First half of Exercise 7(a)…

Question 3: Solve using cross multiplication method.

(i) $ax+by= a-b$   and   $bx-ay = a+b$

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   $ax-by = a^2 - b^2$

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

$\\$

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

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 $x, y \neq 0$

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

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

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

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

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

$\\$

(v) $3x+2y+25=0$   and   $2x+y+10=0$

The given system of equations is

$3x+2y+25=0$

$2x+y+10=0$

By Cross multiplication, we get

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

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

$\Rightarrow x =$ $\frac{-5}{-1}$ $= 5$

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

$\\$

(vi) $ax+by = a^2$   and   $bx+ay=b^2$

The given system of equations may be written as

$ax+by - a^2=0$

$bx+ay-b^2=0$

By Cross multiplication, we get

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

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

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

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

$\\$

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

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

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

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

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

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

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

$\\$

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

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

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

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

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

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

$\Rightarrow x =$ $\frac{a^3-b^3}{a}$

$\Rightarrow y =$ $\frac{a^3-b^3}{a}$

$\\$

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

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

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

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

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

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

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

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

$\\$

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

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

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

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

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

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

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

$\\$

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

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

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

$\\$