Lesson 9 · from §88 of the 1910 text
Systems in Three and Four Variables
Systems in three variables
Consider the equations \begin{aligned} m + n + p &= 6 & (1)\\ 2m + 3n + 4p &= 16 & (2)\\ 3m + 4n + 5p &= 22 & (3)\\ m + 2n + 3p &= 10 & (4)\\ 6m + 9n + 12p &= 48 & (5) \end{aligned} (3) is (1) plus (2); (4) is (2) minus (1); (5) is (2) times 3. So (3), (4), and (5) are derived equations, while (1) and (2) are independent: neither can be derived from the other.
A system of three independent equations of the first degree in three variables, no two being incompatible, has one set of roots and only one.
Example. Solve \begin{cases} m + 6n - 5p = 23 & (1) \\ 3m - 8n + 4p = -1 & (2) \\ 7m - 10n + 10p = 0 & (3) \end{cases}
Eliminate p between (1) and (2): (1)\cdot 4 + (2)\cdot 5 gives 19m - 16n = 87. (6)
Eliminate p again, between (2) and (3): (2)\cdot 5 - (3)\cdot 2 gives m - 20n = -5. (9)
Now (6) and (9) have the same two variables. (6) - (9)\cdot 19: 364n = 182, so n = \frac12. Then from (9), m - 10 = -5 and m = 5. From (1), 5 + 3 - 5p = 23, so p = -3.
Check: 5 + 3 + 15 = 23; 15 - 4 - 12 = -1; 35 - 5 - 30 = 0. ✓
Rule. Decide which variable is most easily eliminated. Using any two equations, eliminate it. With one of those equations and the third, eliminate the same variable again. Solve the two resulting equations in two variables. Substitute the two values in the simplest original equation and solve for the third.
Four or more independent equations in three variables have no common set of roots. In general n + 1 independent linear equations in n variables have no set of roots; n such equations (no two incompatible) have one; n - 1 have an unlimited number. Four equations in four variables are solved by eliminating the same variable three times (first and second, first and third, first and fourth) and solving the three equations that result.
Note. With three axes at right angles, an equation in x, y, z can be pictured as a flat surface, a plane. Two planes may meet in a line; three may meet in a single point, and solving a system of three equations finds that point. Space has only three dimensions, so this picture stops at three variables.
How to type your answers
Type each value in its own box: a whole number, a fraction like -17/5, or a mixed number like 4 1/2.
- If the system is inconsistent (no set of roots), type
nonein every box. - If it has an unlimited number of sets of roots, type
anyin the box for the first letter and give the other letter(s) in terms of it.
Try one:
Quick check
Solve \begin{cases} x + y + z = 6 \\[2.5mm] x - y + z = 2 \\[2.5mm] x + y - z = 0 \end{cases}
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 system.
\begin{cases} m + n - 2p = 13 \\[2.5mm] m - 3n - p = -3 \\[2.5mm] m - n + 4p = -17 \end{cases}
\begin{cases} x + y + 3z = \dfrac72 \\[2.5mm] x - 2y + 4z = 7 \\[2.5mm] 2x - 11y - 24z = 5 \end{cases}
\begin{cases} x + y + z = -1 \\[2.5mm] 3x - y - 5z = 13 \\[2.5mm] 5x + 3y + 2z = 1 \end{cases}
\begin{cases} 2h + 3k - 4l = -26 \\[2.5mm] 3h - k + 27l = 87\tfrac{1}{2} \\[2.5mm] h + 5k + 33l = 74\tfrac{1}{2} \end{cases}
\begin{cases} 2m + 3n - 4p = -3 \\[2.5mm] m + n + 3p = -9 \\[2.5mm] m + 2n - 7p = 6 \end{cases}
\begin{cases} x + 8y + 5z = 1 \\[2.5mm] 3x + 10z + 4y = -5 \\[2.5mm] x + 4z = 0 \end{cases}
\begin{cases} 2h - 3l + 4k - 2 = 0 \\[2.5mm] 3h - 3l - 15 = 0 \\[2.5mm] 7h - 4k - 31 = 0 \end{cases}
\begin{cases} 4r - 10s = 5 \\[2.5mm] 6r - t = 3 \\[2.5mm] 5s + 2t = -\dfrac32 \end{cases}
\begin{cases} 2a_1 - 3a_2 = 4 \\[2.5mm] 3a_1 + a_3 = 5 \\[2.5mm] a_2 - 2a_3 = 2 \end{cases}
\begin{cases} 3r_1 + 5r_2 = 74 \\[2.5mm] r_1 - 2r_3 = -16 \\[2.5mm] 7r_3 - 4r_2 = 44 \end{cases}
\begin{cases} \dfrac1m + \dfrac1n - \dfrac1p = 1 \\[2.5mm] \dfrac1m + \dfrac1n + \dfrac1p = \dfrac23 \\[2.5mm] \dfrac1m - \dfrac1n + \dfrac1p = 0 \end{cases}
\begin{cases} \dfrac2m - \dfrac3n + \dfrac{10}{p} = -3 \\[2.5mm] \dfrac4m + \dfrac5p + \dfrac6n = 15 \\[2.5mm] \dfrac1m - \dfrac1n + \dfrac5p = -\dfrac12 \end{cases}
\begin{cases} \dfrac1A + \dfrac1B = 2 \\[2.5mm] \dfrac1A + \dfrac1C = 3 \\[2.5mm] \dfrac1B + \dfrac1C = 4 \end{cases}
\begin{cases} r + s + t + u = 2.8 \\[2.5mm] r - s + t - u = 7.2 \\[2.5mm] r + 2s + 3t - 5u = 7 \\[2.5mm] r + s - 8t + u = -1.7 \end{cases}
\begin{cases} \dfrac2x + \dfrac3y = 26 \\[2.5mm] \dfrac4y - \dfrac{10}{z} = 3 \\[2.5mm] \dfrac1x + \dfrac5z = \dfrac{25}{2} \end{cases}