Question

x-y=a+b,ax+by=a2-b2 using cross multiplication method

Answer 1

Answer:

x = a and y = -b

Step-by-step explanation:

Given Equations are,

x - y = a+b

x - y - (a+b) = 0 ....................(1)

ax + by = a²-b²

ax + by - (a²-b²) = 0 .......................(2)

here,

$a_1=1\:,\:b_1=-1\:,\:c_1=-(a+b)\:,\:a_2=a\:,\:b_2=b\:and\:c_2=-(a^2-b^2)$

since, $\frac{a_1}{a_2}\neq\frac{a_2}{b_2}$

There exist unique solution,

By cross multiplication method we get,

$\frac{x}{-1\times(-(a^2-b^2))-b\times(-(a+b))}=\frac{y}{-(a+b)\timesa-(-(a^2-b^2))\times1}=\frac{1}{1\timesb-a\times(-1)}$

First Consider,

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

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

$\frac{x}{a(a+b)}=\frac{1}{b+a}$

$x=a$

Secondly consider,

$\frac{y}{-(a+b)\timesa-(-(a^2-b^2))\times1}=\frac{1}{1\timesb-a\times(-1)}$

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

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

$y=-b$

Therefore, x = a and y = -b

Answer 2

Answer:

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