The function \(f(x)\) satisfies \(f(x) + f(x + 2y) = 6x + 6y - 8\)
for all real numbers \(x\) and \(y\) Find the value of \(x\) such that \(f(x) = 0\)
The function \(f(x)\) satisfies \(f(x)+f(x+2y)\ =\ 6x+6y-8\) or all real numbers \(x\) and \(y\) Find the value of \(x\) such that \(f(x)\ =\ 0\)
\(f(x)+f(x+2y)=6x+6y-8\)
\(f(x)+2y\ =\ 6x+6y-8\)
\(f(x)\ =\ 6x+6y-8+2y\)
\(f(x)\ =\ 6x+8y-8\)
\(f(x)\ =\ 8y-8\)
\(f(x)\ =\ 1\)
If im wrong I'm sorry I did this in my head in like three minutes.
Your answer was incorrect but helped me realize that the answer is 4/3. So in an indirect way you got the right result. Thank you for helping
The function \(f(x) \)satisfies \(f(x) + f(x + 2y) = 6x + 6y - 8\)
for all real numbers \(x\) and \(y\)
Find the value of \(x\) such that \(f(x) = 0\)
\(\begin{array}{|rcll|} \hline \mathbf{f(x) + f(x + 2y)} &=& \mathbf{6x + 6y - 8} \quad &| \quad y= 0\\\\ f(x) + f(x + 2\cdot 0) &=& 6x + 6\cdot 0 - 8 \\ f(x) + f(x) &=& 6x - 8 \\ 2f(x)&=& 6x - 8 \quad &| \quad : 2 \\ f(x)&=& 3x - 4 \quad &| \quad f(x) = 0 \\ 0 &=& 3x - 4 \\ 3x &=& 4 \quad &| \quad : 3 \\ \mathbf{x} &=& \mathbf{\dfrac{4}{3}} \\ \hline \end{array}\)