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If $a$ and $b$ are non-zero integers such that $a^2+b^2-4a-2b=0$ and $a^2-b^2=0$ , which of the following could be the value of $(a-b)$ ? $\newline$ (A) $-3$ $\newline$ (B) $2$ $\newline$ (C) $3$ $\newline$ (D) $6$

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Transformations of linear functions

Full solution

Q. If

a

and

b

are non-zero integers such that

a^2+b^2-4a-2b=0

and

a^2-b^2=0

, which of the following could be the value of

(a-b)

?

\newline

(A)

-3

\newline

(B)

2

\newline

(C)

3

\newline

(D)

6

Identify Equations and Relationship: Identify the given equations and the relationship between $a$ and $b$ . We have two equations: $1$ . $a^2 + b^2 − 4a − 2b = 0$ $2$ . $a^2 − b^2 = 0$ The second equation can be factored using the difference of squares. $a^2 − b^2 = (a + b)(a − b) = 0$ Since $a$ and $b$ are non-zero integers, $a + b$ cannot be zero, so $a − b$ must be zero.
Determine Value of a: Determine the value of a in terms of b using the second equation. $\newline$ From $a^2 - b^2 = 0$ , we have: $\newline$ $a - b = 0$ $\newline$ $a = b$
Substitute Value of $a$ : Substitute the value of $a$ into the first equation. $\newline$ Since $a = b$ , we can replace $a$ with $b$ in the first equation: $\newline$ $b^2 + b^2 − 4b − 2b = 0$ $\newline$ Combine like terms: $\newline$ $2b^2 − 6b = 0$
Factor Out $b$ : Factor out $b$ from the equation. $b(2b − 6) = 0$ Since $b$ is a non-zero integer, we can't have $b = 0$ , so we must have: $2b − 6 = 0$
Solve for b: Solve for b. $\newline$ Add $6$ to both sides of the equation: $\newline$ $2b = 6$ $\newline$ Divide both sides by $2$ : $\newline$ $b = 3$
Find Value of a: Since $a = b$ , find the value of $a$ . $\newline$ $a = b = 3$
Calculate $(a - b)$ : Calculate the value of $(a − b)$ . $\newline$ Since $a = b = 3$ , we have: $\newline$ $(a − b) = 3 − 3 = 0$ $\newline$ However, this contradicts the options given, as none of them are zero. We need to recheck our steps for any errors.

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