ExplanationThe key to this tricky-sounding problem is setting up variables correctly and ensuring that you subtract or add appropriately for these variables when representing their ages at different points in time:

L = Lou’s age now

W = Wen’s age now

M = Mildred’s age now

T = Tyla’s age now

Two equations come from the second sentence of the problem:

Equation 1: \((L - 3) = 2(T - 3)\)

Equation 2: \((L - 3) = (M - 3) + 30\)

Another two equations come from the third sentence of the problem:

Equation 3: \(L = W + 47\)

Equation 4: \((W + 4) =\frac{T+4}{2}\)

In order to solve this problem effectively, look for ways to get two of the equations to have the same two variables in them. If you have two equations with only two variables, you can solve for both of those variables. Equation 4 has a W and a T; the only other equation with a T is Equation 1. If you substitute the L in Equation 1 with the W from Equation 3, you will have two equations with just W’s and T’s.

Equation 1: \((L - 3) = 2(T - 3)\)

\((W + 47) – 3 = 2(T - 3)\)

\(W + 44 = 2T - 6\)

\(W + 50 = 2T\)

Equation 4: \((W + 4) =\frac{T+4}{2}\)

\(2W + 8 = T + 4\)

\(2W + 4 = T\)

Now combine the equations to solve for W.

\(W + 50 = 2(2W + 4)\)

\(W + 50 = 4W + 8\)

\(W + 42 = 4W\)

\(42 = 3W\)

\(14 = W\)

Now that you know Wen’s age, you can solve for the rest.

Equation 3: \(L = W + 47\)

\(L = 14 + 47\)

\(L = 61\)

Equation 1: \((L - 3) = 2(T - 3)\)

\((61 - 3) = 2(T - 3)\)

\(58 = 2T - 6\)

\(64 = 2T\)

\(32 = T\)

Equation 2: \((L - 3) = (M - 3) + 30\)

\((61 - 3) = (M - 3) + 30\)

\(58 = M + 27\)

\(31 = M\)

Now that you know that L = 61, W = 14, M = 31, and T = 32, sum them to find the answer:

\(61 + 14 + 31 + 32 = 138\)

_________________

Sandy

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