\left\{ \begin{array} { l } { z = 4 } \\ { 0 = 5 k + z } \end{array} \right.
Solve for z, k
z=4
k=-\frac{4}{5}=-0.8
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\left\{ \begin{array} { l } { z = 4 } \\ { 0 = 5 k + z } \end{array} \right.
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0=5k+4
Consider the second equation. Insert the known values of variables into the equation.
5k+4=0
Swap sides so that all variable terms are on the left hand side.
5k=-4
Subtract 4 from both sides. Anything subtracted from zero gives its negation.
k=-\frac{4}{5}
Divide both sides by 5.
z=4 k=-\frac{4}{5}
The system is now solved.
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