Question

❣____hola____guys____❣
Today's Question?
Q.Solve the following pairs of linear equations by cross-multiplication method:-
(i)x-y=a+b
ax+by=a^2-b^2
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Answer 1

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

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

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

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

$= > x = \frac{−a(a+b)</p><p>}{ −(a+b)</p><p> } = a \: and \: y = \frac{−b(a+b)</p><p>}{−(a+b)</p><p>} = b$

Hence, the solution of the given system of equations is x=a,y=b.

Answer 2

Answer:

Step-by-step explanation:

$x/a+y/b=2;. xb+ay=2ab$…….…..(1)

$ax-by=a^2-b^2$………………………(2)

Multiplying (1)by a &(2) by b

$abx+a^2y=2a^2b$

$abx-b^2y=b(a ^2-b^2)$ subtracting

$a^2y+b^2y=2a^2b-a^2b+b^3$

$y(a^2+b^2)=a^2b+b^3=b(a^2+b^2)$

y=b. Putting value of y in (1)

$x/a+y/b=2;. x/a+b/b=2;. x/a=2–1=1;. x=a$

hope it helps

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