Q. 2 - 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.  

$\left\{\begin{array}{l}3x-4y=4\\ x-3y=\frac{1}{2}\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system is $\left(2,\frac{1}{2}\right)$.

Step 1. Given Information

We are given a system of equations $\left\{\begin{array}{l}3x-4y=4\\ x-3y=\frac{1}{2}\end{array}\right.$.

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

Step 2. Substitute the value of x in the first equation

Solving the second equation $x-3y=\frac{1}{2}$ for $x$ we get

$\begin{array}{rcl}x-3y+3y& =& \frac{1}{2}+3y\\ x& =& 3y+\frac{1}{2}\end{array}$

Now substitute $3y+\frac{1}{2}$ for $x$ in the first equation $3x-4y=4$ and solve for $y$

$\begin{array}{rcl}3\left(3y+\frac{1}{2}\right)-4y& =& 4\\ 9y+\frac{3}{2}-4y& =& 4\\ 5y& =& 4-\frac{3}{2}\\ 5y& =& \frac{5}{2}\\ y& =& \frac{1}{2}\end{array}$

Step 3. Solve for the value of x

In the equation $x=3y+\frac{1}{2}$ substitute $\frac{1}{2}$ for $y$

$$\begin{gathered} x=3·\frac{1}{2}+\frac{1}{2} \\ x=\frac{3}{2}+\frac{1}{2} \\ x=\frac{4}{2} \\ x=2 \end{gathered}$$

So the solution of the system is $\left(2,\frac{1}{2}\right)$.

Step 4. Check the solution

Substitute $2$ for $x$ and $\frac{1}{2}$ for $y$ in both the equations.

$\begin{array}{rcl}3x-4y& =& 4\\ 3·2-4·\frac{1}{2}& =& 4\\ 6-2& =& 4\\ 4& =& 4\\ & & \\ x-3y& =& \frac{1}{2}\\ 2-3·\frac{1}{2}& =& \frac{1}{2}\\ \frac{4-3}{2}& =& \frac{1}{2}\\ \frac{1}{2}& =& \frac{1}{2}\end{array}$

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