Question Solve the system of equations: $x^2 + y^2 = 16$ and $x - y = 2$ .

Studdy Solution

STEP 1

Assumptions
1. We have a system of two non-linear equations.
2. The first equation is a circle with radius 4 centered at the origin: $x^{2}+y^{2}=16$ .
3. The second equation is a line with a slope of 1, where the x-coordinate is always 2 units larger than the y-coordinate: $x-y=2$ .

STEP 2

To solve the system, we can substitute the expression for $x$ from the second equation into the first equation. From the second equation, we can express $x$ as:
$x = y + 2$

STEP 3

Substitute $x = y + 2$ into the first equation:
$(y + 2)^{2} + y^{2} = 16$

STEP 4

Expand the squared term:
$y^{2} + 4y + 4 + y^{2} = 16$

STEP 5

Combine like terms:
$2y^{2} + 4y + 4 = 16$

STEP 6

Subtract 16 from both sides to set the equation to zero:
$2y^{2} + 4y - 12 = 0$

STEP 7

Divide the entire equation by 2 to simplify:
$y^{2} + 2y - 6 = 0$

STEP 8

Factor the quadratic equation:
$(y + 3)(y - 2) = 0$

STEP 9

Set each factor equal to zero and solve for $y$ :
$y + 3 = 0 \quad \text{or} \quad y - 2 = 0$

STEP 10

Solve for $y$ in each case:
$y = -3 \quad \text{or} \quad y = 2$

STEP 11

Now that we have two possible values for $y$ , we can find the corresponding $x$ values by substituting $y$ back into the second equation $x - y = 2$ .
First, let's substitute $y = -3$ :
$x - (-3) = 2$

STEP 12

Solve for $x$ when $y = -3$ :
$x + 3 = 2$
$x = 2 - 3$
$x = -1$

STEP 13

Now, let's substitute $y = 2$ :
$x - 2 = 2$

STEP 14

Solve for $x$ when $y = 2$ :
$x = 2 + 2$
$x = 4$

STEP 15

We have found two solutions for the system of equations:
1. When $y = -3$ , $x = -1$ .
2. When $y = 2$ , $x = 4$ .
So the solutions to the system of equations are:
$(x, y) = (-1, -3) \quad \text{and} \quad (x, y) = (4, 2)$

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