Step by step
The question
Question: solve y = x² − 3x + 2 and y = x − 1.
We call them ① and ②.
We want the points (x, y) that make both true.
① is a parabola, 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.
Both ① and ② say what y is.
So the two right sides are equal.
2 Get 0 on one side
We take x − 1 away from both sides.
Careful!
The answers are points, not just x numbers.
After x, we must find each y.
Now we have x² − 4x + 3 = 0.
3 Factor
We need two numbers.
They multiply to 3, and they add to −4.
They multiply to a plus number, so the two numbers have the same sign.
They add to a minus number, so both are minus.
We list the pairs that make 3, smallest first.
We add each pair.
We need 4.
1 and 3 give 4.
Yes!
With the signs: −1 and −3.
So it is (x − 1)(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 − 1 = 0, we add 1 on both sides.
In x − 3 = 0, we add 3 on both sides.
So x = 1 or x = 3.
5 Find y
Each x goes into the line ②: x = 1 gives y = 0.
And x = 3 gives y = 2.
So the two points are (1, 0) and (3, 2).
6 Check
Let's check with a picture.
We draw both on a grid.
① is a parabola.
② is a line.
They meet at (1, 0) and (3, 2): our answers.
Let's check each point in both equations.
First (1, 0) in ①: the right side is 0, the same as y.
In ②: 0 again.
It works!
Now (3, 2) in ①: the right side is 2, the same as y.
In ②: 2 again.
It works!
Let's look back.
First: put ② into ①.
Then: get 0 on one side.
Then: factor.
Then: one bracket is 0.
Then: find y.
Last: check.