Question

what are the solutions to the simple system of equations below? select all that apply.
$x^2 + y^2 = 41$
$y = x - 1$
options: (5, 4) (-4, -5) (5, -4) no solutions

Explanation:

Step1: Substitute $y=x-1$ into circle equation

Substitute $y = x-1$ into $x^2 + y^2 = 41$:
$$x^2 + (x-1)^2 = 41$$

Step2: Expand and simplify the equation

Expand $(x-1)^2$ and combine terms:
$$x^2 + x^2 - 2x + 1 = 41$$
$$2x^2 - 2x - 40 = 0$$
Divide by 2:
$$x^2 - x - 20 = 0$$

Step3: Factor the quadratic equation

Factor the simplified quadratic:
$$(x-5)(x+4) = 0$$

Step4: Solve for $x$ values

Set each factor equal to 0:
$x-5=0 \implies x=5$; $x+4=0 \implies x=-4$

Step5: Find corresponding $y$ values

For $x=5$: $y=5-1=4$
For $x=-4$: $y=-4-1=-5$

Step6: Verify solutions

Check $(5,4)$: $5^2+4^2=25+16=41$, valid.
Check $(-4,-5)$: $(-4)^2+(-5)^2=16+25=41$, valid.

Answer:

(5, 4), (-4, -5)