Question

Separate 178 into two parts such that the first part is 8 less than twice the second part.

Accepted Answer

Answer on Question#39792 – Math - Algebra

Question.

Separate 178 into two parts such that the first part is 8 less than twice the second part.

Solution:

Suppose that

x + y = 178.

Also we have next condition

x - 2y = 8.

Thus we get next system of equation

\begin{array}{l} \left\{ \begin{array}{l} x + y = 178, \\ x - 2y = 8, \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x + y - (x - 2y) = 178 - 8, \\ x = 8 + 2y, \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 3y = 170, \\ x = 8 + 2y, \end{array} \right. \Rightarrow \left\{ \begin{array}{l} y = 56\frac{2}{3}, \\ x = 8 + 2 \cdot 56\frac{2}{3}, \end{array} \right. \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} y = 56\frac{2}{3}, \\ x = 121\frac{1}{3}. \end{array} \right. \end{array}

Answer:

121\frac{1}{3} \quad \text{and} \quad 56\frac{2}{3}

Question #39792

Expert's answer