∑ Old Books, New Math

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 none in every box.
  • If it has an unlimited number of sets of roots, type any in the box for the first letter and give the other letter(s) in terms of it.

Try one:

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Quick check

Solve \begin{cases} x + y + z = 6 \\[2.5mm] x - y + z = 2 \\[2.5mm] x + y - z = 0 \end{cases}

Exercises

0 of 15 solved
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.

1

\begin{cases} m + n - 2p = 13 \\[2.5mm] m - 3n - p = -3 \\[2.5mm] m - n + 4p = -17 \end{cases}

2

\begin{cases} x + y + 3z = \dfrac72 \\[2.5mm] x - 2y + 4z = 7 \\[2.5mm] 2x - 11y - 24z = 5 \end{cases}

3

\begin{cases} x + y + z = -1 \\[2.5mm] 3x - y - 5z = 13 \\[2.5mm] 5x + 3y + 2z = 1 \end{cases}

4

\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}

5

\begin{cases} 2m + 3n - 4p = -3 \\[2.5mm] m + n + 3p = -9 \\[2.5mm] m + 2n - 7p = 6 \end{cases}

6

\begin{cases} x + 8y + 5z = 1 \\[2.5mm] 3x + 10z + 4y = -5 \\[2.5mm] x + 4z = 0 \end{cases}

7

\begin{cases} 2h - 3l + 4k - 2 = 0 \\[2.5mm] 3h - 3l - 15 = 0 \\[2.5mm] 7h - 4k - 31 = 0 \end{cases}

8

\begin{cases} 4r - 10s = 5 \\[2.5mm] 6r - t = 3 \\[2.5mm] 5s + 2t = -\dfrac32 \end{cases}

9

\begin{cases} 2a_1 - 3a_2 = 4 \\[2.5mm] 3a_1 + a_3 = 5 \\[2.5mm] a_2 - 2a_3 = 2 \end{cases}

10

\begin{cases} 3r_1 + 5r_2 = 74 \\[2.5mm] r_1 - 2r_3 = -16 \\[2.5mm] 7r_3 - 4r_2 = 44 \end{cases}

11

\begin{cases} \dfrac1m + \dfrac1n - \dfrac1p = 1 \\[2.5mm] \dfrac1m + \dfrac1n + \dfrac1p = \dfrac23 \\[2.5mm] \dfrac1m - \dfrac1n + \dfrac1p = 0 \end{cases}

12

\begin{cases} \dfrac2m - \dfrac3n + \dfrac{10}{p} = -3 \\[2.5mm] \dfrac4m + \dfrac5p + \dfrac6n = 15 \\[2.5mm] \dfrac1m - \dfrac1n + \dfrac5p = -\dfrac12 \end{cases}

13

\begin{cases} \dfrac1A + \dfrac1B = 2 \\[2.5mm] \dfrac1A + \dfrac1C = 3 \\[2.5mm] \dfrac1B + \dfrac1C = 4 \end{cases}

14

\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}

15

\begin{cases} \dfrac2x + \dfrac3y = 26 \\[2.5mm] \dfrac4y - \dfrac{10}{z} = 3 \\[2.5mm] \dfrac1x + \dfrac5z = \dfrac{25}{2} \end{cases}