -4, c = -1
So, substituting these values into the formula:
y = (-(-4) ± √((-4)^2 - 4(1)(-1)))/(2(1))
y = (4 ± √(16 + 4))/2
y = (4 ± √20)/2
y = (4 ± 2√5)/2
y = 2 ± √5
Therefore, there are two possible values for y: y = 2 + √5 and y = 2 - √5.
To find the corresponding values of x, substitute these values of y back into the equation x = 4 - y:
For y = 2 + √5:
x = 4 - (2 + √5) = 4 - 2 - √5 = 2 - √5
For y = 2 - √5:
x = 4 - (2 - √5) = 4 - 2 + √5 = 2 + √5
So the solutions to the system of equations are:
x = 2 - √5, y = 2 + √5
x = 2 + √5, y = 2 - √5