Simultaneous Linear Equations — Exercise 5.3

Solve the following systems of simultaneous linear equations by cross-multiplication method :

3x + 2y = 4

8x + 5y = 9

Answer

Given equations can be written as,

3x + 2y - 4 = 0

8x + 5y - 9 = 0

By cross multiplication method,

$\therefore \dfrac{x}{2 \times (-9) - 5 \times (-4)} = \dfrac{y}{(-4) \times 8 - (-9) \times 3} = \dfrac{1}{3 \times 5 - 8 \times 2} \\[1em] \Rightarrow \dfrac{x}{-18 + 20} = \dfrac{y}{-32 + 27} = \dfrac{1}{15 - 16} \\[1em] \Rightarrow \dfrac{x}{2} = \dfrac{y}{-5} = \dfrac{1}{-1} \\[1em] \therefore \dfrac{x}{2} = \dfrac{1}{-1} \text{ and } \dfrac{y}{-5} = \dfrac{1}{-1} \\[1em] \Rightarrow x = -2 \text{ and } y = 5.$

Hence, x = -2 and y = 5.

Solve the following systems of simultaneous linear equations by cross-multiplication method :

3x - 7y + 10 = 0

y - 2x = 3

Answer

Given equations can be written as,

3x - 7y + 10 = 0

-2x + y - 3 = 0

By cross multiplication method,

$\therefore \dfrac{x}{(-7) \times (-3) - 1 \times 10} = \dfrac{y}{10 \times (-2) - (-3) \times 3} = \dfrac{1}{3 \times 1 - (-2) \times (-7)} \\[1em] \Rightarrow \dfrac{x}{21 - 10} = \dfrac{y}{-20 + 9} = \dfrac{1}{3 - 14} \\[1em] \Rightarrow \dfrac{x}{11} = \dfrac{y}{-11} = \dfrac{1}{-11} \\[1em] \therefore \dfrac{x}{11} = \dfrac{1}{-11} \text{ and } \dfrac{y}{-11} = \dfrac{1}{-11} \\[1em] \Rightarrow x = -1 \text{ and } y = 1.$

Hence, x = -1 and y = 1.

Solve the following system of simultaneous linear equations by cross - multiplication method :

2x - 5y = -1

3x + y = 7

Answer

Given equations can be written as,

2x - 5y + 1 = 0

3x + y - 7 = 0

By cross multiplication method,

$\therefore \dfrac{x}{-5 \times (-7) - 1 \times 1} = \dfrac{y}{1 \times 3 - (-7) \times 2} = \dfrac{1}{2 \times 1 - 3 \times -5}\\[1em] \Rightarrow \dfrac{x}{35 - 1} = \dfrac{y}{3 - (-14)} = \dfrac{1}{2 - (-15)}\\[1em] \Rightarrow \dfrac{x}{34} = \dfrac{y}{3 + 14} = \dfrac{1}{2 + 15}\\[1em] \Rightarrow \dfrac{x}{34} = \dfrac{y}{17} = \dfrac{1}{17}\\[1em] \Rightarrow \dfrac{x}{34} = \dfrac{1}{17} \text{ and } \dfrac{y}{17} = \dfrac{1}{17}\\[1em] \Rightarrow x = \dfrac{34}{17} \text{ and }y = \dfrac{17}{17}\\[1em] \Rightarrow x = 2 \text{ and }y = 1.$

Hence, x = 2 and y = 1.

Solve the following system of simultaneous linear equations by cross-multiplication method :

x + 3y + 4 = 0

3x - y = -2

Answer

Given equations can be written as,

x + 3y + 4 = 0

3x - y + 2 = 0

By cross multiplication method,

$\therefore \dfrac{x}{3 \times 2 - (-1) \times 4} = \dfrac{y}{4 \times 3 - 2 \times 1} = \dfrac{1}{1 \times (-1) - 3 \times 3}\\[1em] \Rightarrow \dfrac{x}{6 + 4} = \dfrac{y}{12 - 2} = \dfrac{1}{-1 - 9}\\[1em] \Rightarrow \dfrac{x}{10} = \dfrac{y}{10} = \dfrac{1}{-10}\\[1em] \Rightarrow \dfrac{x}{10} = \dfrac{1}{-10} \text{ and }\dfrac{y}{10} = \dfrac{1}{-10}\\[1em] \Rightarrow x = -\dfrac{10}{10} \text{ and }y = -\dfrac{10}{10}\\[1em] \Rightarrow x = -1 \text{ and }y = -1$

Hence, x = -1 and y = -1.

Solve the following pairs of linear equations by cross-multiplication method:

x - y = a + b

ax + by = a² - b²

Answer

Given equations can be written as,

x - y - (a + b) = 0

ax + by - (a² - b²) = 0

By cross multiplication method,

$\therefore \dfrac{x}{(-1) \times [-(a^2 - b^2)] - b \times [-(a + b)]} = \dfrac{y}{[-(a + b) \times a] - [-(a^2 - b^2) \times 1] } = \dfrac{1}{1 \times b - a \times (-1)} \\[1em] \Rightarrow \dfrac{x}{a^2 - b^2 + ab + b^2} = \dfrac{y}{-a^2 - ab + a^2 - b^2} = \dfrac{1}{b + a} \\[1em] \Rightarrow \dfrac{x}{a^2 + ab} = \dfrac{y}{-ab - b^2} = \dfrac{1}{b + a} \\[1em] \therefore \dfrac{x}{a^2 + ab} = \dfrac{1}{b + a} \text{ and } \dfrac{y}{-b(a + b)} = \dfrac{1}{b + a} \\[1em] \Rightarrow \dfrac{x}{a(a + b)} = \dfrac{1}{b + a} \text{ and } \dfrac{y}{-b(a + b)} = \dfrac{1}{b + a} \\[1em] \Rightarrow x = \dfrac{a(a + b)}{b + a} \text{ and } y = \dfrac{-b(a + b)}{b + a} \\[1em] \Rightarrow x = a \text{ and } y = -b.$

Hence, x = a and y = -b.

Solve the following pairs of linear equations by cross-multiplication method:

2bx + ay = 2ab

bx - ay = 4ab.

Answer

Given equations can be written as,

2bx + ay - 2ab = 0

bx - ay - 4ab = 0

By cross multiplication method,

$\therefore \dfrac{x}{a \times -(4ab) - (-a) \times [-2ab]} = \dfrac{y}{[-(2ab) \times b] - [-(4ab) \times 2b] } = \dfrac{1}{2b \times (-a) - b \times a} \\[1em] \Rightarrow \dfrac{x}{-4a^2b - 2a^2b} = \dfrac{y}{-2ab^2 + 8ab^2} = \dfrac{1}{-2ab - ab} \\[1em] \Rightarrow -\dfrac{x}{6a^2b} = \dfrac{y}{6ab^2} = -\dfrac{1}{3ab} \\[1em] \therefore -\dfrac{x}{6a^2b} = -\dfrac{1}{3ab} \text{ and } \dfrac{y}{6ab^2} = -\dfrac{1}{3ab} \\[1em] \Rightarrow x = \dfrac{6a^2b}{3ab} \text{ and } y = -\dfrac{6ab^2}{3ab} \\[1em] \Rightarrow x = 2a \text{ and } y = -2b.$

Hence, x = 2a and y = -2b.