Q 4.4 - Intermediate Algebra

Question

Solve the system by graphing: $\left\{\begin{array}{l}-x+y=1\\ 3x+2y=12\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system $\left\{\begin{array}{l}-x+y=1\\ 3x+2y=12\end{array}\right.$ is $(2,3)$ and it is found by graphing as,

Step 1. Given Information.

We are given a system of equations $\left\{\begin{array}{l}-x+y=1\\ 3x+2y=12\end{array}\right.$and we need to find its solution by graphing.

Step 2. Graph the first equation.

Solving the equation $-x+y=1$ for $y$ we get,

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

So the slope of the line is 1 and its $y$ intercept is 1. Using this the graph of the equation is given as

Step 3. Graph the second equation on the same rectangular coordinate system.

Solving the equation $3x+2y=12$ for $y$ we get,

$\begin{array}{rcl}3x+2y-3x& =& 12-3x\\ 2y& =& -3x+12\\ \frac{2y}{2}& =& \frac{-3x+12}{2}\\ y& =& -1.5x+6\end{array}$

So the slope of the line is $-1.5$ and its $y$ intercept is 6.

So its graph can be drawn on the same plane as,

Step 4. Identify the solution to the system.

From the graph, it can be seen that the two lines intersect at the point $(2,3)$.

So, the point $(2,3)$ is the solution to the given system.

Step 5. Check the solution

Substitute $2$ for $x$ and $3$ for $y$ in the first equation.

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

It is a true statement.

Again, substitute the values in the second equation.

$\begin{array}{rcl}3x+2y& =& 12\\ 3·2+2·3& =& 12\\ 6+6& =& 12\\ 12& =& 12\end{array}$

This is also a true statement.

So the point satisfies both the equations.