Year 9 · Unit 7 Equations and linear systems · Lesson 6 of 7

Add or subtract until one unknown disappears

Two equations can be added or subtracted like two balances. Make the coefficients of one unknown match, and that unknown cancels.

Number and AlgebraNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. Work out 5 − (−2).
    Answer7: subtracting a negative adds.
  2. Multiply every term of 2x + 3y = 7 by 4.
    Answer8x + 12y = 28.
  3. Solve 6x = 18.
    Answerx = 3.
Words used in this lesson
eliminate
remove an unknown by adding or subtracting. For example: adding 3y and −3y.
coefficient
the number in front of an unknown. For example: in 5y the coefficient is 5.
opposite
the same size with the other sign. For example: +3y and −3y.
multiply through
multiply every term of an equation. For example: 2(x + 2y = 9) is 2x + 4y = 18.
back-substitute
put a found value in to get the other unknown. For example: x = 5, so y = 2.
check
put the pair into both original equations. For example: 5 + 4 = 9 and 15 + 8 = 23.

Take one bill away from the other

A shop in Pune sells jars and packets. One customer pays 19 rupees for 3 jars and 2 packets. Another pays 14 rupees for 3 jars and 1 packet. Both bills have the same jars, so the only difference between them is one packet. Let a jar cost j rupees and a packet cost p rupees.

Predict first: what do you get if you take the second bill away from the first, and what does the result tell you about the price of a packet? Then explore with the figure.

2x + 5y = 18 and 3x + 2y = 16
1
1
subtract
−1x in the result
3y in the result
neither is gonewhat happened

×1 and ×1, then subtract: −x + 3y = 2. Neither is gone.

Show me the steps
  1. 2x + 3y = 12 and 4x − 3y = 6, times 1 and 1: the y terms are opposites, so add. 6x = 18 and x = 3.
  2. 3x + 2y = 17 and 3x − y = 5, times 1 and 1: the x terms match, so subtract. 3y = 12 and y = 4.
  3. 2x + 5y = 18 and 3x + 2y = 16: times 3 and times 2 make 6x in both. Subtract: 11y = 22 and y = 2.
  4. 4x − 3y = 1 and 2x + 5y = 33: times 1 and times 2 make 4x in both. Subtract: −13y = −65 and y = 5.

Elimination:

1. Multiply one or both equations (every term) so that one unknown has the same number in front in both.
2. If the matching terms have the same sign, subtract the equations. If they have opposite signs, add.
3. Solve the equation with one unknown, then put the answer back to find the other.
4. Check in both original equations.

It works because adding or subtracting equal quantities from both sides keeps the balance.

A common mistake: subtracting a negative as if it were a positive. In 5y − (−2y) the answer is 7y, not 3y, because 5 − (−2) = 7. Subtract every term of the second equation, including the total.

Worked examples

The shop: 3j + 2p = 19 and 3j + p = 14. Find the price of each.

  1. The j terms match, so subtract the second from the first: p = 5.
  2. Back into the second: 3j + 5 = 14, so 3j = 9 and j = 3.
  3. Check: 9 + 10 = 19 and 9 + 5 = 14. A jar is 3 rupees and a packet is 5 rupees.

Opposite terms: 5x + 2y = 26 and 3x − 2y = 6.

  1. The y terms are 2y and −2y: add the equations. 8x = 32, so x = 4.
  2. Then 20 + 2y = 26, so y = 3.
  3. Check: 12 − 6 = 6. (4, 3).

Scale one: 3x + 4y = 18 and 2x + 3y = 13.

  1. Multiply the first by 3 and the second by 4, so both have 12y: 9x + 12y = 54 and 8x + 12y = 52.
  2. The y terms match: subtract. x = 2.
  3. Then 6 + 4y = 18, so y = 3. Check: 4 + 9 = 13. (2, 3).

Your go. The first step is given. Solve 4x + y = 19 and 2x + 3y = 17.

  1. Multiply the first equation by 3 so both have 3y: 12x + 3y = 57.
  2. Subtract, solve for x, find y and check.
Check the answer

12x + 3y − (2x + 3y) = 57 − 17, so 10x = 40 and x = 4. Then 16 + y = 19, so y = 3. Check: 8 + 9 = 17. (4, 3).

Check before you go on

1. To eliminate y from 2x + 3y = 11 and 5x − 3y = 10, what do you do?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. x + y = 12 and x − y = 4. Add the equations and find x.

3. 4x + 3y = 27 and 4x − y = 7. Subtract the second equation from the first and find y.

Level 2 · Building

4. x + 2y = 9 and 3x + 4y = 23. Find x.

5. At a market, 2 kg of rice and 3 kg of beans cost 16 dirhams. 4 kg of rice and 1 kg of beans cost 22 dirhams. What does 1 kg of beans cost, in dirhams?

Level 3 · Stretch

6. 3x + 5y = 1 and 2x + 3y = 0. Find x.

7. 5x − 2y = 16 and 3x + 5y = −9. Find y.

Something to think about

Solve 3x + 2y = 13 and x − y = 1 twice: once by substitution, and once by elimination (multiply the second by 2, then add). Do the two routes meet at the same point? Which was shorter?

One way to think about it

Substitution: x = y + 1, so 3(y + 1) + 2y = 13, 5y = 10, y = 2 and x = 3. Elimination: 2x − 2y = 2; adding gives 5x = 15, so x = 3 and y = 2. The same point, (3, 2). Both are quick here; elimination avoids brackets, and substitution is quicker when one equation already says y = …

Exit ticket

8. You solve 3x + 2y = 7 and 5x + 4y = 13 by elimination. Which first step eliminates y?