First half of Exercise 7(a)…

Question 3: Solve using cross multiplication method.

(i) ax+by= a-b    and   bx-ay = a+b

Answer:

The given system of equations may be written as

ax+by -(a-b) = 0

bx-ay - (a+b)= 0

By Cross multiplication, we get

\frac{x}{[ b ] \times [ -(a+b) ] - [-a] \times [ -(a-b) ]} = \frac{-y}{[ a ] \times [ -(a+b) ] - [b] \times [ -(a-b) ]} = \frac{1}{[a] \times [-a]-[b] \times [b]}

\Rightarrow \frac{x}{-ab-b^2-a^2+ab} =  \frac{-y}{-a^2-ab+ab-b^2} = \frac{1}{-a^2-b^2}

\Rightarrow \frac{x}{-b^2-a^2} =  \frac{-y}{-a^2-b^2} = \frac{1}{-(a^2+b^2)}

\Rightarrow \frac{x}{-(a^2+b^2)} =  \frac{-y}{-(a^2+b^2)} = \frac{1}{-(a^2+b^2)}

\Rightarrow x = \frac{-(a^2+b^2)}{-(a^2+b^2)} = 1   and  y = - \frac{-(a^2+b^2)}{-(a^2+b^2)} = -1

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(ii) x+y = (a+b)    and   ax-by = a^2 - b^2

Answer:

The given system of equations may be written as

x+y - (a+b) = 0

ax-by -(a^2 - b^2)= 0

By Cross multiplication, we get

\frac{x}{[ 1 ] \times [ -(a^2 - b^2) ] - [-b] \times [ - (a+b) ]} = \frac{-y}{[ 1 ] \times [ -(a^2 - b^2) ] - [a] \times [ - (a+b) ]} =  \frac{1}{[1] \times [-b]-[1] \times [a]}

\Rightarrow \frac{x}{-a^2+b^2-ab-b^2} =  \frac{-y}{-a^2+b^2+a^2+ab} = \frac{1}{-b-a}

\Rightarrow \frac{x}{-a(a+b)} =  \frac{-y}{b(a+b)} = \frac{1}{-(a+b)}

\Rightarrow x = \frac{-a(a+b)}{-(a+b)} = a   and  y = - \frac{b(a+b)}{-(a+b)} = b

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(iii) ax+by = 1    and   bx+ay = \frac{2ab}{a^2+b^2}

Answer:

The given system of equations may be written as

ax+by - 1 = 0

bx+ay - \frac{2ab}{a^2+b^2} = 0

By Cross multiplication, we get

\frac{x}{[ b ] \times [ -\frac{2ab}{a^2+b^2} ] - [a] \times [ -1 ]} = \frac{-y}{[ a ] \times [ -\frac{2ab}{a^2+b^2} ] - [b] \times [ - 1 ]} =  \frac{1}{[a] \times [a]-[b] \times [b]}

\Rightarrow \frac{x}{-\frac{2ab}{a^2+b^2}+a} =  \frac{-y}{-\frac{2ab}{a^2+b^2}+b} = \frac{1}{a^2-b^2}

\Rightarrow \Big( \frac{x}{\frac{-2ab^2+a^3+ab^2}{a^2+b^2}} \Big) =  \Big( \frac{-y}{\frac{-2a^2b+b^3+a^2b}{a^2+b^2}} \Big) = \frac{1}{a^2-b^2}

\Rightarrow \Big( \frac{x}{\frac{a(a^2-b^2)}{a^2+b^2}} \Big) =  \Big( \frac{-y}{\frac{b(b^2-a^2)}{a^2+b^2}} \Big) = \frac{1}{a^2-b^2}

\Rightarrow x =  \frac{a(a^2-b^2)}{a^2+b^2} \times \frac{1}{a^2-b^2}  = \frac{a}{a^2+b^2}

\Rightarrow y = - \frac{b(b^2-a^2)}{a^2+b^2}  \times \frac{1}{a^2-b^2} = \frac{b}{a^2+b^2} 

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(iv) \frac{a}{x} - \frac{b}{y} = 0    and   \frac{ab^2}{x} + \frac{a^2b}{y} = a^2+b^2 where x, y \neq 0

Answer:

Let \frac{1}{x} =u and \frac{1}{y} = v

The given system of equations may be written as

au-bv + 0 = 0

ab^2u + a^2bv - (a^2+b^2) = 0

By Cross multiplication, we get

\frac{u}{[ -b ] \times [ -(a^2+b^2) ] - [a^2b] \times [ 0 ]} = \frac{-v}{[ a ] \times [ -(a^2+b^2) ] - [ab^2] \times [ 0 ]} =  \frac{1}{[a] \times [a^2b]-[ab^2] \times [-b]}

\Rightarrow \frac{u}{a^2+b^3} =  \frac{-v}{-a^3-ab^2} = \frac{1}{a^3b+ab^3}

\Rightarrow u =  \frac{b(a^2+b^2)}{ab(a^2+b^2)} = \frac{1}{a}  \Rightarrow x = a

\Rightarrow -v =  \frac{-a(a^2+b^2)}{ab(a^2+b^2)} = \frac{-1}{b}  \Rightarrow y = b

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(v) 3x+2y+25=0    and   2x+y+10=0

Answer:

The given system of equations is

3x+2y+25=0

2x+y+10=0

By Cross multiplication, we get

\frac{x}{[ 2 ] \times [ 10 ] - [1] \times [ 25 ]} = \frac{-y}{[3] \times [ 10 ]- [ 2 ] \times [ 25 ]} =  \frac{1}{[3] \times [1]-[2] \times [2]}

\Rightarrow \frac{x}{-5} =  \frac{-y}{-20} = \frac{1}{-1}

\Rightarrow x =  \frac{-5}{-1} = 5

\Rightarrow y =  - ( \frac{-20}{-1} ) = -20

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(vi) ax+by = a^2    and   bx+ay=b^2

Answer:

The given system of equations may be written as

ax+by - a^2=0

bx+ay-b^2=0

By Cross multiplication, we get

\frac{x}{[ b ] \times [ -b^2 ] - [a] \times [-a^2 ]} = \frac{-y}{[b] \times [ -a^2 ]- [ a ] \times [ -b^2 ]} =  \frac{1}{[a] \times [a]-[b] \times [b]}

\Rightarrow \frac{x}{-b^3+a^3} =  \frac{-y}{-ba^2+ab^2} = \frac{1}{a^2-b^2}

\Rightarrow x =  \frac{a^3-b^3}{a^2-b^2} = \frac{(a-b)(a^2+ab+b^2)}{(a+b)(a-b)}  = \frac{a^2+ab+b^2}{a+b}  

\Rightarrow y =  - \frac{-ba^2+ab^2}{a^2-b^2} = \frac{-ab(a-b)}{(a+b)(a-b)}  = \frac{-ab}{a+b}  

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(vii) (a+2b)x+(2a-b)y=2    and   (a-2b)x+(2a+b)y=3

Answer:

The given system of equations may be written as

(a+2b)x+(2a-b)y - 2 = 0

(a-2b)x+(2a+b)y - 3 = 0

By Cross multiplication, we get

\frac{x}{[ 2a-b ] \times [ -3 ] - [2a+b] \times [ -2 ]} = \frac{-y}{[ a+2b ] \times [ -3 ] - [a-2b] \times [ - 2 ]} =  \frac{1}{[a+2b] \times [2a+b]-[a-2b] \times [2a-b]}

\Rightarrow \frac{x}{-6a+3b+4a+2b} =  \frac{-y}{-3a-6b+2a-4b} = \frac{1}{2a^2+5ab+2b^2 - 2a^2 +5ab - 2b^2}

\Rightarrow \frac{x}{-2a+5b} =  \frac{-y}{-a-10b} = \frac{1}{10ab}

\Rightarrow x =  \frac{-2a+5b}{10ab}  

\Rightarrow y = \frac{a+10b}{10ab}

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(viii) x(a-b+ \frac{ab}{a-b} )= y(a+b- \frac{ab}{a+b} )  and   x+y=2a^2

Answer:

We can simplify the equation first:

x(a-b+ \frac{ab}{a-b} )= y(a+b- \frac{ab}{a+b} )

\Rightarrow x ( \frac{a^2+b^2-ab}{a-b} ) - y ( \frac{a^2+b^2 +ab}{a+b} ) +0 = 0

\Rightarrow x( \frac{a^3+b^3}{a^2-b^2} ) - y ( \frac{a^3-b^3}{a^2-b^2}  ) +0 = 0

The given system of equations may be written as

\Rightarrow x( \frac{a^3+b^3}{a^2-b^2} ) - y ( \frac{a^3-b^3}{a^2-b^2}  ) +0 = 0

x+y-2a^2 = 0

By Cross multiplication, we get

\frac{x}{[ -\frac{a^3-b^3}{a^2-b^2} ] \times [-2a^2 ] - [0] \times [ 1 ]} = \frac{-y}{[ \frac{a^3+b^3}{a^2-b^2}] \times [ -2a^2 ] - [0] \times [ 1 ]} =  \frac{1}{[\frac{a^3+b^3}{a^2-b^2}] \times [1]-[-\frac{a^3-b^3}{a^2-b^2}] \times [1]}

\Rightarrow \frac{x}{\frac{(a^3-b^3)(2a^2)}{a^2-b^2} } =  \frac{-y}{\frac{(a^3+b^3)(-2a^2)}{a^2-b^2} } = \frac{1}{\frac{2a^3}{a^2-b^2} }

\Rightarrow \frac{x}{\frac{(a^3-b^3)(2a^2)}{a^2-b^2} } =  \frac{-y}{\frac{(a^3+b^3)(-2a^2)}{a^2-b^2} } = \frac{1}{\frac{2a^3}{a^2-b^2} }

\Rightarrow  \frac{x}{a^3-b^3} = \frac{y}{a^3+b^3} = \frac{1}{a}

\Rightarrow x =  \frac{a^3-b^3}{a}  

\Rightarrow y = \frac{a^3-b^3}{a}  

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(ix) bx+cy = a+b    and   ax( \frac{1}{a-b} - \frac{1}{a+b} ) + cy ( \frac{1}{b-a} - \frac{1}{b+a} ) = \frac{2a}{a+b}

Answer:

First simplify the equation:

ax( \frac{1}{a-b} - \frac{1}{a+b} ) + cy ( \frac{1}{b-a} - \frac{1}{b+a} ) = \frac{2a}{a+b}

\Rightarrow ax ( \frac{a+b-a+b}{a^2-b^2} ) + cy ( \frac{b+a-b+a}{b^2-a^2} ) = \frac{2a}{a+b}

\Rightarrow \frac{2ab}{a^2-b^2} x - \frac{2ac}{a^2-b^2} y = \frac{2a}{a+b}

\Rightarrow \frac{bx}{a-b} - \frac{cy}{a-b} = 1

The given system of equations may be written as

\Rightarrow \frac{bx}{a-b} - \frac{cy}{a-b} - 1 = 0

bx+cy-(a+b) = 0

By Cross multiplication, we get

\frac{x}{[ -\frac{c}{a-b} ] \times [-(a+b) ] - [-1] \times [ c ]} = \frac{-y}{[ \frac{b}{a-b}] \times [ -(a+b) ] - [-1] \times [ b ]} =  \frac{1}{[\frac{b}{a-b}] \times [c]-[-\frac{c}{a-b}] \times [b]}

\Rightarrow \frac{x}{\frac{c(a+b)}{a-b}+c } =  \frac{-y}{\frac{-b(a+b)}{a-b}+b } = \frac{1}{\frac{bc}{a-b} +\frac{bc}{a-b}   }

\Rightarrow \frac{x}{c(a+b)+c(a-b) } =  \frac{-y}{-b(a+b)+b(a-b) } = \frac{1}{2bc}

\Rightarrow  \frac{x}{2ca} = \frac{y}{2b^2} = \frac{1}{2bc}

\Rightarrow x =  \frac{2ca}{2bc} = \frac{a}{b}  

\Rightarrow y =  \frac{2b^2}{2bc} = \frac{b}{c}

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(x) (a-b)x+(a+b)y = 2a^2 - 2b^2    and   (a+b)(x+y) = 4ab

Answer:

First simplify the equations:

(a-b)x+(a+b)y = 2a^2 - 2b^2

\Rightarrow x = \frac{a+b}{a-b} y - 2 \frac{(a-b)(a+b)}{(a-b)} = 0

\Rightarrow x + \frac{a+b}{a-b} y -2(a+b) = 0

Also x + y - \frac{4ab}{a+b} = 0

By Cross multiplication, we get

\frac{x}{[ \frac{a+b}{a-b} ] \times [-\frac{4ab}{a+b}] - [-2(a+b)] \times [ 1 ]} = \frac{-y}{[ -2(a+b)]  \times [ 1 ] - [1] \times [ -\frac{4ab}{a+b}]} =  \frac{1}{[1] \times [1]-[\frac{a+b}{a-b}] \times [1]}

\Rightarrow \frac{x}{\frac{-4ab}{a-b}+2(a+b) } =  \frac{-y}{\frac{4ab}{a+b}-2(a+b) } = \frac{1}{1 -\frac{a+b}{a-b}   }

x = \frac{\frac{-4ab}{a-b}+2(a+b) }{1 -\frac{a+b}{a-b}}  = \frac{-4ab + 2(a^2-b^2)}{a - b - a - b}  = \frac{2ab-a^2+b^2}{b}

y = \frac{-2(a+b) + \frac{4ab}{a+b}}{1 - \frac{a+b}{a-b}} = \frac{-2(a+b)^2+4ab}{a-b-a-b} \times \frac{a-b}{a+b}

=  \frac{-2a^2-2b^2-4ab+4ab}{-2b}   \times \frac{a-b}{a+b}  = \frac{(a^2+b^2)(a-b)}{b(a+b)}

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(xi) \frac{a^2}{x} - \frac{b^2}{y} = 0   and   \frac{a^2b}{x} + \frac{b^2a}{y} = a+b where x, y \neq 0

Answer:

Let \frac{1}{x} = u and \frac{1}{y} = v

Therefore the given set of equations can be written as:

a^2u-b^2v+0 = 0

a^2bu + b^2av-(a+b) = 0

By Cross multiplication, we get

\frac{u}{[-b^2 ] \times [-(a+b)] - [0] \times [ b^2a]} = \frac{-v}{[ a^2]  \times [-(a+b) ] - [0] \times [ a^2b]} =  \frac{1}{[a^2] \times [-(a+b)-[-b^2] \times [a^2b]}

\Rightarrow \frac{u}{b^2(a+b)} =  \frac{-v}{-a^2(a+b) } = \frac{1}{a^3b^2+b^3a^2}

Therefore u = \frac{b^2(a+b)}{ab(a+b)} = \frac{b}{a} \Rightarrow x = \frac{a}{b}

v = \frac{a^2(a+b)}{ab(a+b)} = \frac{a}{b} \Rightarrow y = \frac{b}{a}

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