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Solve for z. Assume the equation has a solution for z. {:[-p*(d+z)=-2z+59],[z=◻]:}

Solve for z z .\newlineAssume the equation has a solution for z z .\newline−p⋅(d+z)=−2z+59z=□ \begin{array}{l} -p \cdot(d+z)=-2 z+59 \\ 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−p⋅(d+z)=−2z+59z=□ \begin{array}{l} -p \cdot(d+z)=-2 z+59 \\ z=\square \end{array}
  1. Identify the system: Identify the system of equations.\newlineWe have the following system of equations:\newline11. −p(d+z)=−2z+59-p(d+z) = -2z + 59\newline22. z=□z = \square (We need to find the value of zz)
  2. Isolate z in the first equation: Isolate z in the first equation.\newlineTo isolate z, we need to distribute the -p across (d+z) and then move all terms containing z to one side of the equation.\newline-p \cdot d - p \cdot z = −2-2z + 5959
  3. Combine like terms: Combine like terms.\newlineAdd pimeszp imes z to both sides of the equation to get all the zz terms on one side.\newline−pimesd=−2z+pimesz+59-p imes d = -2z + p imes z + 59
  4. Factor out z z on the right side: Factor out z z on the right side of the equation.
    −p⋅d=z⋅(−2+p)+59 -p \cdot d = z \cdot (-2 + p) + 59
  5. Isolate zz by subtracting 5959: Isolate zz by subtracting 5959 from both sides and then dividing by (−2+p)(-2 + p).\newline−p⋅d−59=z⋅(−2+p)-p \cdot d - 59 = z \cdot (-2 + p)\newlinez=−p⋅d−59−2+pz = \frac{-p \cdot d - 59}{-2 + p}
  6. Check if the value of p is valid: Check if the value of p is such that the denominator is not zero.\newlineWe must ensure that −2-2 + p \neq 00, otherwise, we cannot divide by zero. Since we are not given a specific value for p, we assume that p \neq 22, which is a condition for the solution to exist.

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