**Linear equations in two unknowns
**
A

**linear equation in two unknowns** is an equation that can be written in the form

$\displaystyle ax + by = c$

with $a$, $b$ and $c$ being real numbers, and with $a$ and $b$ not both zero.

The graph of a linear equation $ax + by = c$ is a straight line: if $b \ne 0$ we can solve the equation for $y$ to get the equation of a straight line in standard form (see the

Chapter summary material on linear functions.). If $b= 0$, then we can solve the equation for $x$ to obtain $x = $ a constant, which gives a vertical line.

**Examples: Linear equations in two unknowns**
The following are linear equations in two unknowns:

$3x + 2y = 4 \qquad$ | |

$x-y = 5$ | |

$2x=0$ | |

The following are

*not* linear equations in two unknowns:

$3x^2 + 2y = 4 \qquad$ | |

$xy = 5$ | |

$0x+0y=1$ | |

To graph the linear equation $3x + 2y = 4$, first solve for $y$ to obtain

so that the slope is $m = -\dfrac{3}{2}$ and the $y$-intercept is $b = 2$. We then obtain the following line by following the instructions in the

summary of linear functions:

**Practice:**

**Practice:**

**Solutions of a linear equation in two unknowns
**
A

**particular solution** of the linear equation $ax + by = c$ is a specific pair of numbers $(x, y)$ that satisfy the equation. If $a$ and $b$ are not both zero, the linear equation $ax + by = c$ has infinitely many such solutions; these solutions are the points on the graph of the equation.

The

**general solution parameterized by $\bold{x}$** of the linear equation $ax + by = c$ has the form $(x, f(x))$, where we solve the given equation for $y$ as a function of $f(x)$ of $x$ if possible (that is, if $b \neq 0$, so that $y$ appears in the equation; otherwise, the general solution cannot be parameterized by $x$). Substituting any number for $x$ then results in a particular solution.

Similarly, the

**general solution parameterized by $\bold{y}$** has the form $(g(y), y)$, where we solve the given equation for $x = g(y)$ if possible. Substituting any number for $y$ then results in a particular solution.

**Examples: Solutions of a linear equation in two unknowns**
The linear equation $3x + 2y = 4$ has a solution $x = 2,\ y = -1,$ or $(x, y) = (2, -1),$ because substituting $x = 2$ and $y = -1$ in the equation makes it a true statement:

There are in fact infinitely many solutions to this equation. Some additional ones are

$(-1,\frac{7}{2}),\ \ (0, 2),\ \ (1, \frac{1}{2}),\ \ (2, -1),\ \ (3, -\frac{5}{2}),\ \ (4, -4),\ \ \cdots \ \ \left(x, \frac{1}{2}(4-3x)\right), \cdots$

The solutions $(-1,\frac{7}{2}),\ \ (0, 2),\ \ (1, \frac{1}{2}),\ \ (2, -1),\ \ (3, -\frac{5}{2}),$ and $(4, -4)$ are examples of particular solutions of $3x + 2y = 4$.

We obtained the $y$-coordinate of the last solution shown by solving the equation $3x + 2y = 4$ for $y$. This is the general solution paramaterized by $x$.

Graphically, the solutions are just the points on the line $3x + 2y = 4:$

**Practice:**

**Solution of a system of two linear equations in two unknowns: Graphical viewpoint
**
A

**solution** of a system of two or more linear equations in $x$ and $y$ is a solution that satisfies all of the equations in the system.

We can solve such a system of equations either

**graphically,** by drawing the graphs and finding where they intersect, or

**algebraically,** by, for example, combining the equations in order to eliminate all but one variable and then solving for that variable.

A system of two linear equations in two unknowns has either:

**A single (or unique) solution.** This happens when the lines corresponding to the two equations are not parallel, so that they intersect at a single point.**No solution.** This happens when the two lines are parallel and different.**An infinite number of solutions.** This occurs when the two equations represent the same straight line. In this case, we represent the solutions by choosing one variable arbitrarily, and solving for the other.

**Examples: Solution of a system of two linear equations in two unknowns: Graphical viewpoint**
**1.** The system of equations

has the unique solution $(-1,3)$ as this is the unique point where the graphs intersect:

**2.** The system of equations

has no solution $(-1,3)$ as the graphs never intersect (the lines are parallel):

**3.** The system of equations

has infinitely many solutions as the graphs are the same line. Two particular solutions are shown on the graph; the general solution is $(x, x+1)$ (solve eiher equation for $y$):

**Practice: Solving graphically:**

**Algebraic solution of a system of two linear equations in two unknowns
**
**Solving algebraically by elimination**
**Note: **To get the elimination method to work nicely we should first ensure that the given system has no decimals or fractions If your system does, then

*before you do anything,* multiply each equation by a suitable integer to get rid of them.

To

**eliminate an unknown** means to obtain an equation in which that unknown does not occur. To eliminate an unknown that occurs in both equations, multiply one or both equations by suitable nonzero constant (if necessary) so that adding the resulting equations cancels out ("eliminates") that unknown. Then solve that equation for the remaining variable. Repeat by eliminating the other variable.

**Examples: Algebraic solution of a system of two linear equations in two unknowns**
Consider the system of equations

We eliminate $x$. Notice that just adding the equations does not eliminate the $x,$ but results in $3x + y = -3.$ However, if we multiply the first equation by $-2$ we get an equivalent system:

$-2x + 4y =16$

$2x + 3y=5$.

*Now* if we add them, the $x$s cancel, and we get

$7y = 21 \implies y = \dfrac{21}{7} = 3.$

We have solved for $y.$

Now that we know the value of $y,$ we

*could* obtain $x$ by substituting this value in one of the original equations, but instead, we can eliminate $y$ starting again with the original system:

Original system:

To eliminate $y,$ we want to make the coefficients of $y$ the same numerically but with opposite sign. We can accomplish this without using fractions by multiplying the first equation by $3$ and the second by $2$:

Now if we add them, the $y$s cancel, and we get:

$7x = -14 \implies y = \dfrac{-14}{7} = -2.$

Solution: $(x, y) = (-2, 3)$

To see what happens when you use the elimination method in a redundant or inconsistent system, consult the

online tutorial.

**Practice: Solving by elimination:**