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Solve for z. Assume the equation has a solution for z. {:[-4z+1=bz+c],[z=◻]:}

Solve for z z .\newlineAssume the equation has a solution for z z .\newline−4z+1=bz+cz=□ \begin{array}{l} -4 z+1=b z+c \\ z=\square \end{array}

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Full solution

Q. Solve for z z .\newlineAssume the equation has a solution for z z .\newline−4z+1=bz+cz=□ \begin{array}{l} -4 z+1=b z+c \\ z=\square \end{array}
  1. Simplify first equation: We are given a system of equations:\newline−4z+1=bz+c-4z + 1 = bz + c\newlinez=□z = \square\newlineWe need to solve for zz. Let's start by simplifying the first equation to isolate zz on one side.
  2. Add 4z4z to both sides: Add 4z4z to both sides of the first equation to move all the zz terms to one side:\newline−4z+4z+1=bz+4z+c-4z + 4z + 1 = bz + 4z + c\newlineThis simplifies to:\newline1=(b+4)z+c1 = (b + 4)z + c
  3. Subtract cc from both sides: Now, subtract cc from both sides to isolate the zz term:\newline1−c=(b+4)z1 - c = (b + 4)z
  4. Divide both sides: To solve for zz, divide both sides by (b+4)(b + 4):1−cb+4=z\frac{1 - c}{b + 4} = z
  5. Final solution: We now have an expression for zz in terms of bb and cc. However, we are also given that zz equals a blank square, which means we cannot determine a numerical value for zz without additional information about bb and cc or the blank square. Therefore, the solution for zz is expressed as:\newlinez=1−cb+4z = \frac{1 - c}{b + 4}

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