1.2 Simultaneous Equations

1. Substitution Method

The most reliable method for solving systems where one equation is linear and the other is non-linear (e.g., quadratic) is substitution.

Step-by-Step Strategy

Rearrange the linear equation to make one variable the subject (e.g., $y = \dots$ or $x = \dots$ ).
Substitute this expression into the non-linear equation. This yields an equation in one variable.
Solve the resulting equation (usually a quadratic).
Substitute back your solutions into the linear equation to find the corresponding values of the other variable.
Write coordinates as pairs $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ .

Worked Examples: Substitution Method

1. Solve

y = x + 1

and

y = x^{2} - 3 x - 4

Substitute

y = x + 1

into the quadratic equation:

x + 1 = x^{2} - 3 x - 4

Rearrange to form a quadratic equation equal to zero:

x^{2} - 4 x - 5 = 0

Factorise:

(x - 5) (x + 1) = 0

So,

x = 5

x = - 1

Substitute back to find

y

When

x = 5, y = 5 + 1 = 6

When

x = - 1, y = - 1 + 1 = 0

Solutions:

(5, 6)

and

(- 1, 0)

2. Solve

2 x + y = 3

and

x^{2} + y^{2} = 5

Rearrange linear equation for

y

y = 3 - 2 x

Substitute into quadratic:

x^{2} + (3 - 2 x)^{2} = 5

Expand:

x^{2} + (9 - 12 x + 4 x^{2}) = 5

Simplify:

5 x^{2} - 12 x + 4 = 0

Factorise:

(5 x - 2) (x - 2) = 0

x = 2

x = 0.4

Find

y

x = 2 \Rightarrow y = 3 - 4 = - 1

x = 0.4 \Rightarrow y = 3 - 0.8 = 2.2

Solutions:

(2, - 1)

and

(0.4, 2.2)

3. Solve

x y = 12

and

x + y = 7

Rearrange

x + y = 7

y = 7 - x

Substitute into

x y = 12

x (7 - x) = 12

7 x - x^{2} = 12 \Rightarrow x^{2} - 7 x + 12 = 0

Factorise:

(x - 3) (x - 4) = 0

x = 3

x = 4

Find

y

x = 3 \Rightarrow y = 4

x = 4 \Rightarrow y = 3

Solutions:

(3, 4)

and

(4, 3)

4. Find the coordinates where

y = 2 x - 5

intersects

y = x^{2} - 4 x + 4

Set equations equal:

x^{2} - 4 x + 4 = 2 x - 5

x^{2} - 6 x + 9 = 0

Factorise:

(x - 3)^{2} = 0

Repeated root:

x = 3

Find

y

y = 2 (3) - 5 = 1

Solution:

(3, 1)

Note: Since there is only one point, the line is a tangent to the curve.

5. Solve

x - 2 y = 1

and

x^{2} - 2 x y + y^{2} = 9

Rearrange linear:

x = 1 + 2 y

Substitute into quadratic:

(1 + 2 y)^{2} - 2 (1 + 2 y) y + y^{2} = 9

Expand:

(1 + 4 y + 4 y^{2}) - (2 y + 4 y^{2}) + y^{2} = 9

Simplify:

1 + 4 y + 4 y^{2} - 2 y - 4 y^{2} + y^{2} = 9

y^{2} + 2 y - 8 = 0

Factorise:

(y + 4) (y - 2) = 0

y = - 4

y = 2

Find

x

y = - 4 \Rightarrow x = 1 - 8 = - 7

y = 2 \Rightarrow x = 1 + 4 = 5

Solutions:

(- 7, - 4)

and

(5, 2)

6. Solve

y = \frac{2}{x}

and

y = x - 1

Equate

y

\frac{2}{x} = x - 1

Multiply by

x

(assuming

x \neq 0

2 = x (x - 1) = x^{2} - x

x^{2} - x - 2 = 0

Factorise:

(x - 2) (x + 1) = 0

x = 2

x = - 1

Find

y

x = 2 \Rightarrow y = 1

x = - 1 \Rightarrow y = - 2

Solutions:

(2, 1)

and

(- 1, - 2)

2. Graphical Interpretation

Solving simultaneous equations algebraically corresponds to finding the points of intersection of their graphs.

2 Distinct Intersections

Discriminant $Δ > 0$

1 Intersection (Tangent)

Discriminant $Δ = 0$

No Intersection

Discriminant $Δ < 0$

When asked to prove that a line is a tangent to a curve, rely on the discriminant of the combined equation. If $b^{2} - 4 a c = 0$ , the line is a tangent.

Worked Examples: Graphical Interpretation

1. Find

k

y = x + k

is a tangent to

y = x^{2}

Substitute:

x^{2} = x + k \Rightarrow x^{2} - x - k = 0

For a tangent,

Δ = b^{2} - 4 a c = 0

(- 1)^{2} - 4 (1) (- k) = 0

1 + 4 k = 0

4 k = - 1 \Rightarrow k = - \frac{1}{4}

The tangent touches

y = x^{2}

once at

(\frac{1}{2}, \frac{1}{4})

2. How many intersections for

y = 2 x + 1

and

y = x^{2} + 3 x + 5

Equate:

x^{2} + 3 x + 5 = 2 x + 1

x^{2} + x + 4 = 0

Calculate Discriminant

Δ

Δ = 1^{2} - 4 (1) (4) = 1 - 16 = - 15

Since

Δ < 0

, there are No Real Roots.

Therefore, the line does not intersect the curve.

3. Find range of

m

for

y = m x + 1

intersecting

y = x^{2} - 2 x + 2

twice.

Equate:

x^{2} - 2 x + 2 = m x + 1

x^{2} - (2 + m) x + 1 = 0

For 2 distinct points,

Δ > 0

(- (2 + m))^{2} - 4 (1) (1) > 0

(2 + m)^{2} - 4 > 0 \Rightarrow (2 + m)^{2} > 4

2 + m > 2

2 + m < - 2

m > 0

m < - 4

4. Show that

y = 4 x - 5

is a tangent to

y = 2 x^{2} - 4 x + 3

Equate:

y = 4 x - 5

and

y = 2 x^{2} - 4 x + 3

2 x^{2} - 4 x + 3 = 4 x - 5

2 x^{2} - 8 x + 8 = 0

2 (x - 2)^{2} = 0

Tangent at

x = 2

. Point:

(2, 3)

5. Line

y = 2 x + k

cuts

x^{2} + y^{2} = 20

. Find range of

k

Substitute

y

x^{2} + (2 x + k)^{2} = 20

x^{2} + 4 x^{2} + 4 k x + k^{2} - 20 = 0

5 x^{2} + 4 k x + (k^{2} - 20) = 0

For cuts,

Δ > 0

. (Tangent if

=

)

(4 k)^{2} - 4 (5) (k^{2} - 20) > 0

16 k^{2} - 20 k^{2} + 400 > 0 \Rightarrow k^{2} < 100

- 10 < k < 10

6. Find intersection of

y = x^{2}

and

y = 2

Equate:

x^{2} = 2

x = \pm \sqrt{2}

Points:

(\sqrt{2}, 2)

and

(- \sqrt{2}, 2)

3. Mixed Practice Questions

1. Two rectangles have lengths differing by

4 cm

and equal widths. Sum of perimeters is

64 cm

. Sum of areas is

120 {cm}^{2}

. Find dimensions.

Let width be

w

. Lengths

l

and

l + 4

Perimeters:

2 (l + w) + 2 (l + 4 + w) = 64

2 l + 2 w + 2 l + 8 + 2 w = 64

4 l + 4 w = 56

l + w = 14

, so

w = 14 - l

Areas:

l w + (l + 4) w = 120

Substitute

w = 14 - l

(14 - l) (l + l + 4) = 120

(14 - l) (2 l + 4) = 120

Divide by

2

(14 - l) (l + 2) = 60

14 l + 28 - l^{2} - 2 l = 60

- l^{2} + 12 l + 28 = 60

l^{2} - 12 l + 32 = 0

Factorise:

(l - 8) (l - 4) = 0

Case 1:

l = 8

. Then

w = 14 - 8 = 6

Dimensions:

8 cm \times 6 cm

and

12 cm \times 6 cm

Case 2:

l = 4

. Then

w = 14 - 4 = 10

Dimensions:

4 cm \times 10 cm

and

8 cm \times 10 cm

Check areas, Case 1:

48 + 72 = 120

. Valid.

Check areas, Case 2:

40 + 80 = 120

. Valid.

Answer:

(l_{1}, l_{2}, w) = (8, 12, 6)

(4, 8, 10)

All measurements are in centimetres.

2. The sum of circumferences of two circles is

44 π cm

and the sum of areas is

250 π {cm}^{2}

. Find radii. (Note: area value adjusted for solvability.)

2 π r_{1} + 2 π r_{2} = 44 π

r_{1} + r_{2} = 22

π r_{1}^{2} + π r_{2}^{2} = 250 π

r_{1}^{2} + r_{2}^{2} = 250

Substitute

r_{2} = 22 - r_{1}

r_{1}^{2} + (22 - r_{1})^{2} = 250

r_{1}^{2} + 484 - 44 r_{1} + r_{1}^{2} = 250

2 r_{1}^{2} - 44 r_{1} + 234 = 0

r_{1}^{2} - 22 r_{1} + 117 = 0

Factorise:

(r_{1} - 9) (r_{1} - 13) = 0

r_{1} = 9, r_{2} = 13

r_{1} = 13, r_{2} = 9

Answer: the radii are

9 cm

and

13 cm

3. Cuboid with dimensions

x

y

, and

x + 2

. Given

x + y = 14

and volume

240 {cm}^{3}

. Find

x

and

y

Volume

V = x y (x + 2) = 240

Substitute

y = 14 - x

x (14 - x) (x + 2) = 240

(14 x - x^{2}) (x + 2) = 240

14 x^{2} + 28 x - x^{3} - 2 x^{2} = 240

- x^{3} + 12 x^{2} + 28 x - 240 = 0

x^{3} - 12 x^{2} - 28 x + 240 = 0

Try integer factors of 240. Let's try

x = 4

64 - 12 (16) - 28 (4) + 240 = 0

x = 4

. Then

y = 14 - 4 = 10

One solution is

(x, y) = (4, 10)

Dimensions:

4 cm

10 cm

, and

6 cm

Check other roots:

(x - 4) (x^{2} - 8 x - 60) = 0

x = 4 \pm 2 \sqrt{19}

Since dimensions must be positive,

x = 4 + 2 \sqrt{19}

is also valid.

Then

y = 10 - 2 \sqrt{19}

Answer:

(x, y) = (4, 10)

, or

(x, y) = (4 + 2 \sqrt{19}, 10 - 2 \sqrt{19})

4. Solid formed by a cone and a hemisphere of radius

r

. Cone height is

h

. Total height is

21 cm

. CSA is

231 π {cm}^{2}

Total height:

h + r = 21 \Rightarrow h = 21 - r

Curved Surface Area (CSA): Cone

π r l

+ Hemisphere

2 π r^{2}

π r \sqrt{r^{2} + h^{2}} + 2 π r^{2} = 231 π

Divide by

π r

\sqrt{r^{2} + h^{2}} + 2 r = \frac{231}{r}

Substitute

h = 21 - r

r \sqrt{r^{2} + (21 - r)^{2}} + 2 r^{2} = 231

Equivalently,

\sqrt{r^{2} + (21 - r)^{2}} = \frac{231}{r} - 2 r

Solving this equation numerically gives

r \approx 7.5627

Then

h = 21 - r \approx 13.4373

Answer:

r \approx 7.56 cm

and

h \approx 13.44 cm

5. Line

y = 3 x - 4

cuts curve

y = x^{2} - 2 x + 1

A

and

B

. Find length

A B

Equate:

x^{2} - 2 x + 1 = 3 x - 4

x^{2} - 5 x + 5 = 0

x = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}

x_{1} \approx 3.618, x_{2} \approx 1.382

Find

y

for each. Let coordinates be

(x_{1}, y_{1})

and

(x_{2}, y_{2})

Length

A B = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

Since

y = 3 x - 4

, the gradient is

3

Length shortcut:

| x_{2} - x_{1} | \sqrt{1 + m^{2}}

| x_{2} - x_{1} | = \sqrt{Δ} / a = \sqrt{5}

A B = \sqrt{5} \times \sqrt{1 + 3^{2}}

A B = \sqrt{50} = 5 \sqrt{2}

Answer:

A B = 5 \sqrt{2}

units.

6. Line

2 x - y = 5

intersects

x^{2} + y^{2} - 4 x + 2 y - 11 = 0

A

and

B

. Find the midpoint.

Rearrange line:

y = 2 x - 5

Substitute:

x^{2} + (2 x - 5)^{2} - 4 x + 2 (2 x - 5) - 11 = 0

x^{2} + 4 x^{2} - 20 x + 25 - 4 x + 4 x - 10 - 11 = 0

5 x^{2} - 20 x + 4 = 0

Midpoint

x

-coordinate is half the sum of roots.

x_{M} = \frac{- b / a}{2} = \frac{20 / 5}{2} = 2

Substitute

x = 2

into the line:

y = 2 (2) - 5 = - 1

Answer: the midpoint is

(2, - 1)

7. Line

x + 2 y = 7

cuts

x^{2} + y^{2} = 13

A

and

B

. Find the equation of the perpendicular bisector.

The perpendicular bisector of any chord of a circle passes through the center.

Circle

x^{2} + y^{2} = 13

has center

(0, 0)

Line

x + 2 y = 7

has gradient

- 0.5

Perpendicular gradient is

2

Equation passing through

(0, 0)

with

m = 2

Answer: the perpendicular bisector is

y = 2 x

8. Line

y = x - 3

intersects

x^{2} - y^{2} = 5

. Point

A

is in the fourth quadrant. Find

A

and

B

Substitute:

x^{2} - (x - 3)^{2} = 5

x^{2} - (x^{2} - 6 x + 9) = 5

6 x - 9 = 5 \Rightarrow 6 x = 14 \Rightarrow x = 7 / 3

Only one intersection point? Wait.

y = 7 / 3 - 3 = 7 / 3 - 9 / 3 = - 2 / 3

Point

(\frac{7}{3}, - \frac{2}{3})

. This is in the 4th quadrant.

Question implies A and B. Maybe

x^{2} - y^{2} = 5

is hyperbola. Line might be tangent?

Let's check algebra:

x^{2} - (x^{2} - 6 x + 9) = 6 x - 9

. Correct.

It is a linear equation, so only one solution. Thus A and B might be same point (tangent) or there is only one intersection.

Answer:

A = B = (\frac{7}{3}, - \frac{2}{3})