Steply

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

Step by step, with the reason for every step · Three equations in x, y, z · Algebra 1

Step by step

The question

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

2x + 3y + z = −1①
x − y − z = 3②

We call them ①, ② and ③.

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

x − y + z = 5③

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 ① and ②

Look at z in ① and ②: z and −z.

They are opposites, so adding removes z.

We write ① and ② under each other: x under x, y under y, z under z.

2x + 3y + z = −1①
x − y − z = 3②

z and −z make zero.

Cross them out.

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

2x + x = 3x.

3y − y = 2y.

−1 + 3 = 2.

2x + x = 3x
3y − y = 2y
−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 ④.

It has no z.

3x + 2y = 2④

2 Remove z from ② and ③

Look at z in ② and ③: −z and z.

They are opposites, so adding removes z.

We write ② and ③ under each other: x under x, y under y, z under z.

x − y − z = 3②
x − y + z = 5③

−z and z make zero.

Cross them out.

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

x + x = 2x.

−y − y = −2y.

3 + 5 = 8.

x + x = 2x
−y − y = −2y
3 + 5 = 8

Now we have 2x − 2y = 8.

We call it ⑤.

It has no z.

2x − 2y = 8⑤

Now ④ and ⑤ are two equations with two letters: x and y.

We solve them like before, by elimination.

We write ④ and ⑤ again, in columns.

3x + 2y = 2④
2x − 2y = 8⑤

3 Add the equations

Look at y: 2y and −2y.

They are opposites.

Added, they make zero.

2y and −2y make zero.

Cross them out.

We add ④ and ⑤: 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 ④.

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 − 6 = 2 − 6

6 and − 6 make zero.

Cross them out.

2 minus 6 is −4.

2 − 6 = −4

Now we have 2y = −4.

2y = −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 = −42

2 divided by 2 is 1.

So just y is left.

−4 divided by 2 is −2.

−42 = −2

So y = −2.

y = −2

6 Find z

We put x = 2 and y = −2 into ②, the easiest one.

2 − (−2) − z = 3

2 − (−2) is 4.

4 − 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 − z − 4 = 3 − 4

4 and − 4 make zero.

Cross them out.

3 minus 4 is −1.

3 − 4 = −1

Now we have −z = −1.

−z = −1

−z means −1 times z.

We want just z.

To remove times −1, we divide by −1.

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

−z−1 = −1−1

−1 divided by −1 is 1.

So just z is left.

−1 divided by −1 is 1.

−1−1 = 1

So z = 1.

z = 1

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

We write it as (2, −2, 1).

(x, y, z) = (2, −2, 1)

7 Check

Let's check in every equation.

We put x = 2, y = −2, z = 1 into ①.

① 2(2) + 3(−2) + 1 = 4 − 6 + 1 = −1

We get −1, the same as the right side.

Now ②.

② 2 − (−2) − 1 = 2 + 2 − 1 = 3

We get 3, the same as the right side.

Now ③.

③ 2 − (−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 ① and ②.

Then: remove z from ② and ③.

Then: add the equations.

Then: find x.

Then: find y.

Then: find z.

Last: check.

Answer
(x, y, z) = (2, −2, 1)

Your turn

Try this one yourself, then check:

Solve 2x + 3y + z = 18, x − y − z = −4, x − y + z = 0
Show the answer
(x, y, z) = (2, 4, 2)

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