Lesson 6 · from §87 of the 1910 text
Literal Equations in Two Variables
Literal equations in two variables
Systems in which the variables have literal coefficients are solved by the same method of addition or subtraction. Treat a, b, c, d (and these letters with subscripts) as known numbers.
Example. Solve \begin{cases} ax + y = c & (1) \\ x + y = d & (2) \end{cases} for x and y.
(1) - (2): ax - x = c - d, or (a - 1)x = c - d, so x = \frac{c - d}{a - 1}. Then from (2), y = d - \dfrac{c - d}{a - 1} = \dfrac{ad - c}{a - 1}.
Check in (2): x + y = \dfrac{c - d + ad - c}{a - 1} = \dfrac{(a - 1)d}{a - 1} = d. ✓
Collect the terms containing the unknown, factor it out, and divide by its (literal) coefficient. Type your answers with * or juxtaposition for multiplication and / for division; write subscripts as a1 or a_1. Any equivalent form is accepted.
Try one:
Quick check
Solve for x and y: \begin{cases} x + y = a \\[2.5mm] x - y = b \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.
Treat a, b, c, d (and these letters with subscripts) as known numbers; solve for the other letters and check. In Exercises 17–20 solve for x and y.
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