Friday, November 14, 2014

Intermediate Algebra, Chapter 3, 3.3, Section 3.3, Problem 32

(a) Write the equation $7x - 3y = 3$ in slope-intercept form.


$
\begin{equation}
\begin{aligned}

7x - 3y =& 3
&& \text{Given equation}
\\
\\
-3y =& -7x + 3
&& \text{Subtract each side by $7x$}
\\
\\
y =& \frac{7}{3}x - \frac{3}{3}
&& \text{Divide each side by $-3$}
\\
\\
y =& \frac{7}{3}x - 1
&&

\end{aligned}
\end{equation}
$



(b) Give the slope of the line.

Here the slope is $\displaystyle \frac{7}{3}$. It can be interpreted as $\displaystyle m = \frac{\text{rise}}{\text{run}} = \frac{7}{3}$

(c) Give the $y$-intercept.

The $y$-intercept is $-1$.

(d) Graph the line.

From $(0,-1)$, move 7 units up and 3 units to the right and plot a second point at $(3,6)$. Draw a line through the two points.

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