# Algebra

## 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.

#### Mini-Lecture.

Below is a mini lecture about linear systems.