3(x + 2) = 3x + k has infinitely many solutions. Find k. Easy

The question

Question: for which k does 3(x + 2) = 3x + k have infinitely many solutions?

Infinitely many solutions means every number works for x.

1 Open the brackets

The 3 in front of the bracket multiplies every part inside.

So we write 3 times each part.

3 times x is 3x.

3 times 2 is 6.

Careful!

The 3 multiplies every part, the 2 too.

3 times 2 is 6, not 2.

2 Compare the x parts

The x parts: 3x on the left and 3x on the right.

3 Take away the x part

We take away 3x on both sides.

They stay equal.

3x and − 3x make zero.

Cross them out.

The x parts are gone.

Now we have 6 = k.

4 Choose k

When the x parts are gone, what is left is either always true or never true.

We want every x to work.

So 6 = k must be always true.

6 = k is true only when k is 6.

So k = 6!

5 Check

Let's check.

We put k = 6 back in.

Try one number, x = 1.

The left side: 3(1 + 2) = 3 × 3 = 9.

The right side: 3 × 1 + 6 = 3 + 6 = 9.

Both sides are 9.

Any other x works the same way: both sides are 3x + 6.

So every x works.

It works!

Let's look back.

First: open the brackets.

Then: compare the x parts.

Then: take away the x part.

Then: choose k.

Last: check.

▶ Watch it solved

2(x - k) = 2x - 8 has infinitely many solutions. Find k. Medium

The question

Question: for which k does 2(x − k) = 2x − 8 have infinitely many solutions?

Infinitely many solutions means every number works for x.

1 Open the brackets

The 2 in front of the bracket multiplies every part inside.

So we write 2 times each part.

2 times x is 2x.

2 times k is 2k.

2 Compare the x parts

The x parts: 2x on the left and 2x on the right.

3 Take away the x part

We take away 2x on both sides.

They stay equal.

2x and − 2x make zero.

Cross them out.

The x parts are gone.

Now we have −2k = −8.

4 Choose k

When the x parts are gone, what is left is either always true or never true.

We want every x to work.

So −2k = −8 must be always true.

So we find the k that makes it true.

5 Remove the −2

−2k means −2 times k.

We want just k.

To remove times −2, we divide by −2.

We do it on both sides, so they stay equal.

−2 divided by −2 is 1.

So just k is left.

−8 divided by −2 is 4.

So k = 4!

6 Check

Let's check.

We put k = 4 back in.

Try one number, x = 1.

The left side: 2(1 − 4) = 2 × (−3) = −6.

The right side: 2 × 1 − 8 = 2 − 8 = −6.

Both sides are −6.

Any other x works the same way: both sides are 2x − 8.

So every x works.

It works!

Let's look back.

First: open the brackets.

Then: compare the x parts.

Then: take away the x part.

Then: choose k.

Then: remove the −2.

Last: check.

▶ Watch it solved

6(x + k) = 6x + 3 has infinitely many solutions. Find k. Harder

The question

Question: for which k does 6(x + k) = 6x + 3 have infinitely many solutions?

Infinitely many solutions means every number works for x.

1 Open the brackets

The 6 in front of the bracket multiplies every part inside.

So we write 6 times each part.

6 times x is 6x.

6 times k is 6k.

2 Compare the x parts

The x parts: 6x on the left and 6x on the right.

3 Take away the x part

We take away 6x on both sides.

They stay equal.

6x and − 6x make zero.

Cross them out.

The x parts are gone.

Now we have 6k = 3.

4 Choose k

When the x parts are gone, what is left is either always true or never true.

We want every x to work.

So 6k = 3 must be always true.

So we find the k that makes it true.

5 Remove the 6

6k means 6 times k.

We want just k.

To remove times 6, we divide by 6.

We do it on both sides, so they stay equal.

6 divided by 6 is 1.

So just k is left.

We can make 3/6 simpler.

It is the same as 1/2.

So k = 1/2!

6 Check

Let's check.

We put k = 1/2 back in.

Try one number, x = 1.

The left side: 6(1 + 1/2) = 6 × 3/2 = 9.

The right side: 6 × 1 + 3 = 6 + 3 = 9.

Both sides are 9.

Any other x works the same way: both sides are 6x + 3.

So every x works.

It works!

Let's look back.

First: open the brackets.

Then: compare the x parts.

Then: take away the x part.

Then: choose k.

Then: remove the 6.

Last: check.

▶ Watch it solved