Given the equations \( y = x^2 - 4x - 5 \) and \( y + x = -1 \), find one point that satisfies both equations.

{ y + x = − 1 y = x 2 − 4 x − 5 ​ { y = − 1 − x y = x 2 − 4 x − 5 ​ x 2 − 4 x − 5 = − 1 − x x 2 − 3 x − 4 = 0 Δ = ( − 3 ) 2 − 4.1. ( − 4 ) = 9 + 15 = 25 x = 2 3 ± 5 ​ x 1 ​ = − 1 → y = − 1 + 1 = 0 x 2 ​ = 4 → y = − 1 − 4 = − 5
We found 2 points that satisfies both equations (4,-5) and (-1,0)

Answered by Ryan2 | 2024-06-10

To find a point that satisfies both given equations, y = x2 - 4x - 5 and y + x = -1, we can substitute the second equation into the first to solve for x.
Step 1: Substitute y from the second equation into the first equation: (y + x) = x2 - 4x - 5 becomes x - 1 = x2 - 4x - 5.
Step 2: Rearrange the equation and solve for x: 0 = x2 - 5x - 4. This is a quadratic equation, which factors to (x - 4)(x - 1) = 0.
Step 3: Solve for x: x = 4 or x = 1. Now we need to find the corresponding y-values for each x. For x = 4, substitute into the second equation y + 4 = -1, giving y = -5. For x = 1, substitute into the second equation y + 1 = -1, giving y = -2.
Step 4: Check the solutions in the original equations: 42 - 4(4) - 5 = -5, and -1 + 4 = -5, confirms that (4, -5) is a solution. 12 - 4(1) - 5 = -2, and -1 + 1 = -2, confirms that (1, -2) is a solution.
Therefore, the points (4, -5) and (1, -2) satisfy both equations.

Answered by qwcat | 2024-06-24

The points ( − 1 , 0 ) and ( 4 , − 5 ) satisfy both equations, and one valid point is ( − 1 , 0 ) .
;

Answered by Ryan2 | 2024-12-23