\left\{ \begin{array} { l } { b = 3 } \\ { 3 k + b = 0 } \end{array} \right.
Solve for b, k
b=3
k=-1
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\left\{ \begin{array} { l } { b = 3 } \\ { 3 k + b = 0 } \end{array} \right.
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3k+3=0
Consider the second equation. Insert the known values of variables into the equation.
3k=-3
Subtract 3 from both sides. Anything subtracted from zero gives its negation.
k=\frac{-3}{3}
Divide both sides by 3.
k=-1
Divide -3 by 3 to get -1.
b=3 k=-1
The system is now solved.
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