Linear Equations

Section 5.3 Linear Equations

Standard Form.

A linear equation is any equation equivalent to one of the form

\begin{equation} Ax+By = C,\tag{5.4} \end{equation}

where \(A\text{,}\) \(B\) and \(C\) are real numbers, and not both of \(A\) and \(B\) are zero. We call (5.4) standard form. For example, \(y = \frac{2}{3}x + 5\) is a linear equation, but not in standard form. Rewriting,

\begin{gather*} y = \frac{2}{3}x + 5\\ 3y = 2x + 15\\ 3y-2x = 15 \end{gather*}

which is now in standard form.

The graph of a linear equation is a line in the coordinate plane.

Typically, but not always, this line will intersect the coordinate axes at an intercept. The \(x\)-intercept is at the point \(\mathcal{P}(x,0)\) on the line with \(y = 0\text{.}\) Similarly, \(y\)-interceptis at the point \(\mathcal{Q}(0,y)\) on the line with \(x = 0\text{.}\) To find these points, substitute zero for the appropriate variable and solve for the other. If a line has two intercepts, this is sufficient to draw the graph.

Example 5.8. Graphing a Line in Standard Form.

Find the intercepts of \(3y-2x = 15\) and sketch its graph.

Solution.

To find the \(x\)-intercept, set \(y=0\) and solve.

\begin{gather*} 3(0)-2x = 15\\ -2x = 15\\ x = -\frac{15}{2} \end{gather*}

So there is an \(x\)-intercept at \(\left(-\frac{15}{2},\, 0)\right)\) .

To find the \(y\)-intercept, set \(x=0\) and solve.

\begin{gather*} 3y-2(0) = 15\\ 3y = 15\\ y = 5 \end{gather*}

So there is a \(y\)-intercept at \(\left(0,5\right)\text{.}\)

Plotting these two points and drawing the line, we obtain its graph.

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