Lesson 1 · from §105, 106 of the 1910 text
One Equation Linear, One Quadratic
Systems of the second degree
The general equation of the second degree in two variables is
ax^2 + by^2 + cxy + dx + ey + f = 0.
To solve a pair of such equations requires the solution of an equation of the fourth degree. Even the far simpler system \begin{cases} x^2 + y = 5 \\ y^2 + x = 3 \end{cases} requires the solution of a biquadratic equation. In fact, only a limited number of systems of equations of the second degree in two variables are solvable by quadratics.
You can always solve graphically, for real roots, any system of quadratic equations with numerical coefficients, as in Chapter XXIV. The algebraic solution of such systems in general must wait for further study of algebra.
One equation linear, the other quadratic
Every system of equations in two variables in which one equation is linear and the other quadratic can be solved by the method of substitution.
Example. Solve the system
\begin{cases} x^2 + y^2 = 5 & (1) \\ x - y = 1 & (2) \end{cases}
| Step | Result | |
|---|---|---|
| Solving (2) for x, | x = 1 + y | (3) |
| Substituting 1 + y for x in (1), | (1 + y)^2 + y^2 = 5 | (4) |
| From (4), | y^2 + y - 2 = 0 | (5) |
Solving (5) by formula,
y = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} = 1 \text{ or } -2.
Substituting 1 for y in (3), x = 1 + 1 = 2. Substituting -2 for y in (3), x = 1 - 2 = -1.
Therefore \begin{cases} x = 2 \\ y = 1 \end{cases} and \begin{cases} x = -1 \\ y = -2 \end{cases} are the two sets of roots.
Check: substituting 2 for x and 1 for y: in (1), 4 + 1 = 5; in (2), 2 - 1 = 1. Substituting -1 for x and -2 for y: in (1), 1 + 4 = 5; in (2), -1 + 2 = 1.
Notice that the roots must be paired: the value 2 of x goes with the value 1 of y, not with -2. Always find the second letter by substituting each value of the first in the linear equation, which gives exactly one partner for each.
The figure shows the circle and the line of the system meeting at the two points A\,(2, 1) and B\,(-1, -2). You can check this in the calculator's Graph tab.
Typing your answers. Give every set of roots as a pair in parentheses, in the order the exercise names, for example (3, 2), (-2, -3). The pairs may come in any order. You may also write x = 3, y = 2; x = -2, y = -3. Radicals are typed sqrt(3), so 2\sqrt3 is 2sqrt(3).
Quick check
Solve the system x^2 + y^2 = 25, x + y = 7. Give both sets of roots.
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 the following systems, pair results, and check each set of roots.
\begin{cases} x + y = 8 \\ x^{2} + y^{2} = 34 \end{cases}
\begin{cases} 2 m + n = 14 \\ m^{2} + 3 m n = 49 \end{cases} Give each set as (m, n).
\begin{cases} 2m^2 + n^2 = 22 \\ m - n\sqrt5 = 0 \end{cases} Give each set as (m, n).
\begin{cases} 4 s + 5 t = 6 \\ s t = -2 \end{cases} Give each set as (s, t).
\begin{cases} x y + 12 = 0 \\ 4 x - 3 y = 30 \end{cases}
\begin{cases} x\sqrt3 + 5y\sqrt3 = -72 \\ xy = -15 \end{cases}
\begin{cases} 3R_1 + 2R_2 = 5 \\ R_1R_2 - 6R_1 = -3 \end{cases} Give each set as (R_1, R_2).
\begin{cases} x y + y^{2} - 5 = 0 \\ x y + 20 = 0 \end{cases}
\begin{cases} h^{2} + k^{2} + k = 10 \\ h + k + 1 = 0 \end{cases} Give each set as (h, k).
\begin{cases} m^{2} + 3 m n + n^{2} = 22 \\ 2 m = n \end{cases} Give each set as (m, n).
\begin{cases} x^{2} + y^{2} + 2 x + 3 y = 10 \\ x - 5 = y \end{cases}
\begin{cases} y - x\sqrt8 = 0 \\ y^2 + x^3 = 9x \end{cases}