4. Lindiwe is twice as old as Preeya and six years older than Sue. Their combined ages total 39 years. How old is Lindiwe?

Let's denote the age of Lindiwe as \( L \), the age of Preeya as \( P \), and the age of Sue as \( S \). Given: 1. Lindiwe is twice as old as Preeya: \( L = 2P \) 2. Lindiwe is six years older than Sue: \( L = S + 6 \) 3. Their combined ages total 39 years: \( L + P + S = 39 \) We can solve this system of equations to find the age of Lindiwe. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}L=2P\\L=S+6\\L+P+S=39\end{array}\right.\) - step1: Substitute the value of \(L:\) \(\left\{ \begin{array}{l}2P=S+6\\2P+P+S=39\end{array}\right.\) - step2: Simplify: \(\left\{ \begin{array}{l}2P=S+6\\3P+S=39\end{array}\right.\) - step3: Solve the equation: \(\left\{ \begin{array}{l}2P=S+6\\S=39-3P\end{array}\right.\) - step4: Substitute the value of \(S:\) \(2P=39-3P+6\) - step5: Simplify: \(2P=45-3P\) - step6: Move the variable to the left side: \(2P+3P=45\) - step7: Add the terms: \(5P=45\) - step8: Divide both sides: \(\frac{5P}{5}=\frac{45}{5}\) - step9: Divide the numbers: \(P=9\) - step10: Substitute the value of \(P:\) \(S=39-3\times 9\) - step11: Calculate: \(S=12\) - step12: Substitute the value of \(P:\) \(L=2\times 9\) - step13: Calculate: \(L=18\) - step14: Calculate: \(\left\{ \begin{array}{l}L=18\\P=9\\S=12\end{array}\right.\) - step15: Check the solution: \(\left\{ \begin{array}{l}L=18\\P=9\\S=12\end{array}\right.\) - step16: Rewrite: \(\left(L,P,S\right) = \left(18,9,12\right)\) The solution to the system of equations is: \( L = 18 \), \( P = 9 \), \( S = 12 \) Therefore, Lindiwe is 18 years old.