Chapter 6.2, Problem 32P

Solve each system by substitution. Tell whether the system has one solution, infinitely many solutions, or no solution.

$4x-y-4=0$

$3x+2y-14=0$

To determine

To solve: The system by substitution and suggest whether the system has one solution, infinitely many solutions, or no solution

  $\begin{array}{l}4x-y-4=0\\ 3x+2y-14=0\end{array}$

Answer to Problem 32P

Therefore the solution is

  $\left(x,y\right)=(2,4)$

The system has one solution, which is consistent and independent.

Explanation of Solution

Given information:

  $\begin{array}{l}4x-y-4=0\\ 3x+2y-14=0\end{array}$

Calculation:

In order to solve two system by substitution, we solve one of the equations for one of

the variables, and plug this into other equation.

  $\begin{array}{l}4x-y-4=0\\ 3x+2y-14=0\end{array}$

First we solve equation $4x-y-4=0$ as follows:

  $y=-4+4x$

Now we substitute $y=-4+4x$ into $3x+2y-14=0$ , we get,

  $\begin{array}{l}3x+2y-14=0\\ \Rightarrow 3x+2(4x-4)-14=0\\ \Rightarrow 11x=22\\ \Rightarrow x=2\end{array}$

By plugging $x=2$ in equation $y=4x-4$ we get

  $\begin{array}{l}y=4x-4\\ \Rightarrow y=4\times 2-4\\ \Rightarrow y=4\end{array}$

Therefore the solution is

  $\left(x,y\right)=(2,4)$

The system has one solution, which is consistent and independent.