Simultaneous Equations – One Linear & One Quadratic

A linear equation is an equation that does not contain any powers higher than 1.

A quadratic equation is an equation with the highest power of 2.

Example 2.2.1:

Solve the following simultaneous equations:

i. 2x + y = 4;       x² + xy = -12
ii. x² + y² = 29;    x + y = 7
iii. x + y = 5;           x² – 2y² = 1
iv. y – 5x = 3;          x² – y² = -45                    (WAEC)

i. 2x + y = 4;       x² + xy = -12

Solution:

(i) Let 

x² + xy = -12  ……….(i)

2x + y = 4 ………..(ii)

Re-write as

y = 4 – 2x    ………….(iii)

Substitute for y in equation (i) 

x² + x(4 – 2x) = -12     

Expand brackets

⇒ x² + 4x – 2x² = -12

⇒ x² – 2x² + 4x + 12 = 0

⇒ -x² + 4x + 12 = 0

multiply through by -1

⇒ x² – 4x -12 = 0 

Factorise

⇒ x² – 6x + 2x -12 = 0 

⇒ x(x – 6) + 2(x -6) = 0 

⇒ (x + 2)(x – 6) =0

∴ x = -2       or       x = 6

Substitute for x in equation (iii)

⇒ y = 4 – 2x    ………….(iii)

When x = -2

⇒ y = 4 – 2(-2)

∴ y = 4 + 4

∴ y = 8

When x = 6

⇒ y = 4 – 2(6)

⇒ y = 4 -12

y = -8

∴  the coordinate (x, y) is given as

(-2,  8)         or        (6, -8)

ii. x² + y² = 29;    x + y = 7

Solution

⇒ x² + y² = 29              ………..(i)

 ⇒ x + y = 7                    …………(ii)

Re-write as      

⇒ y = 7 – x   ……… (iii)

Substitute for y in equation  (i)

⇒ x² + (7 – x)²         ………….(iv)

⇒ x² + 49 – 14x + x² = 29

⇒ 2x² – 14x + 20 = 0

Divide both sides by 2

⇒ x² – 7x + 10 = 0

⇒ (x – 2)(x – 5) = 0

∴ x = 2    or     x = 5

Substitute for x in equation (iii)

y = 7 – 2           or           y = 7 – 5

y = 5          or            y = 2

When x = 2, y = 5     or            when x = 5, y = 2

When x = 2, y = 5      or            when x = 5, y = 2

⇒ (2, 5) (5, 2)

iii. x + y = 5;           x² – 2y² = 1 

Solution

⇒ x² – 2y² = 1             ……….(i)

⇒ x + y = 5                     ………(ii)

⇒ x = 5 – y                           ……………(iii)

Substitute for x in equation (i)

⇒(5 – y)² – 2y² = 1

⇒ 25 – 10y + y² – 2y² = 1

⇒ 25 – 10y – y² = 1

⇒ – y² – 10 + 25 – 1 = 0

⇒ – y² – 10 + 24 = 0

multiply through by -1

⇒ y² + 10y – 24 = 0

factorise

⇒ y² + 12y – 2y – 24 = 0

⇒ y(y + 12) – 2(y + 12) = 0

⇒ (y – 2) (y + 12) = 0

∴ y = 2       or           y = -12

Substitute for y in equation (iii)

⇒ x = 5 – y ……………(iii)

When y = 2  

⇒ x = 5 – 2

 ⇒ x = 3       

∴ (x, y) =  (3, 2)

When y = -12

⇒ x = 5 – (-12)

⇒ x = 5 + 12

⇒ x = 17       

∴ (x, y) =  (17,  -12)

iv. y – 5x = 3;          x² – y² = -45 

Solution    

x² – y² = -45       ………..(i)

y – 5x = 3             …………(ii)

y = 5x + 3            …………(iii)

Substitute for y in equation (i) 

⇒ x² – (5x + 3)² = -45

⇒ x² – [25x² + 30x + 9] = -45

⇒ x² – 25x² – 30x – 9 = -45

⇒ – 24x² – 30x – 9 + 45 = 0

⇒ – 24x² – 30x + 36 = 0

multiply through by -1

⇒ 24x² + 30x – 36 = 0

⇒ 4x² + 5x – 6 = 0

factorise

(4x -3)(x + 2) = 0

4x – 3 = 0         or        x + 2 = 0

4x = 3            or         x = -2

x = \( \frac {3}{4} \)             or           x = -2

Substitute for x in equation (iii)

y = 5x + 3 …………(iii)

When x = -2

y = 5(-2) + 3

= -10 + 3

y = -7

When x =\( \frac {3}{4} \),    y = \(\scriptsize 5 \left( \frac {3}{4}\right) \:+ \:3 \\=\frac{15}{4}\: + \: \frac {3}{1} \\ = \frac{15 \:+\: 12}{4} \\ = \frac{27}{4}\\ = \scriptsize 6 \normalsize \frac{3}{4} \)

Answer:    (-2,  -7)       or        \(\left(\frac {3}{4}, \: \scriptsize 6 \normalsize \frac{3}{4} \right ) \)