Algebra - Simultaneous Equations

Simultaneous Linear Systems in Two and Three Variables

A linear equation in two variables represents a line, while a linear equation in three variables, $Ax + By + Cz = D$, represents a flat plane in three-dimensional space. Solving three simultaneous equations means finding the single point $(x, y, z)$ where all three planes intersect.

To solve a three-variable system, eliminate the same variable twice using two different pairs of equations. This produces a pair of equations containing only two variables, which you can then solve using standard elimination or substitution.

Take the system:

1. $x + y + z = 6$
2. $5x + 3y - 2z = 5$
3. $3x - 7y + z = -8$

Eliminating $z$ first is efficient. Multiply equation (1) by $2$ and add it to equation (2) to get $7x + 5y = 17$. Next, subtract equation (3) from equation (1) to eliminate $z$ again, giving $-2x + 8y = 14$. Solving this two-variable system gives $y = 2$ and $x = 1$. Substituting both values back into $x + y + z = 6$ produces $z = 3$. Always substitute all three answers back into the unused original equations to check your arithmetic.

Simultaneous Equations: One Linear and One Non-Linear

When one equation is linear and the other contains higher powers or product terms like $x^2$, $y^2$, or $xy$, always use the substitution method. Geometrically, this finds the coordinates where a line intersects a curve such as a circle, ellipse, or parabola.

Rearrange the linear equation to make one variable the subject, choosing whichever variable has a coefficient of $\pm 1$ to avoid cumbersome fractions. Then substitute this expression directly into the non-linear equation. Never try to substitute from the non-linear equation into the linear one.

For instance, given $x + 4y = 1$ and $2x^2 + 3xy = 35$, isolate $x$ in the linear relation to obtain $x = 1 - 4y$. Substituting this into the non-linear equation gives $2(1 - 4y)^2 + 3(1 - 4y)y = 35$. Expanding and collecting terms leads to the quadratic $20y^2 - 13y - 33 = 0$. Factoring gives $(20y - 33)(y + 1) = 0$, so $y = -1$ or $y = \frac{33}{20}$. Substituting each value back into $x = 1 - 4y$ yields the coordinate pairs $(5, -1)$ and $\left(-\frac{28}{5}, \frac{33}{20}\right)$.