Q. 6 - 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. 

$\left\{\begin{array}{l}2x+3y-13=0\\ 3x-2y=0\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the given system is $\left(2,3\right)$.

Step 1. Given Information

We are given a system of equations $\left\{\begin{array}{l}2x+3y-13=0\\ 3x-2y=0\end{array}\right.$.

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

Step 2. Solve the equations for y

Multiply both sides of the first equation $2x+3y-13=0$ by $3$.

$\begin{array}{rcl}3(2x+3y-13)& =& 3·0\\ 6x+9y-39& =& 0\end{array}$

Multiply both sides of the second equation $3x-2y=0$ by $3$

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

Now subtract the third equation and fourth equation.

$\begin{array}{rcl}6x+9y-39-(6x-4y)& =& 0-0\\ 6x+9y-39-6x+4y& =& 0\\ 13y-39& =& 0\\ 13y& =& 39\\ y& =& 3\end{array}$

Step 3. Solve the equation for x

Substitute $3$ for $y$ in the second equation and solve for $x$.

$\begin{array}{rcl}3x-2y& =& 0\\ 3x-2·3& =& 0\\ 3x-6& =& 0\\ 3x& =& 6\\ x& =& 2\end{array}$

So the solution is given by the ordered pair $\left(2,3\right)$.

Step 4. Check the solution

Substitute $2$ for $x$ and $3$ for $y$ in both the equations.

$\begin{array}{rcl}2x+3y-13& =& 0\\ 2·2+3·3-13& =& 0\\ 4+9-13& =& 0\\ 13-13& =& 0\\ 0& =& 0\\ & & \\ 3x-2y& =& 0\\ 3·2-2·3& =& 0\\ 6-6& =& 0\\ 0& =& 0\end{array}$

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