Lesson 6 · from §81 of the 1910 text
Solving Two Equations Graphically
Solving two equations graphically
Graph x + 2y = 8 and 2x - y = 6 on the same axes. The lines cross at the point (4, 2):
Since (4, 2) is on both graphs, it satisfies both equations: 4 + 2\cdot 2 = 8 and 2\cdot 4 - 2 = 6. ✓
So the graphical solution of two linear equations is to plot both graphs and read the x and y of the point where they cross. Since two straight lines can cross in only one point, only one pair of values of x and y satisfies a pair of linear equations.
Lines that never meet
Graph x + y = 4 and x + y = 6. The lines are parallel:
No pair of numbers can add to both 4 and 6, so the pair has no solution. Such equations are called inconsistent. Type none for these.
Graphical answers can be off by a little. Always verify by substituting into both equations.
Try one:
Quick check
Solve graphically: x + y = 5, x - y = 1.
From the history of algebra
René Descartes (1596–1650), the French philosopher, made one of the most important discoveries in mathematics: that algebraic equations can be pictured geometrically. Never strong in health, he nevertheless served in several military campaigns as a young man, and it is said that during a weary winter in camp in Austria he first conceived these ideas. He saw that a letter (a coördinate) could stand for a negative as well as a positive number, which helped convince mathematicians that negative numbers are real numbers and useful. He also introduced our way of writing exponents, and the custom of using the last letters of the alphabet for unknowns and the first letters for known numbers. He would have written x^3 - 8x + 16 = 40 as x^3 * - 8x + 16 \propto 40: the asterisk showed a missing power, and though the sign = was already in use, he did not accept it.
Exercises
How to type math
Use ^ for powers (a^2), / for fractions ((a+b)/2), sqrt(x) for square roots, and cbrt(x) for cube roots. You can leave out the multiplication sign: 3ab means 3 \cdot a \cdot b. Mixed numbers like 4 1/2 work too. The preview under each box shows how your answer was read.
Solve each pair graphically: graph both equations on paper, read off the point where the lines cross, and verify by substituting. If the lines never meet, type none in both boxes.
\begin{cases} x + y = 6 \\ x + 2 y = 8 \end{cases}
\begin{cases} x - y = 4 \\ 2 x + y = 5 \end{cases}
\begin{cases} x + y = 6 \\ x - y = 4 \end{cases}
\begin{cases} x + y + 8 = 0 \\ x - y - 2 = 0 \end{cases}
\begin{cases} 2 x + y + 2 = 0 \\ 2 x + 2 y + 7 = 0 \end{cases}
\begin{cases} 4 x - y = -1 \\ x - 2 y = 5 \end{cases}
\begin{cases} 2 x + 5 y = 10 \\ x + 4 y = 8 \end{cases}
\begin{cases} x + 6 y = 12 \\ 2 x + 3 y = 12 \end{cases}
\begin{cases} x + 2 y = 10 \\ x = 3 y \end{cases}
\begin{cases} x + y + 6 = 0 \\ x = y \end{cases}
\begin{cases} x + y = 4 \\ x + y = 6 \end{cases}
\begin{cases} x + 2 y = 4 \\ x + 2 y = 10 \end{cases}
\begin{cases} 2 x - y = 6 \\ 4 x - 2 y = 8 \end{cases}
\begin{cases} x + y = 6 \\ x - 4 = 0 \end{cases}
\begin{cases} x - y = 7 \\ y + 2 = 0 \end{cases}
\begin{cases} x = 2 y \\ x - 4 = 0 \end{cases}
\begin{cases} 3 x - y = 0 \\ 3 x + y = 0 \end{cases}
\begin{cases} x - 3 = 0 \\ y + 2 = 0 \end{cases}