Chapter 6.4, Problem 31HP

To determine

To write a system of equations with the given conditions.

Answer to Problem 31HP

  $\begin{array}{l}3x+y=15\\ x+y=5\end{array}$

Explanation of Solution

Given:

Multiplying one equation by -3 and adding the two equations gives the solution.

Calculation:

Consider an equation in random. For example, $3x+y=15$

The condition is given that multiplying one equation by -3 and adding the two equations gives the solution.

Therefore, let the other equation be $x+y=5$ .

Now multiply $x+y=5$ with -3 $\Rightarrow -3x-3y=-15$

Add the above equation with the first equation.

  $\begin{array}{l}\Rightarrow (3x+y)+(-3x-3y)=-15+15\\ \Rightarrow -2y=0\\ \Rightarrow y=0\end{array}$

Substitute the value of y in $3x+y=15$ ,

  $\begin{array}{l}\Rightarrow 3x+0=15\\ \Rightarrow 3x=15\\ \Rightarrow x=5\end{array}$

Conclusion:

Therefore, the system of equations is: $3x+y=15$ and $x+y=5.$