Simultaneous Equations (Two Unknowns)

Each value of x has its own value of y: substitute it into one of the original equations.

1. Substitute each x into y=5-2x
2. x=2: y=5-4=1
3. x=-4: y=5+8=13
4. Check in y=x^2-3: 2^2-3=1 ✓ and (-4)^2-3=16-3=13 ✓
5. The y values are not 0: the solutions are where the two graphs meet, not where they cross the x-axis.
6. Answer: x=2,\ y=1 and x=-4,\ y=13

Write (x+1)^2 as (x+1)(x+1) and expand all four terms before simplifying.

1. x^2+(x+1)^2=61
2. (x+1)^2=(x+1)(x+1)=x^2+2x+1 (not x^2+1: you can't square term by term)
3. x^2+x^2+2x+1=61
4. 2x^2+2x-60=0
5. Divide by 2: x^2+x-30=0
6. Answer: x^2+x-30=0

A solution of both equations is a point that lies on both graphs, so look for where they cross.

Part (a)

1. The solutions are the points where the line meets the circle.
2. The line meets the circle on the y-axis at (0,\,-5)
3. Answer: x=0,\ y=-5

Part (b)

1. The other crossing point is at (4,\,3)
2. Check: 4^2+3^2=16+9=25 ✓ and 2(4)-5=3 ✓
3. Answer: x=4,\ y=3

The first equation tells you what y is, so replace y in the second equation with 3x-7.

x=

1. Substitute y=3x-7 into the second equation: 2x+5(3x-7)=33
2. 2x+15x-35=33
3. 17x=68
4. Answer: x=4

y=

1. y=3(4)-7=5
2. Check: 2(4)+5(5)=8+25=33 ✓
3. Answer: y=5

Let P be (x,\,y) and multiply the matrix by the column vector \begin{pmatrix}x\\y\end{pmatrix} to get two equations.

1. Let P=(x,\,y)
2. \begin{pmatrix}3&-2\\1&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}3x-2y\\x+4y\end{pmatrix}
3. So 3x-2y=13 and x+4y=9
4. Double the first: 6x-4y=26
5. Add to x+4y=9: 7x=35, so x=5
6. 5+4y=9, so y=1
7. Answer: P=(5,\,1)

Multiply the first equation by 2 so the y terms can be eliminated by adding.

1. The point where two lines meet satisfies both equations, so solve them simultaneously.
2. Double the first equation: 10x+4y=6
3. Add to 3x-4y=20: 13x=26, so x=2
4. Substitute into 5x+2y=3: 10+2y=3, so 2y=-7 and y=-3.5
5. Check in the second equation: 3(2)-4(-3.5)=6+14=20 ✓
6. Answer: P=(2,\,-3.5)

Clear the fractions first: multiply every term of each equation by the lowest common multiple of its denominators.

x=

1. Multiply the first equation by 12: 4(x+2)+3y=24, so 4x+3y=16
2. Multiply the second equation by 5: 2x+y=10, so y=10-2x
3. Substitute: 4x+3(10-2x)=16
4. 4x+30-6x=16, so -2x=-14
5. Answer: x=7

y=

1. y=10-2(7)=-4
2. Check: \dfrac{7+2}{3}+\dfrac{-4}{4}=3-1=2 ✓
3. Answer: y=-4

Write two equations using f for the cost of a flat white and c for a croissant, then double the second so the f terms match.

1. Let a flat white cost £f and a croissant £c
2. 4f+3c=14.10 and 2f+5c=12.30
3. Double the second: 4f+10c=24.60
4. Subtract the first: 7c=10.50, so c=1.50
5. 2f+5(1.50)=12.30, so 2f=4.80
6. Answer: £2.40

Substitute each point's x and y into y=ax^2+bx to get two equations in a and b. Answer part (a) before attempting part (b).

Part (a)

1. Substitute (2,\,10): 4a+2b=10, so 2a+b=5
2. Substitute (-1,\,7): a-b=7
3. Add the two equations: 3a=12
4. Answer: a=4

Part (b)

1. a-b=7 with a=4: 4-b=7
2. Check: 4(2)^2-3(2)=16-6=10 ✓ and 4(-1)^2-3(-1)=4+3=7 ✓
3. Answer: b=-3

Set the two expressions for y equal to each other and rearrange to a quadratic equal to 0. Answer part (a) before attempting part (b).

Part (a)

1. At the points where they meet, the y values are equal: 2x^2-3x-1=x+5
2. 2x^2-4x-6=0
3. Divide by 2: x^2-2x-3=0
4. (x+1)(x-3)=0, so x=-1 or x=3
5. The smaller is x=-1, and y=-1+5=4
6. Answer: (-1,\,4)

Part (b)

1. From part (a), the other point has x=3
2. y=3+5=8
3. Check: 2(9)-9-1=8 ✓
4. Answer: (3,\,8)

Treat k like a number: eliminate y exactly as you would if the right-hand sides were 9 and 8.

x=

1. Double the first equation: 10x+4y=18k
2. Add 3x-4y=8k: 13x=26k
3. Answer: x=2k

y=

1. Substitute x=2k into 5x+2y=9k: 10k+2y=9k
2. 2y=-k
3. Answer: y=-\dfrac{k}{2}

Make y the subject of the line, substitute into the circle and expand the squared bracket in full.

1. From the line: y=10-3x
2. Substitute: x^2+(10-3x)^2=10
3. (10-3x)(10-3x)=100-60x+9x^2
4. x^2+100-60x+9x^2=10, so 10x^2-60x+90=0
5. Divide by 10: x^2-6x+9=0
6. (x-3)^2=0, so x=3 (one repeated root, as the line is a tangent)
7. y=10-3(3)=1
8. Answer: P=(3,\,1)

The first equation already gives x in terms of y, so substitute it into the second and solve the quadratic in y.

x=

1. Substitute x=y^2-4 into x+2y=4: y^2-4+2y=4
2. y^2+2y-8=0
3. (y-2)(y+4)=0, so y=2 or y=-4
4. y=2 is the solution given, so the other has y=-4
5. x=(-4)^2-4=16-4
6. Answer: x=12

y=

1. From the working for x: y=-4
2. Check: 12+2(-4)=12-8=4 ✓
3. Answer: y=-4

Substitute y=2x+1 into xy=10 to get a quadratic in x, then pick the solution that fits the condition.

1. Substitute: x(2x+1)=10
2. 2x^2+x-10=0
3. (2x+5)(x-2)=0, so x=-\dfrac{5}{2} or x=2
4. x=2 gives y=5 (both positive), so use x=-\dfrac{5}{2}
5. y=2\left(-\dfrac{5}{2}\right)+1=-5+1=-4
6. Check: \left(-\dfrac{5}{2}\right)(-4)=10 ✓
7. Answer: \left(-\dfrac{5}{2},\,-4\right)

Substitute n=3 and n=6 into an^2+bn to get two equations in a and b.

1. n=3: 9a+3b=12, so 3a+b=4
2. n=6: 36a+6b=78, so 6a+b=13
3. Subtract: 3a=9, so a=3
4. 3(3)+b=4, so b=-5
5. nth term =3n^2-5n
6. 10th term =3(100)-5(10)=300-50
7. Answer: 250

Multiply the first equation by 2 so it has no fractions, then make x the subject of xy=-9 and substitute. Answer part (a) before attempting part (b).

Part (a)

1. Clear the fraction: multiply the first equation by 2 to get 4x+6y=15
2. From xy=-9: x=-\dfrac{9}{y}
3. Substitute: -\dfrac{36}{y}+6y=15
4. Multiply by y: -36+6y^2=15y
5. 6y^2-15y-36=0, and dividing by 3: 2y^2-5y-12=0
6. (2y+3)(y-4)=0, so y=-\dfrac{3}{2} or y=4
7. y=-\dfrac{3}{2} gives x=-9\div\left(-\dfrac{3}{2}\right)=6
8. Answer: x=6,\ y=-\dfrac{3}{2}

Part (b)

1. From part (a), the other solution has y=4
2. x=-9\div4=-\dfrac{9}{4}
3. Check: 2\left(-\dfrac{9}{4}\right)+3(4)=-\dfrac{9}{2}+12=\dfrac{15}{2} ✓
4. Answer: x=-\dfrac{9}{4},\ y=4

Write one equation from the perimeter and one from Pythagoras' theorem, then substitute. Answer part (a) before attempting part (b).

Part (a)

1. Let the other two sides be a cm and b cm
2. Perimeter: a+b+29=70, so a+b=41
3. Pythagoras: a^2+b^2=29^2=841
4. Substitute b=41-a: a^2+(41-a)^2=841
5. a^2+1681-82a+a^2=841
6. 2a^2-82a+840=0, so a^2-41a+420=0
7. (a-20)(a-21)=0, so a=20 or a=21
8. If a=20 then b=21, and if a=21 then b=20
9. Answer: 20 cm and 21 cm

Part (b)

1. The smallest angle is opposite the shortest side, 20 cm
2. \tan\theta=\dfrac{20}{21}
3. \theta=\tan^{-1}\left(\dfrac{20}{21}\right)=43.602\ldots^\circ
4. Answer: 43.6^\circ

Differentiate to find the gradient of the tangent at A; the normal's gradient is the negative reciprocal. Answer part (a) before attempting part (b).

Part (a)

1. y=3^2-4(3)+7=4, so A=(3,\,4)
2. \dfrac{dy}{dx}=2x-4, which is 2 at x=3
3. The normal is perpendicular, so its gradient is -\dfrac{1}{2}
4. y-4=-\dfrac{1}{2}(x-3)
5. Multiply by 2: 2y-8=-x+3
6. Answer: x+2y=11

Part (b)

1. From the normal: y=\dfrac{11-x}{2}
2. Substitute into the curve: \dfrac{11-x}{2}=x^2-4x+7
3. Multiply by 2: 11-x=2x^2-8x+14
4. 2x^2-7x+3=0
5. (2x-1)(x-3)=0, so x=\dfrac{1}{2} or x=3
6. x=3 is A, so B has x=\dfrac{1}{2}
7. y=\dfrac{11-0.5}{2}=5.25
8. Answer: B=\left(\dfrac{1}{2},\,\dfrac{21}{4}\right)

Find the equation of L first, then make x the subject (to avoid fractions) and substitute into the circle equation.

Part (a)

1. Use y-y_1=m(x-x_1): y-5=-\dfrac{1}{2}(x-5)
2. Multiply by 2: 2y-10=-(x-5)=-x+5
3. Rearrange: x+2y=15
4. Answer: x+2y=15

Part (b)

1. From L: x=15-2y
2. Substitute into the circle: (15-2y)^2+y^2=170
3. (15-2y)(15-2y)=225-60y+4y^2, so 225-60y+4y^2+y^2=170
4. 5y^2-60y+55=0, and dividing by 5: y^2-12y+11=0
5. (y-1)(y-11)=0, so y=1 or y=11
6. y=1 gives x=15-2=13; y=11 gives x=15-22=-7
7. A has the positive x-coordinate.
8. Answer: A=(13,\,1)

Part (c)

1. From part (b), the other solution is y=11, x=-7.
2. Check: (-7)^2+11^2=49+121=170 ✓ and -7+2(11)=15 ✓
3. Answer: B=(-7,\,11)

Write the circle as (x-a)^2+(y-b)^2=r^2, substitute for y and expand each squared bracket in full. Answer part (a) before attempting part (b).

Part (a)

1. The circle is (x-3)^2+(y+1)^2=25
2. Substitute y=2x-3: y+1=2x-2
3. (x-3)^2+(2x-2)^2=25
4. x^2-6x+9+4x^2-8x+4=25
5. Answer: 5x^2-14x-12=0

Part (b)

1. Use the quadratic formula: x=\dfrac{14\pm\sqrt{(-14)^2-4(5)(-12)}}{2(5)}
2. x=\dfrac{14\pm\sqrt{436}}{10}
3. x=3.488\ldots or x=-0.688\ldots
4. A has the negative x-coordinate: x=-0.688\ldots
5. y=2(-0.688\ldots)-3=-4.376\ldots
6. Answer: A=(-0.69,\,-4.38)

Part (c)

1. B has x=3.488\ldots and y=2(3.488\ldots)-3=3.976\ldots
2. Horizontal distance: 3.488\ldots-(-0.688\ldots)=4.176\ldots
3. Vertical distance: 3.976\ldots-(-4.376\ldots)=8.352\ldots
4. AB=\sqrt{4.176\ldots^2+8.352\ldots^2}=9.338\ldots
5. Answer: 9.34