Question

solve the system.
y = -x + 9
y = x² + 3
select all points that are a solution to the system.
a. (-3, 12)
b. (-3, 0)
c. (2, 7)
d. (2, 0)
e. (-2, 7)
f. (3, 6)

Explanation:

Step1: Substitute \( y \) from first equation into second

Since \( y = -x + 9 \) and \( y = x^2 + 3 \), set equal: \( -x + 9 = x^2 + 3 \)

Step2: Rearrange into quadratic equation

\( x^2 + x - 6 = 0 \) (by moving all terms to one side: \( x^2 + x + 3 - 9 = 0 \))

Step3: Factor the quadratic

\( (x + 3)(x - 2) = 0 \) (find two numbers that multiply to -6 and add to 1: 3 and -2)

Step4: Solve for \( x \)

\( x + 3 = 0 \Rightarrow x = -3 \); \( x - 2 = 0 \Rightarrow x = 2 \)

Step5: Find corresponding \( y \) values

For \( x = -3 \): \( y = -(-3) + 9 = 12 \) (point: (-3, 12))
For \( x = 2 \): \( y = -2 + 9 = 7 \) (point: (2, 7)) Wait, but option E is (-2,7). Wait, maybe I made a mistake. Wait, let's check option E: x=-2. Let's test x=-2 in both equations. First equation: y = -(-2)+9=11? Wait no, wait original equations: y=-x+9. If x=-2, y=2+9=11? But option E is (-2,7). Wait, maybe I factored wrong. Wait, let's redo step 2. Wait, original substitution: -x +9 = x² +3 → x² +x -6=0? Wait, -x +9 -x² -3=0 → -x² -x +6=0 → multiply by -1: x² +x -6=0. Yes. Wait, but when x=-2, let's plug into quadratic: (-2)² + (-2) -6 = 4 -2 -6 = -4 ≠0. Wait, maybe the user's selected options have a mistake? Wait, no, let's check each option:

Option A: (-3,12). Plug into first equation: y=-(-3)+9=12 ✔️. Second equation: y=(-3)² +3=12 ✔️. So A is correct.

Option E: (-2,7). First equation: y=-(-2)+9=11≠7. Second equation: y=(-2)² +3=7. Wait, so (-2,7) satisfies second equation but not first? Wait, that can't be. Wait, maybe I messed up the substitution. Wait, original equations: y=-x+9 and y=x²+3. So to be a solution, must satisfy both. Let's check E: (-2,7). First equation: 7 = -(-2)+9 → 7=2+9=11? No. Second equation: 7=(-2)²+3=7 ✔️. So E is not a solution? But the user has A and E checked. Wait, maybe I made a mistake in the quadratic. Wait, let's re-express the system:

Wait, maybe the user's equations are different? Wait, the given equations: y = -x +9 and y = x² +3. Let's solve again.

Set -x +9 = x² +3 → x² +x -6=0 → roots x=2 and x=-3. So solutions are (2,7) and (-3,12). Now check option C: (2,7) – first equation: y=-2+9=7 ✔️, second: y=4+3=7 ✔️. So C is correct. Option E: (-2,7) – first equation: y=2+9=11≠7. Second equation: y=4+3=7. So E is not a solution. Wait, maybe the user's selected options are wrong, but according to the problem, we need to find all solutions. Wait, maybe I made a mistake in the quadratic. Wait, let's use quadratic formula: x = [-1 ±√(1 +24)]/2 = [-1 ±5]/2. So x=(4)/2=2, x=(-6)/2=-3. So only x=2 and x=-3. So solutions are (2,7) and (-3,12). So correct options are A and C. But the user has A and E checked. Maybe a typo in the problem. But according to the equations, the solutions are (-3,12) and (2,7). So option A: (-3,12) is correct. Option C: (2,7) is correct. Let's check option C: (2,7). First equation: y=-2+9=7 ✔️. Second equation: y=4+3=7 ✔️. So C is correct. Option E: (-2,7) – first equation: y=2+9=11≠7. So E is incorrect. Wait, maybe the first equation was y = -x + 7? Then x=-2: y=2+7=9≠7. No. Alternatively, maybe the second equation is y = x² - 3? Then for x=-2, y=4-3=1≠7. No. Alternatively, maybe the first equation is y = -x + 5? Then x=-2: y=2+5=7. Then second equation: y=(-2)² +3=7. Then that would work. But the given equations are y=-x+9 and y=x²+3. So based on given equations, the solutions are (-3,12) and (2,7). So correct options are A and C. But the user's selected A and E. Maybe a mistake. But according to the problem, we need to solve the system. So the correct solutions are (-3,12) and (2,7), so options A and C. Wait, but let's recheck the user's options:

A. (-3,12) – correct.

C. (2,7) – correct.

E. (-2,7) – incorrect.

Wait, maybe I made a mistake in the sign. Wait, first equation: y = -x +9. If x=2, y=7 ✔️. x=-3, y=12 ✔️. So the correct solutions are A and C. But the user has A and E selected. Maybe the problem has a typo, but based on the equations, the solutions are (-3,12) and (2,7), so options A and C.

Answer:

A. (-3, 12), C. (2, 7)