Step by step
The question
Question: solve y = 2x + 5 and x² + y² = 10.
We call them ① and ②.
We want the points (x, y) that make both true.
② is a circle, and ① is a line.
Not both lines, so this is called a nonlinear system.
1 Put ① into ②
In ①, y is already alone: y = 2x + 5.
So in ②, we write (2x + 5) in place of y.
The brackets keep it together.
2 Open the brackets
(2x + 5) squared is (2x + 5) times (2x + 5).
Each part times each part, then like terms: 4x² + 20x + 25.
Careful!
Squaring a bracket is not squaring each part.
(2x + 5) squared has a middle part too: 20x.
So we get x² + 4x² + 20x + 25 = 10.
Like terms together: x² and 4x² make 5x².
We want 0 on one side.
We take away 10 on both sides.
Now we have 5x² + 20x + 15 = 0.
Every number can be divided by 5.
We divide both sides by 5.
0 divided by 5 is still 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 plus number, so both are plus.
We list the pairs that make 3, smallest first.
We add each pair.
We need 4.
1 and 3 give 4.
Yes!
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 take away 1 on both sides.
In x + 3 = 0, we take away 3 on both sides.
So x = −1 or x = −3.
5 Find y
Each x goes into the line ①: x = −3 gives y = −1.
And x = −1 gives y = 3.
So the two points are (−3, −1) and (−1, 3).
6 Check
Let's check with a picture.
We draw both on a grid.
② is a circle around (0, 0).
① is a line.
They meet at (−3, −1) and (−1, 3): our answers.
Let's check each point in both equations.
First (−3, −1) in ②: x² + y² is 10.
True!
In ①: −1 again.
It works!
Now (−1, 3) in ②: x² + y² is 10.
True!
In ①: 3 again.
It works!
Let's look back.
First: put ① into ②.
Then: open the brackets.
Then: factor.
Then: one bracket is 0.
Then: find y.
Last: check.