Question

what values of y and z make (\triangle rstcong\triangle dbc)?

Explanation:

Step1: Set up equations for corresponding sides

Since \(\triangle{RST}\cong\triangle{DBC}\), corresponding sides are equal.
For \(ST\) and \(BC\): \(10z=z + 36\)
For \(RT\) and \(DC\): \(y+3z + 11=z+6y-46\)

Step2: Solve for \(z\)

From \(10z=z + 36\), subtract \(z\) from both sides:
\(10z-z=z + 36-z\)
\(9z=36\)
Divide both sides by \(9\): \(z=\frac{36}{9}=4\)

Step3: Substitute \(z = 4\) into the second equation

Substitute \(z = 4\) into \(y+3z + 11=z+6y-46\)
\(y+3\times4 + 11=4+6y-46\)
\(y + 12+11=6y-42\)
\(y+23=6y-42\)

Step4: Solve for \(y\)

Subtract \(y\) from both sides: \(y - y+23=6y-y-42\)
\(23 = 5y-42\)
Add \(42\) to both sides: \(23+42=5y-42+42\)
\(65=5y\)
Divide both sides by \(5\): \(y=\frac{65}{5}=13\)

Answer:

\(y = 13\), \(z = 4\)