Simultaneous Equations (Three Unknowns)

Subtract term by term, taking care with the signs: subtracting -4y is the same as adding 4y.

1. x: 3x-x=2x
2. y: 2y-(-4y)=2y+4y=6y
3. z: -z-(-z)=0, so z is eliminated
4. Right-hand side: 10-(-2)=12
5. Answer: 2x+6y=12

Both of her equations in two unknowns must be missing the same letter: which letter has already gone from 4x+y=9?

1. 4x+y=9 has no z, so Jo must eliminate z again
2. She needs a different pair of equations that includes (3)
3. 2\times(2)-(3): \;4x-5y=3, which has only x and y
4. Eliminating y or x instead, or substituting into (1), leaves an equation with z in it
5. Answer: Eliminate z from equations (2) and (3)

The last equation has only one unknown, so solve it first and substitute into the other two.

1. From 3z=6: \;z=2
2. Substitute: \;x+y=3 and 2x-y=9
3. Add these: \;3x=12, so x=4
4. 4+y=3
5. Answer: y=-1

Look at the coefficients of x and z in the first two equations: one operation removes both.

1. Add the first two equations
2. x and -x cancel, and 2z and -2z cancel
3. 4y=12
4. Answer: y=3
5. (Check: x=2, z=3 completes the solution, and 4+3-3=4 in the third equation.)

Add all three equations together and look at the coefficient of each letter.

1. Add the three equations.
2. a: 4+1+2=7, b: 1+3+3=7, c: 2+3+2=7
3. So 7a+7b+7c=13+8+7=28
4. Divide by 7: \;a+b+c=4
5. (Check: a=2, b=-1, c=3 satisfies all three, and 2-1+3=4.)

Two of the equations contain only x and y: solve those as a pair first.

x=

1. The first two equations have only x and y
2. Multiply the first by 3: \;6x+3y=24
3. Add the second: \;7x=35
4. Answer: x=5

y=

1. Substitute x=5 into 2x+y=8: \;10+y=8
2. Answer: y=-2

z=

1. Substitute into x+y+2z=9: \;5-2+2z=9
2. 2z=6
3. Answer: z=3

Write three equations, one of them using the angle sum of a triangle.

1. a+b+c=180, b=a+15, c=a+b-30
2. Substitute a+b=c+30 into the first: \;2c+30=180, so c=75
3. Then a+b=105 and b=a+15
4. 2a+15=105, so a=45 and b=60
5. Answer: the smallest angle is 45^\circ

Let a pen cost p pence, a ruler r pence and an eraser e pence, and write one equation for each fact.

1. Let the costs in pence be p, r and e.
2. p+r+e=115, 2p+r=160, p=4e
3. Substitute p=4e: \;5e+r=115 and 8e+r=160
4. Subtract: \;3e=45, so e=15
5. r=115-5(15)=40
6. (Check: p=60; 60+40+15=115 and 120+40=160.)
7. A ruler costs 40p

Use one equation to write z in terms of x, then another to write y in terms of x, and substitute both into the remaining equation.

x=

1. Each equation has only two unknowns
2. From the third: \;z=5-2x
3. Substitute into the second: \;y+20-8x=-11, so y=8x-31
4. Substitute into the first: \;3x-16x+62=10
5. -13x=-52
6. Answer: x=4

y=

1. y=8x-31=32-31
2. Answer: y=1

z=

1. z=5-2x=5-8
2. Answer: z=-3
3. Check in y+4z=-11: \;1-12=-11 ✓

One equation has no b, so eliminate b from the other two and solve the result with it.

a=

1. Label the equations (1), (2), (3). Equation (3) has no b
2. Eliminate b from (1) and (2): (1)+3\times(2)
3. 5a+5c=20, so a+c=4
4. Solve with (3): a=4-c gives 12-3c+4c=13, so c=1
5. Answer: a=3

b=

1. Substitute a=3, c=1 into (2): \;3-b+2=7
2. Answer: b=-2

c=

1. From a+c=4 and (3): \;c=1
2. Check in (1): \;6-6-1=-1 ✓

Eliminate the same letter from two different pairs of equations; y is easiest here.

x=

1. Label the equations (1), (2), (3). Eliminate y twice
2. (1)+2\times(2): \;5x+7z=9 (4)
3. (2)+(3): \;5x+z=-3 (5)
4. (4) - (5): \;6z=12, so z=2
5. In (5): \;5x+2=-3
6. Answer: x=-1

y=

1. Substitute into (1): \;-1+2y+2=7
2. 2y=6
3. Answer: y=3

z=

1. From (4) - (5): \;z=2
2. Check in (2): \;-2-3+6=1 ✓ and (3): \;-3+3-4=-4 ✓

First rearrange every equation so that the x, y and z terms are on the left and the number is on the right.

x=

1. Rearrange each equation into the form ax+by+cz=d
2. 2x+y-z=9 (1), \;-x+3y+2z=1 (2), \;x+y+z=4 (3)
3. (1)+(3): \;3x+2y=13 (4)
4. (2)-2\times(3): \;-3x+y=-7 (5)
5. (4) + (5): \;3y=6, so y=2
6. In (4): \;3x+4=13
7. Answer: x=3

y=

1. From (4) + (5): \;3y=6
2. Answer: y=2

z=

1. In (3): \;3+2+z=4
2. Answer: z=-1
3. Check in 3y=x-2z+1: \;6=3+2+1 ✓

Multiply each equation that has fractions by a number that clears all of its denominators.

x=

1. Clear the fractions: multiply the first equation by 6 and the second by 4
2. 3x+2y+6z=18 (1), \;x-4y+2z=20 (2), \;2x+y-z=3 (3)
3. (1)+6\times(3): \;15x+8y=36 (4)
4. (2)+2\times(3): \;5x-2y=26 (5)
5. (4) +\,4\times(5): \;35x=140
6. Answer: x=4

y=

1. In (5): \;20-2y=26
2. -2y=6
3. Answer: y=-3

z=

1. In (3): \;8-3-z=3
2. Answer: z=2
3. Check in the first equation: \;2-1+2=3 ✓

Equation (2) has +c, so use it with each of the other two equations to eliminate c.

a=

1. Label the equations (1), (2), (3). Eliminate c using (2)
2. (1)+3\times(2): \;13a-10b=-38 (4)
3. (3)-5\times(2): \;-13a+26b=78 (5)
4. (4) + (5): \;16b=40, so b=\dfrac52
5. In (4): \;13a-25=-38, so 13a=-13
6. Answer: a=-1

b=

1. From (4) + (5): \;16b=40
2. b=\dfrac{40}{16}
3. Answer: b=\dfrac52

c=

1. In (2): \;-3-10+c=-11
2. Answer: c=2
3. Check in (3): \;-2+15+10=23 ✓

Put in the values of x, y and z: this gives three linear equations whose unknowns are a, b and c.

a=

1. Substitute x=2, y=-1, z=3 to get equations in a, b and c
2. 2a-b+3c=-1 (1), \;-3a+2b-c=0 (2), \;-a+3b+2c=7 (3)
3. From (2): \;c=2b-3a
4. In (1): \;-7a+5b=-1 (4)
5. In (3): \;-7a+7b=7, so b=a+1 (5)
6. Substitute (5) into (4): \;-2a+5=-1
7. Answer: a=3

b=

1. b=a+1
2. Answer: b=4

c=

1. c=2b-3a=8-9
2. Answer: c=-1
3. Check: 3(2)+4(-1)+(-1)(3)=-1 ✓

Label the equations, then remove the same letter from two different pairs. Here y is the easiest to eliminate because equation (1) has -y.

x=

1. Label the equations (1), (2), (3).
2. Eliminate y: 3\times(1)+(2): \;14x+12z=-11 (4)
3. Eliminate y: 2\times(1)+(3): \;14x+6z=-2 (5)
4. (4) - (5): \;6z=-9, so z=-\dfrac32
5. Substitute into (5): \;14x-9=-2, so 14x=7 and x=\dfrac12

y=

1. Using x=\dfrac12 and z=-\dfrac32:
2. Substitute into (1): \;2-y-3=-5
3. y=4

z=

1. From (4) - (5): \;6z=-9, so z=-\dfrac32
2. Check in (2): \;1+12-9=4 ✓ and (3): \;3+8-3=8 ✓

Let the three prices be letters and write one equation for each bullet point, then eliminate the same letter twice. Answer part (a) before attempting part (c).

Part (a)

1. Let the prices in pounds be A, C and S
2. 2A+3C+S=56.5 (1), \;A+2C+2S=45.5 (2), \;3A+2C+S=61.5 (3)
3. (3)-(1): \;A-C=5 (4)
4. 2\times(1)-(2): \;3A+4C=67.5 (5)
5. Substitute A=C+5 into (5): \;7C+15=67.5, so C=7.5
6. A=7.5+5
7. Answer: £12.50

Part (b)

1. From part (a): \;7C=52.5
2. Answer: £7.50
3. (A senior ticket costs £9: check 25+22.50+9=56.50 ✓)

Part (c)

1. Let n be the number of adult tickets, so 30-n are child tickets
2. 12.5n+7.5(30-n)=255
3. 5n+225=255
4. 5n=30
5. Answer: 6 adult tickets

Substitute n=3, n=5 and n=6 into an^2+bn+c to get three equations in a, b and c. Answer part (a) before attempting part (b).

Part (a)

1. Substitute n=3, 5 and 6:
2. 9a+3b+c=10 (1), \;25a+5b+c=32 (2), \;36a+6b+c=49 (3)
3. (2)-(1): \;16a+2b=22, so 8a+b=11 (4)
4. (3)-(2): \;11a+b=17 (5)
5. (5) - (4): \;3a=6, so a=2 and b=-5
6. In (1): \;18-15+c=10, so c=7
7. Answer: 2n^2-5n+7

Part (b)

1. 2n^2-5n+7=235
2. 2n^2-5n-228=0
3. (2n+19)(n-12)=0
4. n must be a positive integer, so n=12
5. Answer: the 12th term

Use the factor theorem to turn each fact into a linear equation in p, q and r. Answer part (a) before attempting part (b).

Part (a)

1. Factor theorem: \mathrm{f}(2)=0 and \mathrm{f}(-1)=0
2. \mathrm{f}(2)=0: \;4p+2q+r=-16 (1)
3. \mathrm{f}(-1)=0: \;p-q+r=2 (2)
4. \mathrm{f}(1)=-6: \;p+q+r=-8 (3)
5. (3)-(2): \;2q=-10, so q=-5
6. (1)-(3): \;3p+q=-8, so 3p=-3 and p=-1
7. In (3): \;-1-5+r=-8, so r=-2
8. Answer: \mathrm{f}(x)=2x^3-x^2-5x-2

Part (b)

1. (x-2)(x+1)=x^2-x-2 is a factor
2. 2x^3-x^2-5x-2=(x^2-x-2)(2x+1)
3. So (x-2)(x+1)(2x+1)=0
4. Answer: x=2, \;x=-1, \;x=-\dfrac12

Differentiate, then use the stationary point twice (the gradient is 0 there and the point is on the curve) and the other point once. Answer part (a) before attempting part (b).

Part (a)

1. \dfrac{dy}{dx}=3x^2+2ax+b
2. Stationary at x=-1: \;3-2a+b=0 (1)
3. (-1,\ 10) on the curve: \;a-b+c=11 (2)
4. (2,\ -17) on the curve: \;4a+2b+c=-25 (3)
5. (3)-(2): \;3a+3b=-36, so a+b=-12
6. From (1): \;b=2a-3, so 3a-3=-12 and a=-3
7. b=-9 and, from (2), c=11+3-9=5
8. Answer: y=x^3-3x^2-9x+5

Part (b)

1. \dfrac{dy}{dx}=3x^2-6x-9=3(x+1)(x-3)
2. The other stationary point is where x=3
3. y=27-27-27+5=-22
4. Answer: (3,\ -22)