Section 5.6
Intersections of Lines — Related Topics
A third method to determine the point of intersection of two lines is by graphing the linear equations and reading the intersection point from the graph. When the intersection is at a point with integer coordinates, the exact answer can be determined with a careful graph. Otherwise, the best that can be done is an estimate of the point of intersection.
Another kind of problem is one of determining a simultaneous solution of two equations in two unknowns where the variables are in the denominators of the terms. Although the equations no longer represent lines in a coordinate plane, they can be reinterpreted to make the problem familiar.
Examples:
a. Draw a graph to determine the solution of the system $\left\{ \,\begin{matrix}
3x+2y=6\,; \\
y=-\tfrac{1}{2}x-1\,. \\
\end{matrix} \right.$
The first line has an $x$-intercept of $2$ and a $y$-intercept of $3$. Its graph is labeled (A) in the figure below.
The second line has a $y$-intercept of $-1$ and a slope $m=-\tfrac{1}{2}$. Its graph is labeled (B) in the figure below.
The lines intersect at the point $(4,-3)$.
b. Solve the system of equations $\left\{ \,\begin{matrix}
\tfrac{3}{x}+\,\tfrac{2}{y}=1\,; \\
\tfrac{5}{x}+\tfrac{7}{y}=9\,. \\
\end{matrix} \right.$
If we make a symbolic substitution, this system takes on a more familiar look. Let $P=\tfrac{1}{x}$ and $Q=\tfrac{1}{y}$. Then the system of equations becomes: $\left\{ \,\begin{matrix}
3P+2Q=1\,; \\
5P+7Q=9\,. \\
\end{matrix} \right.$
This is a standard "two equations in two unknowns" problem. The solution is $P=-1\text{ and }Q=2$. Then, since $P=\tfrac{1}{x}$, we have $x=-1$; and since $Q=\tfrac{1}{y}$, we have $y=\tfrac{1}{2}$.
Solving linear systems by graphing: Solving Linear Systems by Graphing
Exercises:
Draw graphs of the following pairs of lines to determine their intersection point.
1. $\left\{ \,\begin{matrix}
3x-2y=6 \\
y=\tfrac{1}{2}x-1 \\
\end{matrix} \right.$ 2. $\left\{ \,\begin{matrix}
7x+5y=3 \\
y=-2x \\
\end{matrix} \right.$
3. $\left\{ \,\begin{matrix}
y-3=-2(x+3) \\
y=\tfrac{1}{2}x+2 \\
\end{matrix} \right.$ 4. $\left\{ \,\begin{matrix}
3x-5y=-30 \\
x-5y=-20 \\
\end{matrix} \right.$
5. $\left\{ \,\begin{matrix}
y=\tfrac{2}{3}x-4 \\
y=-2x+4 \\
\end{matrix} \right.$ 6. $\left\{ \,\begin{matrix}
y=-\tfrac{4}{3}x+8 \\
y=\tfrac{1}{3}x-2 \\
\end{matrix} \right.$
7. $\left\{ \,\begin{matrix}
y-6=-\tfrac{1}{3}(x+2) \\
y=x \\
\end{matrix} \right.$ 8. $\left\{ \,\begin{matrix}
y-5=-\tfrac{1}{3}(x+3) \\
y=2x-3 \\
\end{matrix} \right.$
Estimate the solution of each of the following systems of equations by drawing a graph.
9. $\left\{ \,\begin{matrix}
10x-3y=1 \\
5x+6y=3 \\
\end{matrix} \right.$ 10. $\left\{ \,\begin{matrix}
12x-8y=2 \\
6x+4y=3 \\
\end{matrix} \right.$
11. $\left\{ \,\begin{matrix}
3x-6y=-7 \\
3x+2y=8 \\
\end{matrix} \right.$ 12. $\left\{ \,\begin{matrix}
x-y=4 \\
x+y=1 \\
\end{matrix} \right.$
Solve the following systems of equations for $x$ and $y$.
13. $\left\{ \,\begin{matrix}
\tfrac{10}{x}-\tfrac{3}{y}=1 \\
\tfrac{5}{x}+\tfrac{6}{y}=3 \\
\end{matrix} \right.$ 14. $\left\{ \,\begin{matrix}
\tfrac{12}{x}-\tfrac{8}{y}=2 \\
\tfrac{6}{x}+\tfrac{4}{y}=3 \\
\end{matrix} \right.$
15. $\left\{ \,\begin{matrix}
\tfrac{6}{x}-\tfrac{2}{y}=6 \\
\tfrac{2}{x}+\tfrac{1}{y}=7 \\
\end{matrix} \right.$ 16. $\left\{ \,\begin{matrix}
\tfrac{2}{x}-\tfrac{1}{y}=3 \\
\tfrac{3}{x}+\tfrac{2}{y}=8 \\
\end{matrix} \right.$
17. $\left\{ \,\begin{matrix}
\tfrac{6}{x}+\tfrac{5}{y}=2 \\
\tfrac{4}{x}+\tfrac{1}{y}=6 \\
\end{matrix} \right.$ 18. $\left\{ \,\begin{matrix}
\tfrac{1}{x}+\tfrac{1}{y}=\tfrac{1}{6} \\
\tfrac{2}{x}+\tfrac{3}{y}=0 \\
\end{matrix} \right.$
19. $\left\{ \,\begin{matrix}
\tfrac{3}{x}+\tfrac{4}{y}=1 \\
\tfrac{4}{x}+\tfrac{5}{y}=2 \\
\end{matrix} \right.$ 20. $\left\{ \,\begin{matrix}
\tfrac{3}{x}+\tfrac{4}{y}=5 \\
\tfrac{4}{x}+\tfrac{5}{y}=6 \\
\end{matrix} \right.$