To find the solutions to the system of equations represented by:
- \( y = x^2 - 3x - 4 \) (a quadratic equation)
- \( y = -x - 1 \) (a linear equation)
we need to determine the points at which these two curves intersect.
To do this algebraically, we can set the two equations equal to each other:
\[ x^2 - 3x - 4 = -x - 1 \]
Next, we rearrange the equation to set it to 0:
\[ x^2 - 3x + x - 4 + 1 = 0 \]
Simplifying gives:
\[ x^2 - 2x - 3 = 0 \]
Now we can factor the quadratic:
\[ (x - 3)(x + 1) = 0 \]
Setting each factor to zero gives us:
- \( x - 3 = 0 \) → \( x = 3 \)
- \( x + 1 = 0 \) → \( x = -1 \)
Now, we substitute these \( x \)-values back into either original equation to find the corresponding \( y \)-values. Let's use the linear equation \( y = -x - 1 \).
For \( x = 3 \):
\[ y = -3 - 1 = -4 \] → The point is \( (3, -4) \).
For \( x = -1 \):
\[ y = -(-1) - 1 = 0 \] → The point is \( (-1, 0) \).
Thus, the solutions to the system of equations are the points \( (-1, 0) \) and \( (3, -4) \).
The correct answer is:
(−1,0) and (3,−4)
open paren negative 1 comma 0 close paren and open paren 3 comma negative 4 close paren.
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