Math 1324, Intro Math for Business and Social Sciences CRN 1330 Caitlin Intro Homework: Section 2.1 (F2O) Score: 0.5 of 2 of 16 (2 complete) 2.1.5 Determine whether (3,4) is a solution of (* = 6)2 – (y +6)2. Since the expression (x-6)^2 – (y + 6)^2 evaluated at (3,4) is [7 (34). Solution: None Options: 1. None 2. None 3. None 10. None 13. None 16. None
The Correct Answer and Explanation is:
Let’s carefully evaluate whether the point (3,4)(3, 4) is a solution to the expression: (x−6)2−(y+6)2(x – 6)^2 – (y + 6)^2
Step 1: Substitute (x,y)=(3,4)(x, y) = (3, 4) into the expression
(3−6)2−(4+6)2=(−3)2−(10)2(3 – 6)^2 – (4 + 6)^2 = (-3)^2 – (10)^2 =9−100=−91= 9 – 100 = -91
Step 2: Check what the problem is asking
It asks whether (3,4)(3, 4) is a solution of the expression (x−6)2−(y+6)2(x – 6)^2 – (y + 6)^2. But a point is only a solution if the equation is equal to some value.
However, the problem seems to be missing an equation (like “is this equal to 0?” or “= -91?”), so we will assume it is asking whether (3,4)(3, 4) satisfies the expression being equal to 0: (x−6)2−(y+6)2=0(x – 6)^2 – (y + 6)^2 = 0
Substitute again: (3−6)2−(4+6)2=9−100=−91≠0(3 – 6)^2 – (4 + 6)^2 = 9 – 100 = -91 \neq 0
So, (3, 4) is not a solution.
Final Answer:
(3, 4) is NOT a solution.
Explanation:
To determine whether a point like (3,4)(3, 4) is a solution to a mathematical expression or equation, we need to substitute the values of x=3x = 3 and y=4y = 4 into the expression and evaluate the result. In this problem, the expression given is: (x−6)2−(y+6)2(x – 6)^2 – (y + 6)^2
This is a difference of squares, a common algebraic structure with the general form A2−B2A^2 – B^2, which factors into (A−B)(A+B)(A – B)(A + B). However, factoring is not necessary here—we simply evaluate it directly at the point.
Plugging in x=3x = 3 and y=4y = 4: (3−6)2−(4+6)2=(−3)2−(10)2=9−100=−91(3 – 6)^2 – (4 + 6)^2 = (-3)^2 – (10)^2 = 9 – 100 = -91
If we were checking whether this expression is equal to 0 at this point (as typically done in solving equations), then −91≠0-91 \neq 0, so the point does not satisfy the equation. That means it is not a solution.
In conclusion, determining if a point is a solution involves plugging in the coordinates and checking whether the resulting value satisfies the equation. Since the result was −91-91 and not 0, the point (3,4)(3, 4) is not a solution to the equation.
