Simultaneous Equations

Put each pair of values into both equations, not just the first one.

Adding the two equations makes the y terms cancel.

Both equations have one y, so subtracting one equation from the other makes the y terms cancel.

Look at the signs in front of 2y in each equation: should you add or subtract?

Replace y in the second equation with x + 1.

Multiply the first equation by 2 so that both equations have 2y.

Part (a)

1. Multiply the first equation by 2: 4x+2y=26.
2. Subtract 3x+2y=20: (4x+2y)-(3x+2y)=26-20, so x=6.

Part (b)

1. Put x=6 into 2x+y=13: 12+y=13, so y=1.
2. Check: 3\times6+2\times1=18+2=20.

Multiply the first equation by 3 so that both equations have 3x.

Part (a)

1. Multiply the first equation by 3: 3x+6y=30.
2. Subtract 3x+5y=27: (3x+6y)-(3x+5y)=30-27, so y=3.
3. Put y=3 into x+2y=10: x+6=10, so x=4.

Part (b)

1. y=3, found in part (a).
2. Check: 3\times4+5\times3=12+15=27.

Both equations have +2y, so subtract one equation from the other.

Part (a)

1. Subtract the second equation from the first: (4x+2y)-(x+2y)=9-3, so 3x=6.
2. x=2.

Part (b)

1. Put x=2 into x+2y=3: 2+2y=3, so 2y=1.
2. y=0.5.
3. Check: 4\times2+2\times0.5=8+1=9.

The y terms are +2y and −2y. Adding the equations makes them cancel.

Part (a)

1. Add the equations: (3x+2y)+(x-2y)=5+7, so 4x=12.
2. x=3.

Part (b)

1. Put x=3 into 3x+2y=5: 9+2y=5, so 2y=-4.
2. y=-2.
3. Check: 3-2\times(-2)=3+4=7.

Rewrite the first equation with the x term first. Then multiply the second equation by 3 so that the y terms will cancel.

Part (a)

1. Write the first equation as 2x+3y=13.
2. Multiply the second equation by 3: 3x-3y=12.
3. Add the two equations: the y terms cancel, so 5x=25.
4. x=5.

Part (b)

1. Put x=5 into x-y=4: 5-y=4, so y=1.
2. Check: 3\times1+2\times5=3+10=13.

Multiply the second equation by 3. Then the y terms are −3y and +3y.

Part (a)

1. Multiply the second equation by 3: 15x+3y=39.
2. Add 2x-3y=-5: 17x=34, so x=2.

Part (b)

1. Put x=2 into 5x+y=13: 10+y=13, so y=3.
2. Check: 2\times2-3\times3=4-9=-5.

Compare the two orders: what is the difference between them, and what does it cost?

Part (a)

1. The first order has 2 more drinks than the second.
2. So 2 drinks cost £12.30 − £8.10 = £4.20.
3. One drink costs £4.20 ÷ 2 = £2.10

Part (b)

1. 2 sandwiches and 1 drink cost £8.10, so 2 sandwiches cost £8.10 − £2.10 = £6.00
2. One sandwich costs £6.00 ÷ 2 = £3.00
3. Check: 2 × £3.00 + 3 × £2.10 = £6.00 + £6.30 = £12.30

Let the long tile be L cm and the short tile be S cm, and write an equation for each row.

Part (a)

1. The rows give 3L+2S=64 and L+4S=48.
2. Multiply the first equation by 2: 6L+4S=128.
3. Subtract L+4S=48: 5L=80, so L=16.
4. A long tile is 16 cm.

Part (b)

1. Put L=16 into L+4S=48: 16+4S=48, so 4S=32.
2. S=8, so a short tile is 8 cm.
3. Check: 3\times16+2\times8=48+16=64.

Test each pair in both equations. Take care with the negative numbers.

1. x=-2, y=5: 2x+y=-4+5=1 is true, but x-3y=-2-15=-17, not 11.
2. x=5, y=-2: 2x+y=10-2=8, not 1.
3. x=-3, y=2: 2x+y=-6+2=-4, not 1.
4. x=2, y=-3: 2x+y=4-3=1 and x-3y=2+9=11. Both are true.
5. So x=2 and y=-3.

When you multiply an equation, every term on both sides must be multiplied.

Part (a)

1. Multiplying x+y=8 by 3 gives 3x+3y=24.
2. Ella multiplied the left-hand side but not the 8 on the right-hand side.

Part (b)

1. The correct equation is 3x+3y=24.
2. Subtract it from 4x+3y=29: x=5.

Part (c)

1. Put x=5 into x+y=8: y=3.
2. Check: 4\times5+3\times3=20+9=29.

Multiply the first equation by 3 so that the y terms are −3y and +3y.

Part (a)

1. Multiply the first equation by 3: 12x-3y=39.
2. Add 2x+3y=-4: 14x=39+(-4)=35.
3. x=35\div14=2.5.

Part (b)

1. Put x=2.5 into 4x-y=13: 10-y=13, so y=-3.
2. Check: 2\times2.5+3\times(-3)=5-9=-4.

Let a be the number of adult tickets and c the number of child tickets. Write one equation for the number of tickets and one for the money.

Part (a)

1. Number of tickets: a+c=150.
2. Money: 8a+5c=1020.
3. Multiply the first equation by 5: 5a+5c=750.
4. Subtract from 8a+5c=1020: 3a=270, so a=90.
5. 90 adult tickets were sold.

Part (b)

1. c=150-90=60.
2. Check: 90\times8+60\times5=720+300=1020.

Find the cost of one bag of crisps and one bottle of water first.

1. Let a bag of crisps cost c pounds and a bottle of water cost w pounds.
2. 2c+w=2.30 and c+3w=3.40.
3. Multiply the first equation by 3: 6c+3w=6.90.
4. Subtract c+3w=3.40: 5c=3.50, so c=0.70.
5. Then w=2.30-2\times0.70=0.90.
6. 3c+2w=3\times0.70+2\times0.90=2.10+1.80=3.90
7. The total cost is £3.90

Work in pence. Write one equation for the number of coins and one for their value.

1. Let Jo have t twenty-pence coins and f fifty-pence coins.
2. Number of coins: t+f=24. Value in pence: 20t+50f=810.
3. Multiply the first equation by 20: 20t+20f=480.
4. Subtract from 20t+50f=810: 30f=330, so f=11.
5. Then t=24-11=13. Check: 13\times20+11\times50=260+550=810.
6. Jo has 11 fifty-pence coins.

Let the call-out fee be £c and the hourly charge be £h. Write an equation for each job.

Part (a)

1. c+2h=110 and c+5h=215.
2. Subtract the first equation from the second: 3h=105, so h=35.
3. The plumber charges £35 for each hour.

Part (b)

1. c+2\times35=110, so c=110-70=40.
2. The call-out fee is £40

Part (c)

1. Take away the call-out fee: £180 − £40 = £140
2. £140 ÷ £35 = 4
3. The job took 4 hours.