Lesson 8, Topic 2
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Simultaneous Equations – One Linear & One Quadratic

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Example 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)

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 + 4x + 12 = 0

x2 – 4x -12 = 0  (Factorise)

(x + 2)(x – 6) =0

x = -2       or       x = 6

Substitute for x in equation (iii)

When x = -2; y= 4 – 2(-2)

i.e.     y = 4 + 4

y = 8

When x = 6;  y = 4 – 2(6)

y = 4 -12

= -8

∴      the coordinate (x, y) is given as

(-2,  8)         or        (6, -8)

(ii)    x2+ y2 = 29              ………..(i)

x + y = 7                    …………(ii)

Re-write as       y = 7 – x   ……… (iii)

Substitute for y in equation  (i)

i.e.          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)

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

y2 + 10y – 24 = 0

(y – 2) (y + 12) = 0

y = 2       or           y = -12

Substitute for y in equation (iii)

When y = 2,    x = 5 – 2

x = 3        i.e.   (3, 2)

When y = -12,   x = 5 – (-12)

x = 5 + 12

x = 17        i.e.   (17,  -12)

(iv)       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

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)

When x = -2, y = 5(-2) + 3

= -10 + 3

Y=y = -7

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

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

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