Steply

Three equations in x, y, z

Algebra 1 ยท 20 problems, easy to harder

โ–ถ Watch the first one solved

How we do it

  1. Remove z from โ‘  and โ‘ก
  2. Remove z from โ‘ก and โ‘ข
  3. Make the y numbers opposite
  4. Add the equations
  5. Find x
  6. Find y
  7. Find z
  8. Check

The steps for Solve x + y + z = 6, 2x + 3y - z = 5, 3x - 2y + z = 19. Every solution names its steps, so you always know what comes next.

Worked examples

Solve x + y - z = 0, x - y + 3z = 8, x - y - z = -4 Easy

The question

Question: solve x + y โˆ’ z = 0, x โˆ’ y + 3z = 8 and x โˆ’ y โˆ’ z = โˆ’4.

x + y โˆ’ z = 0โ‘ 
x โˆ’ y + 3z = 8โ‘ก

We call them โ‘ , โ‘ก and โ‘ข.

We want the x, y and z that make all three true.

x โˆ’ y โˆ’ z = โˆ’4โ‘ข

The plan: remove one letter twice.

Then we have two equations with two letters, and we know how to solve those.

1 Remove y from โ‘  and โ‘ก

Look at y in โ‘  and โ‘ก: y and โˆ’y.

They are opposites, so adding removes y.

We write โ‘  and โ‘ก under each other: x under x, y under y, z under z.

x + y โˆ’ z = 0โ‘ 
x โˆ’ y + 3z = 8โ‘ก

y and โˆ’y make zero.

Cross them out.

We add: the x parts, the z parts, and the numbers.

x + x = 2x.

โˆ’z + 3z = 2z.

0 + 8 = 8.

x + x = 2x
โˆ’z + 3z = 2z
0 + 8 = 8

Careful!

We add every part of both equations: each letter, and the numbers on the right too.

Careful: add every part, the numbers too

Now we have 2x + 2z = 8.

We call it โ‘ฃ.

It has no y.

2x + 2z = 8โ‘ฃ

2 Remove y from โ‘  and โ‘ข

Look at y in โ‘  and โ‘ข: y and โˆ’y.

They are opposites, so adding removes y.

We write โ‘  and โ‘ข under each other: x under x, y under y, z under z.

x + y โˆ’ z = 0โ‘ 
x โˆ’ y โˆ’ z = โˆ’4โ‘ข

y and โˆ’y make zero.

Cross them out.

We add: the x parts, the z parts, and the numbers.

x + x = 2x.

โˆ’z โˆ’ z = โˆ’2z.

0 + (โˆ’4) = โˆ’4.

x + x = 2x
โˆ’z โˆ’ z = โˆ’2z
0 + (โˆ’4) = โˆ’4

Now we have 2x โˆ’ 2z = โˆ’4.

We call it โ‘ค.

It has no y.

2x โˆ’ 2z = โˆ’4โ‘ค

Now โ‘ฃ and โ‘ค are two equations with two letters: x and z.

We solve them like before, by elimination.

We write โ‘ฃ and โ‘ค again, in columns.

2x + 2z = 8โ‘ฃ
2x โˆ’ 2z = โˆ’4โ‘ค

3 Add the equations

Look at z: 2z and โˆ’2z.

They are opposites.

Added, they make zero.

2z and โˆ’2z make zero.

Cross them out.

We add โ‘ฃ and โ‘ค: the x parts together, and the numbers together.

2x + 2x = 4x.

8 + (โˆ’4) = 4.

2x + 2x = 4x
8 + (โˆ’4) = 4

Now we have 4x = 4.

4x = 4

4 Find x

4x means 4 times x.

We want just x.

To remove times 4, we divide by 4.

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

4x4 = 44

4 divided by 4 is 1.

So just x is left.

On the right too: 4 divided by 4 is 1.

44 = 1

So x = 1.

x = 1

5 Find z

We put x = 1 into โ‘ฃ.

2 times 1 is 2.

2(1) + 2z = 8

Now we have 2 + 2z = 8.

2 + 2z = 8

We want z by itself.

The 2 is in the way.

To remove 2, we take away 2.

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

2 + 2z โˆ’ 2 = 8 โˆ’ 2

2 and โˆ’ 2 make zero.

Cross them out.

8 minus 2 is 6.

8 โˆ’ 2 = 6

Now we have 2z = 6.

2z = 6

2z means 2 times z.

We want just z.

To remove times 2, we divide by 2.

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

2z2 = 62

2 divided by 2 is 1.

So just z is left.

6 divided by 2 is 3.

62 = 3

So z = 3.

z = 3

6 Find y

We put x = 1 and z = 3 into โ‘ , the easiest one.

1 โˆ’ 3 + y = 0

1 โˆ’ 3 is โˆ’2.

โˆ’2 + y = 0

We want y by itself.

The โˆ’2 is in the way.

To remove โˆ’2, we add 2.

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

โˆ’2 + y + 2 = 0 + 2

โˆ’2 and + 2 make zero.

Cross them out.

0 plus 2 is 2.

0 + 2 = 2

So y = 2.

y = 2

So x = 1, y = 2 and z = 3, all at the same time.

We write it as (1, 2, 3).

(x, y, z) = (1, 2, 3)

7 Check

Let's check in every equation.

We put x = 1, y = 2, z = 3 into โ‘ .

โ‘  1 + 2 โˆ’ 3 = 0

We get 0, the same as the right side.

Now โ‘ก.

โ‘ก 1 โˆ’ 2 + 3(3) = 1 โˆ’ 2 + 9 = 8

We get 8, the same as the right side.

Now โ‘ข.

โ‘ข 1 โˆ’ 2 โˆ’ 3 = โˆ’4

We get โˆ’4, the same as the right side.

All of them work.

So the answer is right!

Let's look back.

First: remove y from โ‘  and โ‘ก.

Then: remove y from โ‘  and โ‘ข.

Then: add the equations.

Then: find x.

Then: find z.

Then: find y.

Last: check.

Answer
(x, y, z) = (1, 2, 3)

โ–ถ Watch it solved

Solve 2x + 3y + z = -1, x - y - z = 3, x - y + z = 5 Medium

The question

Question: solve 2x + 3y + z = โˆ’1, x โˆ’ y โˆ’ z = 3 and x โˆ’ y + z = 5.

2x + 3y + z = โˆ’1โ‘ 
x โˆ’ y โˆ’ z = 3โ‘ก

We call them โ‘ , โ‘ก and โ‘ข.

We want the x, y and z that make all three true.

x โˆ’ y + z = 5โ‘ข

The plan: remove one letter twice.

Then we have two equations with two letters, and we know how to solve those.

1 Remove z from โ‘  and โ‘ก

Look at z in โ‘  and โ‘ก: z and โˆ’z.

They are opposites, so adding removes z.

We write โ‘  and โ‘ก under each other: x under x, y under y, z under z.

2x + 3y + z = โˆ’1โ‘ 
x โˆ’ y โˆ’ z = 3โ‘ก

z and โˆ’z make zero.

Cross them out.

We add: the x parts, the y parts, and the numbers.

2x + x = 3x.

3y โˆ’ y = 2y.

โˆ’1 + 3 = 2.

2x + x = 3x
3y โˆ’ y = 2y
โˆ’1 + 3 = 2

Careful!

We add every part of both equations: each letter, and the numbers on the right too.

Careful: add every part, the numbers too

Now we have 3x + 2y = 2.

We call it โ‘ฃ.

It has no z.

3x + 2y = 2โ‘ฃ

2 Remove z from โ‘ก and โ‘ข

Look at z in โ‘ก and โ‘ข: โˆ’z and z.

They are opposites, so adding removes z.

We write โ‘ก and โ‘ข under each other: x under x, y under y, z under z.

x โˆ’ y โˆ’ z = 3โ‘ก
x โˆ’ y + z = 5โ‘ข

โˆ’z and z make zero.

Cross them out.

We add: the x parts, the y parts, and the numbers.

x + x = 2x.

โˆ’y โˆ’ y = โˆ’2y.

3 + 5 = 8.

x + x = 2x
โˆ’y โˆ’ y = โˆ’2y
3 + 5 = 8

Now we have 2x โˆ’ 2y = 8.

We call it โ‘ค.

It has no z.

2x โˆ’ 2y = 8โ‘ค

Now โ‘ฃ and โ‘ค are two equations with two letters: x and y.

We solve them like before, by elimination.

We write โ‘ฃ and โ‘ค again, in columns.

3x + 2y = 2โ‘ฃ
2x โˆ’ 2y = 8โ‘ค

3 Add the equations

Look at y: 2y and โˆ’2y.

They are opposites.

Added, they make zero.

2y and โˆ’2y make zero.

Cross them out.

We add โ‘ฃ and โ‘ค: the x parts together, and the numbers together.

3x + 2x = 5x.

2 + 8 = 10.

3x + 2x = 5x
2 + 8 = 10

Now we have 5x = 10.

5x = 10

4 Find x

5x means 5 times x.

We want just x.

To remove times 5, we divide by 5.

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

5x5 = 105

5 divided by 5 is 1.

So just x is left.

10 divided by 5 is 2.

105 = 2

So x = 2.

x = 2

5 Find y

We put x = 2 into โ‘ฃ.

3 times 2 is 6.

3(2) + 2y = 2

Now we have 6 + 2y = 2.

6 + 2y = 2

We want y by itself.

The 6 is in the way.

To remove 6, we take away 6.

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

6 + 2y โˆ’ 6 = 2 โˆ’ 6

6 and โˆ’ 6 make zero.

Cross them out.

2 minus 6 is โˆ’4.

2 โˆ’ 6 = โˆ’4

Now we have 2y = โˆ’4.

2y = โˆ’4

2y means 2 times y.

We want just y.

To remove times 2, we divide by 2.

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

2y2 = โˆ’42

2 divided by 2 is 1.

So just y is left.

โˆ’4 divided by 2 is โˆ’2.

โˆ’42 = โˆ’2

So y = โˆ’2.

y = โˆ’2

6 Find z

We put x = 2 and y = โˆ’2 into โ‘ก, the easiest one.

2 โˆ’ (โˆ’2) โˆ’ z = 3

2 โˆ’ (โˆ’2) is 4.

4 โˆ’ z = 3

We want z by itself.

The 4 is in the way.

To remove 4, we take away 4.

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

4 โˆ’ z โˆ’ 4 = 3 โˆ’ 4

4 and โˆ’ 4 make zero.

Cross them out.

3 minus 4 is โˆ’1.

3 โˆ’ 4 = โˆ’1

Now we have โˆ’z = โˆ’1.

โˆ’z = โˆ’1

โˆ’z means โˆ’1 times z.

We want just z.

To remove times โˆ’1, we divide by โˆ’1.

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

โˆ’zโˆ’1 = โˆ’1โˆ’1

โˆ’1 divided by โˆ’1 is 1.

So just z is left.

โˆ’1 divided by โˆ’1 is 1.

โˆ’1โˆ’1 = 1

So z = 1.

z = 1

So x = 2, y = โˆ’2 and z = 1, all at the same time.

We write it as (2, โˆ’2, 1).

(x, y, z) = (2, โˆ’2, 1)

7 Check

Let's check in every equation.

We put x = 2, y = โˆ’2, z = 1 into โ‘ .

โ‘  2(2) + 3(โˆ’2) + 1 = 4 โˆ’ 6 + 1 = โˆ’1

We get โˆ’1, the same as the right side.

Now โ‘ก.

โ‘ก 2 โˆ’ (โˆ’2) โˆ’ 1 = 2 + 2 โˆ’ 1 = 3

We get 3, the same as the right side.

Now โ‘ข.

โ‘ข 2 โˆ’ (โˆ’2) + 1 = 2 + 2 + 1 = 5

We get 5, the same as the right side.

All of them work.

So the answer is right!

Let's look back.

First: remove z from โ‘  and โ‘ก.

Then: remove z from โ‘ก and โ‘ข.

Then: add the equations.

Then: find x.

Then: find y.

Then: find z.

Last: check.

Answer
(x, y, z) = (2, โˆ’2, 1)

โ–ถ Watch it solved

Solve x + y + z = 6, 2x + 3y - z = 5, 3x - 2y + z = 19 Harder

The question

Question: solve x + y + z = 6, 2x + 3y โˆ’ z = 5 and 3x โˆ’ 2y + z = 19.

x + y + z = 6โ‘ 
2x + 3y โˆ’ z = 5โ‘ก

We call them โ‘ , โ‘ก and โ‘ข.

We want the x, y and z that make all three true.

3x โˆ’ 2y + z = 19โ‘ข

The plan: remove one letter twice.

Then we have two equations with two letters, and we know how to solve those.

1 Remove z from โ‘  and โ‘ก

Look at z in โ‘  and โ‘ก: z and โˆ’z.

They are opposites, so adding removes z.

We write โ‘  and โ‘ก under each other: x under x, y under y, z under z.

x + y + z = 6โ‘ 
2x + 3y โˆ’ z = 5โ‘ก

z and โˆ’z make zero.

Cross them out.

We add: the x parts, the y parts, and the numbers.

x + 2x = 3x.

y + 3y = 4y.

6 + 5 = 11.

x + 2x = 3x
y + 3y = 4y
6 + 5 = 11

Careful!

We add every part of both equations: each letter, and the numbers on the right too.

Careful: add every part, the numbers too

Now we have 3x + 4y = 11.

We call it โ‘ฃ.

It has no z.

3x + 4y = 11โ‘ฃ

2 Remove z from โ‘ก and โ‘ข

Look at z in โ‘ก and โ‘ข: โˆ’z and z.

They are opposites, so adding removes z.

We write โ‘ก and โ‘ข under each other: x under x, y under y, z under z.

2x + 3y โˆ’ z = 5โ‘ก
3x โˆ’ 2y + z = 19โ‘ข

โˆ’z and z make zero.

Cross them out.

We add: the x parts, the y parts, and the numbers.

2x + 3x = 5x.

3y โˆ’ 2y = y.

5 + 19 = 24.

2x + 3x = 5x
3y โˆ’ 2y = y
5 + 19 = 24

Now we have 5x + y = 24.

We call it โ‘ค.

It has no z.

5x + y = 24โ‘ค

Now โ‘ฃ and โ‘ค are two equations with two letters: x and y.

We solve them like before, by elimination.

We write โ‘ฃ and โ‘ค again, in columns.

3x + 4y = 11โ‘ฃ
5x + y = 24โ‘ค

3 Make the y numbers opposite

We want to remove y.

In โ‘ฃ it is 4y, in โ‘ค it is y.

We make them opposite: 4y and โˆ’4y.

Then adding removes y.

We multiply every part of โ‘ค by โˆ’4.

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

โˆ’4 ร— (5x + y) = โˆ’4 ร— 24โ‘ค ร— โˆ’4

โˆ’4 times 5x is โˆ’20x.

โˆ’4 times y is โˆ’4y.

โˆ’4 times 24 is โˆ’96.

โˆ’4 ร— 5x = โˆ’20x, โˆ’4 ร— y = โˆ’4y, โˆ’4 ร— 24 = โˆ’96

We call it โ‘ฅ.

โˆ’20x โˆ’ 4y = โˆ’96โ‘ฅ

4 Add the equations

We write โ‘ฃ and โ‘ฅ under each other: x under x, y under y.

3x + 4y = 11โ‘ฃ
โˆ’20x โˆ’ 4y = โˆ’96โ‘ฅ

Look at y: 4y and โˆ’4y.

They are opposites.

Added, they make zero.

4y and โˆ’4y make zero.

Cross them out.

We add โ‘ฃ and โ‘ฅ: the x parts together, and the numbers together.

3x โˆ’ 20x = โˆ’17x.

11 + (โˆ’96) = โˆ’85.

3x โˆ’ 20x = โˆ’17x
11 + (โˆ’96) = โˆ’85

Now we have โˆ’17x = โˆ’85.

โˆ’17x = โˆ’85

5 Find x

โˆ’17x means โˆ’17 times x.

We want just x.

To remove times โˆ’17, we divide by โˆ’17.

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

โˆ’17xโˆ’17 = โˆ’85โˆ’17

โˆ’17 divided by โˆ’17 is 1.

So just x is left.

โˆ’85 divided by โˆ’17 is 5.

โˆ’85โˆ’17 = 5

So x = 5.

x = 5

6 Find y

We put x = 5 into โ‘ค.

5 times 5 is 25.

5(5) + y = 24

Now we have 25 + y = 24.

25 + y = 24

We want y by itself.

The 25 is in the way.

To remove 25, we take away 25.

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

25 + y โˆ’ 25 = 24 โˆ’ 25

25 and โˆ’ 25 make zero.

Cross them out.

24 minus 25 is โˆ’1.

24 โˆ’ 25 = โˆ’1

So y = โˆ’1.

y = โˆ’1

7 Find z

We put x = 5 and y = โˆ’1 into โ‘ , the easiest one.

5 + (โˆ’1) + z = 6

5 + (โˆ’1) is 4.

4 + z = 6

We want z by itself.

The 4 is in the way.

To remove 4, we take away 4.

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

4 + z โˆ’ 4 = 6 โˆ’ 4

4 and โˆ’ 4 make zero.

Cross them out.

6 minus 4 is 2.

6 โˆ’ 4 = 2

So z = 2.

z = 2

So x = 5, y = โˆ’1 and z = 2, all at the same time.

We write it as (5, โˆ’1, 2).

(x, y, z) = (5, โˆ’1, 2)

8 Check

Let's check in every equation.

We put x = 5, y = โˆ’1, z = 2 into โ‘ .

โ‘  5 + (โˆ’1) + 2 = 5 โˆ’ 1 + 2 = 6

We get 6, the same as the right side.

Now โ‘ก.

โ‘ก 2(5) + 3(โˆ’1) โˆ’ 2 = 10 โˆ’ 3 โˆ’ 2 = 5

We get 5, the same as the right side.

Now โ‘ข.

โ‘ข 3(5) โˆ’ 2(โˆ’1) + 2 = 15 + 2 + 2 = 19

We get 19, the same as the right side.

All of them work.

So the answer is right!

Let's look back.

First: remove z from โ‘  and โ‘ก.

Then: remove z from โ‘ก and โ‘ข.

Then: make the y numbers opposite.

Then: add the equations.

Then: find x.

Then: find y.

Then: find z.

Last: check.

Answer
(x, y, z) = (5, โˆ’1, 2)

โ–ถ Watch it solved

Practice

Try each one, then open the answer. Stuck? Open the problem to watch every step.

Easy

  1. x + y - z = 0, x - y + 3z = 8, x - y - z = -4
    Answer
    (x, y, z) = (1, 2, 3)
  2. x - y - z = -3, x - y + z = 3, x + 3y + z = 15
    Answer
    (x, y, z) = (3, 3, 3)
  3. x + 3y + z = 12, x - y - z = 0, x - y + z = 8
    Answer
    (x, y, z) = (5, 1, 4)
  4. x - y + z = 4, x + 3y - z = 10, x - y - z = -2
    Answer
    (x, y, z) = (4, 3, 3)
  5. x + y - z = 3, x - y - z = 1, x + y + 3z = -1
    Answer
    (x, y, z) = (1, 1, โˆ’1)
  6. x + y + 3z = 16, x + y - z = 8, x - y - z = -2
    Answer
    (x, y, z) = (5, 5, 2)
  7. x - y + 3z = 7, x + y - z = -1, x - y - z = 3
    Answer
    (x, y, z) = (2, โˆ’2, 1)
  8. x + 3y + z = 3, x - y + z = -1, x - y - z = 3
    Answer
    (x, y, z) = (2, 1, โˆ’2)

Medium

  1. 2x + 3y + z = -1, x - y - z = 3, x - y + z = 5
    Answer
    (x, y, z) = (2, โˆ’2, 1)
  2. 2x + y + 3z = 18, x + y - z = 1, x - y - z = 3
    Answer
    (x, y, z) = (5, โˆ’1, 3)
  3. x - y - z = 0, x + y - z = 10, x + y + 3z = 6
    Answer
    (x, y, z) = (4, 5, โˆ’1)
  4. x + y + z = 7, x - y + z = 3, x + y - z = 5
    Answer
    (x, y, z) = (4, 2, 1)
  5. x - 2y + z = -3, 2x + y - z = 13, x + y + z = 3
    Answer
    (x, y, z) = (4, 2, โˆ’3)
  6. 2x + y - z = -3, x + y + 3z = 18, x - y - z = -10
    Answer
    (x, y, z) = (โˆ’1, 4, 5)
  7. x + 2y + z = 3, 2x - y - z = 2, x + y + z = 4
    Answer
    (x, y, z) = (2, โˆ’1, 3)
  8. x - y - z = -9, x - y + z = -1, x + 3y + z = 15
    Answer
    (x, y, z) = (โˆ’1, 4, 4)

Harder

  1. x + y + z = 6, 2x + 3y - z = 5, 3x - 2y + z = 19
    Answer
    (x, y, z) = (5, โˆ’1, 2)
  2. 2x - y + z = 3, x + 2y - z = 8, x + y + 2z = 3
    Answer
    (x, y, z) = (3, 2, โˆ’1)
  3. x + 3y + z = 10, 2x - y + z = 2, x + y - z = -2
    Answer
    (x, y, z) = (0, 2, 4)
  4. x + y + z = 2, x - y + z = 4, 3x + y + 3z = 10
    Answer
    0 = 2