Solve x^2 + y^2 = 25 and y = x + 1 Easy

The question

Question: solve x² + y² = 25 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 = 25.

Like terms together: x² and x² make 2x².

We want 0 on one side.

We take away 25 on both sides.

Now we have 2x² + 2x − 24 = 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 −12, and they add to 1.

They multiply to a minus number, so one number is plus and one is minus.

They add to a plus number, so the bigger one is plus.

We list the pairs that make 12, smallest first.

We take the smaller from the bigger.

We need 1.

3 and 4 give 1.

Yes!

With the signs: −3 and 4.

So it is (x − 3)(x + 4) = 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 − 3 = 0, we add 3 on both sides.

In x + 4 = 0, we take away 4 on both sides.

So x = 3 or x = −4.

5 Find y

Each x goes into the line ②: x = −4 gives y = −3.

And x = 3 gives y = 4.

So the two points are (−4, −3) and (3, 4).

6 Check

Let's check with a picture.

We draw both on a grid.

① is a circle around (0, 0) with radius 5.

② is a line.

They meet at (−4, −3) and (3, 4): our answers.

Let's check each point in both equations.

First (−4, −3) in ①: x² + y² is 25.

True!

In ②: −3 again.

It works!

Now (3, 4) in ①: x² + y² is 25.

True!

In ②: 4 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.

▶ Watch it solved

Solve y = x^2 + 1 and y = 2x Medium

The question

Question: solve y = x² + 1 and y = 2x.

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 = 2x.

Both ① and ② say what y is.

So the two right sides are equal.

2 Get 0 on one side

We take 2x away from both sides.

Careful!

The answers are points, not just x numbers.

After x, we must find each y.

Now we have x² − 2x + 1 = 0.

3 Factor

We need two numbers.

They multiply to 1, and they add to −2.

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 1, smallest first.

We add each pair.

We need 2.

1 and 1 give 2.

Yes!

With the signs: −1 and −1.

So it is (x − 1)(x − 1) = 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.

The two brackets are the same.

In x − 1 = 0, we add 1 on both sides.

Both brackets give the same answer.

So there is only one answer: x = 1.

5 Find y

Each x goes into the line ②: x = 1 gives y = 2.

Only one x, so only one point: (1, 2).

The line just touches the parabola.

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, 2): our answer.

Let's check each point in both equations.

First (1, 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.

▶ Watch it solved

Solve x^2 + y^2 = 25 and y = 2x - 5 Harder

The question

Question: solve x² + y² = 25 and y = 2x − 5.

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 = 2x − 5.

So in ①, 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 = 25.

Like terms together: x² and 4x² make 5x².

We want 0 on one side.

We take away 25 on both sides.

Now we have 5x² − 20x = 0.

Every number can be divided by 5.

We divide both sides by 5.

0 divided by 5 is still 0.

3 Factor

There is no plain number.

Both parts have x in them.

So we take x out.

So it is x(x − 4) = 0.

4 One bracket is 0

x times the bracket is 0.

That only happens if one of them is 0.

We make each part 0.

Two small equations.

x = 0 is done already.

In x − 4 = 0, we add 4 on both sides.

So x = 0 or x = 4.

5 Find y

Each x goes into the line ②: x = 0 gives y = −5.

And x = 4 gives y = 3.

So the two points are (0, −5) and (4, 3).

6 Check

Let's check with a picture.

We draw both on a grid.

① is a circle around (0, 0) with radius 5.

② is a line.

They meet at (0, −5) and (4, 3): our answers.

Let's check each point in both equations.

First (0, −5) in ①: x² + y² is 25.

True!

In ②: −5 again.

It works!

Now (4, 3) in ①: x² + y² is 25.

True!

In ②: 3 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.

▶ Watch it solved