Department of Mathematics


Worked Examples

Systems of Equations

When solving a linear system Each equation is a sum of terms, and each term is either a constant, or the product of a constant and (the first power of) a single variable. of two equations and two unknowns, it is equivalent to trying to find a point of intersection between two lines in a Cartesian plane. Therefore there are three possibilities:
1. Exactly one solution (i.e. the lines intersect)
2. Infinitely many solutions (i.e. the lines coincide, meaning one line is a multiple of the other)
3. No solution (i.e. the lines are parallel and consequently do not intersect)
There are several ways to solve such systems. Below are descriptions of three methods.

Method 1: Substitution

Step 1: Choose one equation to isolate a variable (i.e solve for one variable in terms of the other).

Consider the following system: $$\begin{array}{ll} y-\frac{2}{3}x=-\frac{1}{3} & (E_1) \\ 3x+4y=5 & (E_2) \end{array}$$ Here, $(E_1)$ and $(E_2)$ are used to denote Equation 1 and Equation 2, respectively.
From $(E_1)$, we have: $ \qquad y=\frac{2}{3}x-\frac{1}{3}$

Step 2: Substitute the isolated expression from Step 1 into the remaining equation. (This will lead to having one equation and one unknown to solve for!)

Substitute $\; y=\frac{2}{3}x-\frac{1}{3}$ into $(E_2)$, and solve for $x$: $$\begin{align*} 3x+4\left( \frac{2}{3}x-\frac{1}{3} \right) &= 5 \\ \frac{9}{3}x+\frac{8}{3}x-\frac{4}{3} &= 5 \\ 9x+8x-4 &= 15 \\ 17x-4 &= 15 \\ 17x &= 19 \\ x &= \frac{19}{17} \end{align*}$$ Step 3: Back-substitute (i.e. substitute the value found in Step 2 back into the isolated expression from Step 1!)

Recall that $y=\frac{2}{3}x-\frac{1}{3}$. So, we substitute the $x$-value found to find $y$: $$\begin{align*} y &= \frac{2}{3}\left(\frac{19}{17}\right) - \frac{1}{3} \\ y &= \frac{38}{51} - \frac{1}{3} \\ y &= \frac{38-17}{51} \\ y &= \frac{21}{51} = \frac{7}{17} \end{align*}$$ So, this system has exactly one solution: $ \quad \left(\frac{19}{17}, \frac{7}{17} \right)$

Method 2: Elimination

Step 1: Identify which variable you want to eliminate, and then multiply one equation by a number that will cause that variable to cancel out when added to the other equation.

Consider the following system: $$\begin{array}{ll} 3x-y=2 & (E_1) \\ -6x+2y = -4 & (E_2) \end{array}$$ Suppose we want to eliminate the $x$, which can be done by multiplying $(E_1)$ by $2$ (then adding the result to $(E_2)$). Our new, equivalent system becomes: $$\begin{align*} 6x-2y&=4 \\ -6x+2y&=-4 \end{align*}$$ Step 2: Add the equations in order to cancel the desired variable and solve for the one remaining. $$\frac{\begin{align*} 6x-2y&=4 \\ -6x+2y&=-4 \end{align*}}{0x+0y=0} $$ Because we got $0 = 0$ (or in general: $a=a$) as our new equation, we can conclude that the original equations must coincide, putting us in the case of infinitely many solutions.

So, we let one of the variables, such as $x$, be a parameter: $$\textrm{Let } x=t, \textrm{ where }t \in \mathbb{R} $$ Then, we solve for the remaining variable in terms of that parameter:
From $(E_1)$, we have that: $$ y=3x-2 \Rightarrow y = 3t-2 $$ So, our final solution is $(t, 3t-2)$, where $\; t \in \mathbb{R}$.

Note: If we had gotten an equation involving $y$ (after cancelling $x$ in Step 2), we would have been able to isolate and solve for it. Then, we would have proceeded to Step 3, below.

Step 3: Back-substitute the value found in Step 2 (if any) into $(E_1)$ or $(E_2)$ (whichever is more convenient), and solve for the remaining variable.

Method 3: Equating Equations

Step 1: Choose a variable to isolate in both $(E_1)$ and $(E_2)$.

Consider the following system: $$\begin{array}{ll} y-2x=-6 & (E_1) \\ 4x-2y=10 & (E_2) \end{array}$$ Suppose we isolate $y$ (since it’s already missing a coefficient in $(E_1)$).
From $(E_1)$: $$y=-6+2x$$
From $(E_2)$: $$\begin{align*} 2y &= 4x-10 \\ y &= 2x-5 \end{align*}$$
Step 2: Equate the isolated expressions from $(E_1)$ and $(E_2)$. We can do this because we are looking for a point ofintersection, which means that the $x$- and $y$-values in each line are equal at this point (by definition)! This will again result in one equation and one unknown.

$$\begin{align*}−6 + 2x = 2x − 5 \\ \Rightarrow −6+ \cancel{2x}=\cancel{2x}−5\end{align*}$$ But this implies that $−6 = −5$, which is a contradiction!
Therefore, this system has no solution (i.e. the lines do not intersect).

Note: If we had been able to solve for $x$ we would have proceeded to Step 3, below.

Step 3: Back-substitute into the isolated expressions (found in Step 1) to solve for the remaining variable.


Below is a mini lecture about linear systems.