Solve for \(c\) in \(2d+c+4=0\).
Solve for \(c\).
\[2d+c+4=0\]
Subtract \(2d\) from both sides.
\[c+4=-2d\]
Subtract \(4\) from both sides.
\[c=-2d-4\]
\[c=-2d-4\]
Substitute \(c=-2d-4\) into \(5c=1-3d\).
Start with the original equation.
\[5c=1-3d\]
Let \(c=-2d-4\).
\[5(-2d-4)=1-3d\]
\[5(-2d-4)=1-3d\]
Solve for \(d\) in \(5(-2d-4)=1-3d\).
Solve for \(d\).
\[5(-2d-4)=1-3d\]
Expand.
\[-10d-20=1-3d\]
Add \(10d\) to both sides.
\[-20=1-3d+10d\]
Simplify \(1-3d+10d\) to \(1+7d\).
\[-20=1+7d\]
Subtract \(1\) from both sides.
\[-20-1=7d\]
Simplify \(-20-1\) to \(-21\).
\[-21=7d\]
Divide both sides by \(7\).
\[-\frac{21}{7}=d\]
Simplify \(\frac{21}{7}\) to \(3\).
\[-3=d\]
Switch sides.
\[d=-3\]
\[d=-3\]
Substitute \(d=-3\) into \(c=-2d-4\).
Start with the original equation.
\[c=-2d-4\]
Let \(d=-3\).
\[c=-2\times -3-4\]
Simplify.
\[c=2\]
\[c=2\]
Therefore,
\[\begin{aligned}&c=2\\&d=-3\end{aligned}\]
c=2;d=-3
