We want to write a system of linear equations in which the solution is a point, (3,-1). To do so, we need to find lines that go through this point. Let's start with Equation 1. Let the slope of this line be 3.
y= 3x+b
Now we will substitute ( 3, -1) into the equation to find y-intercept b.
The value of b is -10. Let's complete the first equation.
y=3x+( -10) ⇒ y=3x-10
Now let's graph this line in a coordinate plane. We will do this by creating a table of values.
x
y=3x-10
y
(x,y)
0
y=3( 0)-10
-10
( 0, -10)
2
y=3( 2)-10
-4
( 2, -4)
4
y=3( 4)-10
2
( 4, 2)
Let's plot the points and connect them with a line.
Now let's find the second line that passes through (3,-1). To do this, let's choose a different slope for the second line. Let the slope be -1.
y= -1x+b ⇒ y=- x+b
Now we will substitute the point ( 3, -1) into the above equation to find the y-intercept b.
The value of b is 2. Now we are ready to complete the second equation and write our example system.
y=3x-10 & (I) y=- x+ 2 & (II)
Let's add the second line to our graph. We will again use a table of values to find points that lie on the line.
x
y=- x+2
y
(x,y)
-1
y=-( -1)+2
3
( -1, 3)
0
y=-( 0)+2
2
( 0, 2)
1
y=- 1+2
1
( 1, 1)
Let's plot the points and connect them with a line.
Notice that this is only an example solution, as we may think of infinitely many other systems that will have a solution of (3,-1).
