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.

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

Short answer

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

Step-by-step solution

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 role="math" $y$

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

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.