Step by step
The question
Question: solve x² + y² = 13 and y = x − 1.
We call them ① and ②.
We want the points (x, y) that make both true.
① is a circle, and ② is a line.
Not both lines, so this is called a nonlinear system.
1 Put ② into ①
In ②, y is already alone: y = x − 1.
So in ①, we write (x − 1) in place of y.
The brackets keep it together.
2 Open the brackets
(x − 1) squared is (x − 1) times (x − 1).
Each part times each part, then like terms: x² − 2x + 1.
Careful!
Squaring a bracket is not squaring each part.
(x − 1) squared has a middle part too: −2x.
So we get x² + x² − 2x + 1 = 13.
Like terms together: x² and x² make 2x².
We want 0 on one side.
We take away 13 on both sides.
Now we have 2x² − 2x − 12 = 0.
Every number can be divided by 2.
We divide both sides by 2.
0 divided by 2 is still 0.
3 Factor
We need two numbers.
They multiply to −6, and they add to −1.
They multiply to a minus number, so one number is plus and one is minus.
They add to a minus number, so the bigger one is minus.
We list the pairs that make 6, smallest first.
We take the smaller from the bigger.
We need 1.
2 and 3 give 1.
Yes!
With the signs: 2 and −3.
So it is (x + 2)(x − 3) = 0.
4 One bracket is 0
Two numbers multiply to 0.
That only happens if one of them is 0.
So one bracket must be 0.
We make each bracket 0.
Two small equations.
In x + 2 = 0, we take away 2 on both sides.
In x − 3 = 0, we add 3 on both sides.
So x = −2 or x = 3.
5 Find y
Each x goes into the line ②: x = −2 gives y = −3.
And x = 3 gives y = 2.
So the two points are (−2, −3) and (3, 2).
6 Check
Let's check with a picture.
We draw both on a grid.
① is a circle around (0, 0).
② is a line.
They meet at (−2, −3) and (3, 2): our answers.
Let's check each point in both equations.
First (−2, −3) in ①: x² + y² is 13.
True!
In ②: −3 again.
It works!
Now (3, 2) in ①: x² + y² is 13.
True!
In ②: 2 again.
It works!
Let's look back.
First: put ② into ①.
Then: open the brackets.
Then: factor.
Then: one bracket is 0.
Then: find y.
Last: check.