Chapter 6 of 8

All 8 chapters
  1. 1Harder linear equations, readunderstood
  2. 2Changing the subject, readunderstood
  3. 3Inequalities, readunderstood
  4. 4Equations of straight lines, readunderstood
  5. 5Two equations: adding or taking away, readunderstood
  6. 6Two equations: substitution, readunderstood
  7. 7Two equations on a graph, readunderstood
  8. 8Word problems and regions, readunderstood

M4.8.6 Chapter 6

Two equations: substitution

One equation says y = 2x + 1. How do I use that in 3x + y = 11?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Solve two equations by putting one into the other
  • Check an answer in both equations

Leila has two clues about two numbers x and y. Clue one: y = 2x + 1. Clue two: 3x + y = 11. The first clue already says what y is, in terms of x.

To substitute is to put one thing in place of another. If y is the same as 2x + 1, then wherever y appears you can write 2x + 1 instead. Put it in brackets, so it stays one whole thing.

Draw it first

Put it in

  1. A column of equations with arrows between them. The first is 3x + y = 11. Step 2 adds an arrow with the note put 2x + 1 in for y, and the equation 3x + (2x + 1) = 11. Step 3 adds an arrow with the note collect the x terms, and 5x + 1 = 11. Step 4 adds an arrow with the note take 1 from both sides, and 5x = 10.3x + y = 11

    1 Start with the second equation, 3x + y = 11.

  2. A column of equations with arrows between them. The first is 3x + y = 11. Step 2 adds an arrow with the note put 2x + 1 in for y, and the equation 3x + (2x + 1) = 11. Step 3 adds an arrow with the note collect the x terms, and 5x + 1 = 11. Step 4 adds an arrow with the note take 1 from both sides, and 5x = 10.3x + y = 113x + y = 113x + (2x + 1) = 11put 2x + 1 in for y

    2 The first equation says y is 2x + 1. Put 2x + 1 in brackets in place of y.

  3. A column of equations with arrows between them. The first is 3x + y = 11. Step 2 adds an arrow with the note put 2x + 1 in for y, and the equation 3x + (2x + 1) = 11. Step 3 adds an arrow with the note collect the x terms, and 5x + 1 = 11. Step 4 adds an arrow with the note take 1 from both sides, and 5x = 10.3x + y = 113x + y = 113x + (2x + 1) = 11put 2x + 1 in for y3x + y = 113x + (2x + 1) = 115x + 1 = 11put 2x + 1 in for ycollect the x terms

    3 Now only x is left. Collect the x terms: 3x + 2x = 5x.

  4. A column of equations with arrows between them. The first is 3x + y = 11. Step 2 adds an arrow with the note put 2x + 1 in for y, and the equation 3x + (2x + 1) = 11. Step 3 adds an arrow with the note collect the x terms, and 5x + 1 = 11. Step 4 adds an arrow with the note take 1 from both sides, and 5x = 10.3x + y = 113x + y = 113x + (2x + 1) = 11put 2x + 1 in for y3x + y = 113x + (2x + 1) = 115x + 1 = 11put 2x + 1 in for ycollect the x terms3x + y = 113x + (2x + 1) = 115x + 1 = 115x = 10put 2x + 1 in for ycollect the x termstake 1 from both sides

    4 Take 1 from both sides to get 5x = 10. Then x = 2.

Once x = 2, find y from the first equation: y = 2 × 2 + 1 = 5. The answer is a pair: x = 2 and y = 5.

Worked example

Solve y = 2x + 1 and 3x + y = 11.

  1. Put 2x + 1 in place of y in 3x + y = 11: 3x + (2x + 1) = 11.
  2. Collect the x terms: 5x + 1 = 11, so 5x = 10.
  3. Divide both sides by 5: x = 2.
  4. Find y from y = 2x + 1: y = 2 × 2 + 1 = 5.
  5. Check in both equations. In y = 2x + 1: 2 × 2 + 1 = 5. In 3x + y = 11: 3 × 2 + 5 = 11.

Answer: x = 2, y = 5

Why check in both equations? A wrong pair can fit one equation and miss the other. The pair x = 3, y = 7 fits the first, but not the second.

A table with the head pair, (1), (2). Rows: x = 2, y = 5, 5 = 5, 11 = 11; x = 3, y = 7, 7 = 7, 16 ≠ 11.pair(1)(2)x = 2, y = 55 = 511 = 11x = 3, y = 77 = 716 ≠ 11
(1) is y = 2x + 1 and (2) is 3x + y = 11. Only the first pair fits both.

Often neither equation gives a letter on its own. Then first make a letter the subject of one equation. This is the skill from the second chapter of this book.

A column of equations with arrows between them. The first is 3x − y = 5. An arrow with the note add y to both sides leads to 3x = 5 + y. An arrow with the note take 5 from both sides leads to 3x − 5 = y.3x − y = 53x = 5 + y3x − 5 = yadd y to both sidestake 5 from both sides
The y had a minus sign in front, so add y to both sides first. Now y = 3x − 5.

Worked example

Solve x + 2y = 11 and 3x − y = 5.

  1. Make y the subject of the second equation: y = 3x − 5.
  2. Put 3x − 5 in place of y in the first: x + 2(3x − 5) = 11.
  3. Expand the bracket: x + 6x − 10 = 11, so 7x = 21.
  4. Divide both sides by 7: x = 3.
  5. Find y: y = 3 × 3 − 5 = 4.
  6. Check in the first equation: 3 + 2 × 4 = 11. In the second: 3 × 3 − 4 = 5.

Answer: x = 3, y = 4

Worked example

Solve x = y + 4 and 2x + 3y = 18.

  1. Put y + 4 in place of x in the second equation: 2(y + 4) + 3y = 18.
  2. Expand and collect: 2y + 8 + 3y = 18, so 5y + 8 = 18 and 5y = 10.
  3. Divide both sides by 5: y = 2.
  4. Find x from x = y + 4: x = 2 + 4 = 6.
  5. Check in both equations. In x = y + 4: 6 = 2 + 4. In 2x + 3y = 18: 2 × 6 + 3 × 2 = 18.

Answer: x = 6, y = 2

See it three ways

With things

Plot both rules on squared paper. First write 3x + y = 11 as y = −3x + 11 by taking 3x from both sides. Look for the one point that is on both lines.

As a picture

A graph with the x axis x and the y axis y. A line named y = 2x + 1 and a line named y = −3x + 11 cross at the point P (2, 5).P (2, 5)y = 2x + 1y = −3x + 1112345246810120xy

In words

The lines cross at one point. That point is on both lines, so both equations are true there. It is x = 2 and y = 5.

In symbols

3x + (2x + 1) = 11 gives x = 2. Then y = 2 × 2 + 1 = 5.

Bars show a substitution well. Say two numbers add to 13, and x is 3 more than y. The bar for x is the bar for y with 3 added. So x can be replaced by y + 3.

Two bars. The top bar has a part marked y, 5 long, and a part marked 3, with a bracket labelled x. The bottom bar is one part marked y, 5 long, with a bracket labelled y.y3xyy
x is y and 3 more, so x = y + 3.

Try it

A number puzzle

You need: paper and a pencil.

  1. Two numbers add to 13. One is 3 more than the other. Write x = y + 3 and x + y = 13.
  2. Put y + 3 in place of x in the second equation. Solve it for y.
  3. Find x, then check both clues with the two numbers.

What you will see: You get y = 5 and x = 8. The numbers 8 and 5 add to 13, and 8 is 3 more than 5.

If you cannot try it, this is what happens.
A column of equations with arrows between them. The first is x + y = 13. An arrow with the note put y + 3 for x leads to (y + 3) + y = 13. An arrow with the note collect leads to 2y + 3 = 13.x + y = 13(y + 3) + y = 132y + 3 = 13put y + 3 for xcollect

A wrong idea many people have

To put y = x + 2 into 10 − y, write 10 − x + 2.

Why it feels right: You just write the new letters where y was, so it feels like no more is needed.

The test that breaks it: Take x = 1, so y = 3. Then 10 − y = 7. But 10 − x + 2 = 11. With brackets, 10 − (x + 2) = 7.

What is true: Put the whole expression in brackets. The minus sign then applies to every term inside.

A wrong idea many people have

It is enough to check an answer in one equation.

Why it feels right: You used both equations to find it, so it feels as if both must fit.

The test that breaks it: x = 3 and y = 7 fit y = 2x + 1, because 2 × 3 + 1 = 7. But 3 × 3 + 7 = 16 in the second equation, not 11.

What is true: An answer must fit both equations. Check it in each one.

Key words

substitute
to put one expression in place of a letter that it equals.

Back to your guess

The answer is B: Put 2x + 1 in place of y: 3x + (2x + 1) = 11.

The equation says y is the whole of 2x + 1, so the whole of it goes in place of y. That gives 3x + (2x + 1) = 11, which has only x in it. Then x = 2 and y = 5.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. Put y = 3x into x + y = 12. What do you get?

    Two equations labelled (1) and (2): y = 3x and x + y = 12.y = 3x(1)x + y = 12(2)
    Tap your answer.

    1. Yes, that is right. The letter y is replaced by 3x. So x + y = 12 becomes x + 3x = 12, which is 4x = 12, and x = 3.

    2. Not this one. You put 3 in place of y. The equation says y is 3x, so the whole of 3x goes in. Go to M4.8.6. Try another.

    3. Not this one. You put 3x in place of the x. The equation y = 3x tells you what to put in place of y. Go to M4.8.6. Try another.

    4. Not this one. The 3x went to the wrong side. It must replace y, and the 12 stays where it is. Go to M4.8.6. Try another.

    See the answer

    x + 3x = 12

    The letter y is replaced by 3x. So x + y = 12 becomes x + 3x = 12, which is 4x = 12, and x = 3.

    • If you chose B: You put 3 in place of y. The equation says y is 3x, so the whole of 3x goes in. Go to M4.8.6.
    • If you chose C: You put 3x in place of the x. The equation y = 3x tells you what to put in place of y. Go to M4.8.6.
    • If you chose D: The 3x went to the wrong side. It must replace y, and the 12 stays where it is. Go to M4.8.6.

  2. Put y = x + 2 into 3x + 2y = 14. Which is right?

    Tap your answer.

    1. Not this one. The 2 in front of y multiplies the whole of x + 2. Use a bracket: the 2 times x and the 2 times 2. Go to M4.8.6. Try another.

    2. Yes, that is right. Put x + 2 in brackets in place of y. The bracket stays because the 2 in front multiplies all of it. Then 5x + 4 = 14.

    3. Not this one. You put 2 in place of y. The equation y = x + 2 says that y is the whole of x + 2. Go to M4.8.6. Try another.

    4. Not this one. The right side 14 is not changed. Only y is replaced. Go to M4.8.6. Try another.

    See the answer

    3x + 2(x + 2) = 14

    Put x + 2 in brackets in place of y. The bracket stays because the 2 in front multiplies all of it. Then 5x + 4 = 14.

    • If you chose A: The 2 in front of y multiplies the whole of x + 2. Use a bracket: the 2 times x and the 2 times 2. Go to M4.8.6.
    • If you chose C: You put 2 in place of y. The equation y = x + 2 says that y is the whole of x + 2. Go to M4.8.6.
    • If you chose D: The right side 14 is not changed. Only y is replaced. Go to M4.8.6.

  3. Solve y = x + 3 and 2x + y = 12. What are x and y?

    Tap your answer.

    1. Not this one. The first equation says y = x + 3, so y is not the same as x. Find y by putting x = 3 into it. Go to M4.8.6. Try another.

    2. Yes, that is right. Put x + 3 in place of y: 2x + (x + 3) = 12, so 3x = 9 and x = 3. Then y = 3 + 3 = 6. Check: 2 × 3 + 6 = 12.

    3. Not this one. This pair fits y = x + 3, but 2 × 4 + 7 = 15, not 12. Check in both equations. Go to M4.8.6. Try another.

    4. Not this one. Putting in gives 3x + 3 = 12, so 3x = 9. You must divide by 3: x = 3, not 9. Go to M4.8.6. Try another.

    See the answer

    x = 3, y = 6

    Put x + 3 in place of y: 2x + (x + 3) = 12, so 3x = 9 and x = 3. Then y = 3 + 3 = 6. Check: 2 × 3 + 6 = 12.

    • If you chose A: The first equation says y = x + 3, so y is not the same as x. Find y by putting x = 3 into it. Go to M4.8.6.
    • If you chose C: This pair fits y = x + 3, but 2 × 4 + 7 = 15, not 12. Check in both equations. Go to M4.8.6.
    • If you chose D: Putting in gives 3x + 3 = 12, so 3x = 9. You must divide by 3: x = 3, not 9. Go to M4.8.6.

  4. Which pair fits both x + y = 10 and x − y = 4?

    Tap your answer.

    1. Not this one. It fits x + y = 10, but 6 − 4 = 2, not 4. A pair must fit both equations. Go to M4.8.6. Try another.

    2. Not this one. It fits x − y = 4, but 8 + 4 = 12, not 10. A pair must fit both equations. Go to M4.8.6. Try another.

    3. Yes, that is right. 7 + 3 = 10 and 7 − 3 = 4, so the pair fits both equations.

    4. Not this one. It fits x + y = 10, but 5 − 5 = 0, not 4. A pair must fit both equations. Go to M4.8.6. Try another.

    See the answer

    x = 7, y = 3

    7 + 3 = 10 and 7 − 3 = 4, so the pair fits both equations.

    • If you chose A: It fits x + y = 10, but 6 − 4 = 2, not 4. A pair must fit both equations. Go to M4.8.6.
    • If you chose B: It fits x − y = 4, but 8 + 4 = 12, not 10. A pair must fit both equations. Go to M4.8.6.
    • If you chose D: It fits x + y = 10, but 5 − 5 = 0, not 4. A pair must fit both equations. Go to M4.8.6.

  5. You will solve x + y = 8 and 2x + 3y = 21 by substitution. Which first step makes a letter the subject?

    Tap your answer.

    1. Yes, that is right. Take x from both sides of x + y = 8. That gives y = 8 − x. Then put 8 − x in place of y in the other equation.

    2. Not this one. Take x from both sides of x + y = 8. That leaves y = 8 − x, with the x taken away from 8. Go to M4.8.2. Try another.

    3. Not this one. In 2x + 3y = 21 the x has a 2 in front. You would take 3y away, then divide by 2. Go to M4.8.2. Try another.

    4. Not this one. There is a plus sign between x and y, so you take y away. Dividing is for products. Go to M4.8.2. Try another.

    See the answer

    y = 8 − x

    Take x from both sides of x + y = 8. That gives y = 8 − x. Then put 8 − x in place of y in the other equation.

    • If you chose B: Take x from both sides of x + y = 8. That leaves y = 8 − x, with the x taken away from 8. Go to M4.8.2.
    • If you chose C: In 2x + 3y = 21 the x has a 2 in front. You would take 3y away, then divide by 2. Go to M4.8.2.
    • If you chose D: There is a plus sign between x and y, so you take y away. Dividing is for products. Go to M4.8.2.

  6. Put y = x + 4 into 12 − y. What do you get?

    Tap your answer.

    1. Not this one. The minus sign must apply to the whole of x + 4. Use a bracket: 12 − (x + 4), which is 12 − x − 4. Go to M4.8.6. Try another.

    2. Not this one. The sign in front of y is a minus, and it stays a minus. Go to M4.8.6. Try another.

    3. Not this one. You have joined x and 4 as if both were x terms. They are different terms, and cannot be added. Go to M4.8.6. Try another.

    4. Yes, that is right. The whole of x + 4 goes in place of y, in brackets. Then 12 − (x + 4) = 12 − x − 4 = 8 − x.

    See the answer

    12 − (x + 4) = 8 − x

    The whole of x + 4 goes in place of y, in brackets. Then 12 − (x + 4) = 12 − x − 4 = 8 − x.

    • If you chose A: The minus sign must apply to the whole of x + 4. Use a bracket: 12 − (x + 4), which is 12 − x − 4. Go to M4.8.6.
    • If you chose B: The sign in front of y is a minus, and it stays a minus. Go to M4.8.6.
    • If you chose C: You have joined x and 4 as if both were x terms. They are different terms, and cannot be added. Go to M4.8.6.

  7. Solve y = 2x + 3 and 4x + y = 15. Give x first, then y.

    See the answer

    x = 2, y = 7

    Put 2x + 3 in place of y: 4x + (2x + 3) = 15, so 6x = 12 and x = 2. Then y = 2 × 2 + 3 = 7.

  8. Solve x = y + 5 and 2x + y = 22. Give x first, then y.

    See the answer

    x = 9, y = 4

    Put y + 5 in place of x: 2(y + 5) + y = 22, so 3y + 10 = 22 and y = 4. Then x = 4 + 5 = 9.

  9. From an earlier chapter

    Solve x + y = 12 and x − y = 4. Give x first, then y.

    See the answer

    x = 8, y = 4

    Add the equations: 2x = 16, so x = 8. Then 8 + y = 12, so y = 4.

Keep it

Copy this into your exercise book.

Put one equation into the other, in brackets. Solve for one letter, find the other, then check in both equations.

Draw: Two equations, an arrow with the words put in, and the new equation with only one letter.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

No internet tomorrow? Save the whole book as one file (989 KB). It opens on any phone.