Question

in the context of solving systems of equations, what does the point of intersection represent?
a. the average of the equations
b. the sum of the equations
c. the difference of the equations
d. the solution to both equations

for the equations $y = -x$ and $y = x^2 - 2x + 1$, what is the solution?
a. $(2, 1)$ and $(-1,4)$
b. $(3, 0)$
c. no solution
d. $(1, 2)$

why is it necessary to check both $x$-values in the linear equation after solving a system algebraically?
a. to confirm that the discriminant is positive
b. to verify whether the solution is real or imaginary
c. to ensure that there are no errors in calculation
d. to find the corresponding $y$-values for the points of intersection

Explanation:

First Question:

In a system of equations, the point of intersection of their graphs (or the solution to the system) is the set of values that satisfy all equations in the system. So it represents the solution to both equations. Options a, b, c are incorrect as average, sum, or difference of equations don't relate to the intersection's meaning.

Step1: Set the equations equal

Since \( y = -x \) and \( y = x^2 - 2x + 1 \), set \( -x = x^2 - 2x + 1 \).

Step2: Rearrange into quadratic equation

Rearrange to \( x^2 - 2x + 1 + x = 0 \) → \( x^2 - x + 1 = 0 \)? Wait, no, wait: \( -x = x^2 - 2x + 1 \) → \( x^2 - 2x + 1 + x = 0 \) → \( x^2 - x + 1 = 0 \)? Wait, discriminant \( D = (-1)^2 - 4(1)(1) = 1 - 4 = -3 < 0 \). Wait, maybe I made a mistake. Wait, original equations: \( y = -x \) and \( y = x^2 - 2x + 1 \). Let's substitute \( y \): \( -x = x^2 - 2x + 1 \) → \( x^2 - 2x + 1 + x = 0 \) → \( x^2 - x + 1 = 0 \). Discriminant is negative, but the options have real solutions. Wait, maybe I misread the second equation. Wait, maybe it's \( y = x^2 - 2x - 1 \)? No, the user wrote \( y = x^2 - 2x + 1 \). Wait, but the options include (2,1) and (-1,4). Let's test option a: For (2,1): In \( y = -x \), 1 = -2? No. Wait, ( -1,4): 4 = -(-1) = 1? No. Wait, option b: (3,0): 0 = -3? No. Option d: (1,2): 2 = -1? No. Wait, maybe there's a typo, but among the options, let's re - check the equation. Wait, maybe the second equation is \( y = x^2 - 2x - 1 \)? No, the user's equation is \( y = x^2 - 2x + 1 \). Wait, maybe I made a mistake in substitution. Wait, \( y = -x \), so substitute into \( y = x^2 - 2x + 1 \): \( -x = x^2 - 2x + 1 \) → \( x^2 - x + 1 = 0 \). Discriminant is negative, meaning no real solutions. But option c is "No solution". Wait, maybe that's the case. Wait, but let's check again. \( x^2 - x + 1 = 0 \), discriminant \( D = 1 - 4 = -3 < 0 \), so no real solutions. So the answer should be c? But wait, the options: a. (2,1) and (-1,4); b. (3,0); c. No solution; d. (1,2). So based on calculation, discriminant is negative, so no real solution.

Step3: Check discriminant

Discriminant \( D = b^2 - 4ac \) for \( ax^2 + bx + c = 0 \). Here, \( a = 1 \), \( b = -1 \), \( c = 1 \), so \( D = (-1)^2 - 4(1)(1) = -3 < 0 \), so no real solutions.

When solving a system algebraically (e.g., substituting to solve a quadratic and linear system), after finding \( x \)-values, we check them in the linear equation to find the corresponding \( y \)-values (since the solution is a point \((x,y)\)). Option a: Discriminant check is for quadratics, not for checking in linear equation. Option b: Checking in linear equation doesn't verify real/imaginary (that's from discriminant). Option c: Checking in linear equation isn't mainly for error in calculation (though it can catch some, but the main reason is to find \( y \)-values). Option d: Correct, as we need the \( y \)-values to get the points of intersection (the solutions are \((x,y)\) pairs).

Answer:

d. The solution to both equations

Second Question: