# How do you write a system of equations with the solution (4,-3)?

Mar 28, 2015

We'll make a linear system (a system of linear equations) whose only solution in $\left(4 , - 3\right)$.
First note that there are several (or many) ways to do this. We'll look at two ways:

Standard Form Linear Equations

A linear equation can be written in several forms. "Standard Form" is $a x + b y = c$ where $a$, $b$ and $c$ are constants (numbers).

We want to make two equations that
(i) have this form,
(ii) do not have all the same solutions (the equations are not equivalent), and
(iii) $\left(4 , - 3\right)$ is a solution to both.

$a x + b y = c$. We want $a$, $b$ and $c$ so that

$a \left(4\right) + b \left(- 3\right) = c$ (This will make (i) and (iii) true.)

Choose $a$, $b$ and $c$ the make the equation true.
How? Choose two of the and find the third.

Example: If we make $a = 1$ and $b = 1$, then because we have
$1 \left(4\right) + 1 \left(- 3\right) = c$ , we can see that we need $c = 1$.

One equation of my system will be $x + y = 1$

Now in order to satisfy (ii) My second equations need to not be a multiple of the first.
If I used $2 x + 2 y = 2$, it would share, not only $\left(4 , - 3\right)$, but every solution.

T make sure that we do not get a multiple, my second choice for $a$ and $b$ will not be a constant multiple of the first choice. (Not $a = 3$, and $b = 3$. and not $a = 5$, and $b = 5$. an so on.)

I want to keep this example simple, so I'll keep $a = 1$ and choose a different $b$, then I'll find (calculate) the $c$ I need.

Let's use $a = 1$ and $b = 2$. This makes $1 \left(4\right) + 2 \left(- 3\right) = c$, so $c = 4 - 6 = - 2$

My second equation is $x + 2 x = - 2$

My system is:
$x + y = 1$
$x + 2 x = - 2$

We can check that $\left(4 , - 3\right)$ solves both equations, but the equations are not equivalent. ($\left(1 , 0\right)$ solves the first, but not the second.)

Intersecting Lines

A different way of thinking about the question is much more geometrical.

We want two different lines through the point $\left(4 , - 3\right)$

(i) lines (ii) distinct lines (iii) through the point $\left(4 , - 3\right)$

We'll make sure we have lines.
If the equations of the lines have different slope, then we can be certain that the lines are distinct. (that we really have 2 different lines, not just two equations for the same line.)
So we'll make sure the slopes are different.

There are still several ways to think about how to do this

First Method:
Use slope form or point-slope form for the equation of a line.

$\frac{y - \left(- 3\right)}{x - 4} = m$ or $\left(y - \left(- 3\right)\right) = m \left(x - 4\right)$

Choose two different $m$'s and simplify, if you think you should.

$m = 1$ leads to $y = x - 7$
$m = 2$ leads to $y = 2 x - 11$
$m = - 5$ leads to $y = - 5 x + 17$

Second method:
Use slope intercept form $y = m x + b$ where we want $- 3 = m \left(4\right) + b$

Choose two different $m$'s and find the associated $b$'s

$m = 1$ leads to $y = x - 7$
$m = 2$ leads to $y = 2 x - 11$
$m = - 5$ leads to $y = - 5 x + 17$