Q. 3 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 1–10, solve each system of equations using the method of substitution or the method of elimination. If the system has no solution, say that it is inconsistent. 

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

Step-by-Step Solution

Answer

The solution of the system is $(2,-1)$.

Step 1. Given Information

We are given a system of equations $\left\{\begin{array}{l}x-2y-4=0\\ 3x+2y-4=0\end{array}\right.$.

We need to solve the system of using the method of substitution or the method of elimination. 

Step 2. Eliminate y and solve for x

Add the two equations and eliminate $y$.

$\begin{array}{rcl}x-2y-4+3x+2y-4& =& 0\\ 4x-8& =& 0\end{array}$

Now solve the equation for $x$

$\begin{array}{rcl}4x-8+8& =& 0+8\\ 4x& =& 8\\ x& =& 2\end{array}$

Step 3. Solve for y

Substitute $2$ for $x$ in the first equation and solve for $y$.

$\begin{array}{rcl}x-2y-4& =& 0\\ 2-2y-4& =& 0\\ -2y-2& =& 0\\ -2y& =& 2\\ y& =& -1\end{array}$

So the solution is given by the ordered pair $\left(2,-1\right)$.

Step 4. Check the solution

Substitute $2$ for $x$ and $-1$ for $y$ in both the equations.

$\begin{array}{rcl}x-2y-4& =& 0\\ 2-2·(-1)-4& =& 0\\ 2+2-4& =& 0\\ 4-4& =& 0\\ 0& =& 0\\ & & \\ 3x+2y-4& =& 0\\ 3·2+2·(-1)-4& =& 0\\ 6-2-4& =& 0\\ 6-6& =& 0\\ 0& =& 0\end{array}$

So both the equations are satisfied. Thus the found solution is correct.