If f(x)=ax+b,where a and b are integers,f(-1) | vedclass

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(A) Given,$f(x) = ax + b$,where $a$ and $b$ are integers.We are given $f(-1) = -5$ and $f(4) = 3$.Substituting $x = -1$ into the function:$f(-1) = a(-

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$2, -3$

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(A) Given,$f(x) = ax + b$,where $a$ and $b$ are integers.We are given $f(-1) = -5$ and $f(4) = 3$.Substituting $x = -1$ into the function:$f(-1) = a(-1) + b = -5$$-a + b = -5$  (Equation $i$)Substituting $x = 4$ into the function:$f(4) = a(4) + b = 3$$4a + b = 3$  (Equation $ii$)Subtracting Equation $i$ from Equation $ii$:$(4a + b) - (-a + b) = 3 - (-5)$$4a + b + a - b = 3 + 5$$5a = 8$Since $a$ must be an integer,there is a discrepancy in the provided value $f(4)=3$. If we assume $f(3)=3$ as per standard problem types:$f(3) = 3a + b = 3$ (Equation $ii$)Subtracting Equation $i$ from Equation $ii$:$(3a + b) - (-a + b) = 3 - (-5)$$4a = 8 \implies a = 2$Substituting $a = 2$ into Equation $i$:$-2 + b = -5 \implies b = -3$Thus,$a = 2$ and $b = -3$.

If $f(x)=ax+b$,where $a$ and $b$ are integers,$f(-1)=-5$ and $f(4)=3$,then $a$ and $b$ are respectively

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