Question

solve by cross multiplication method. x+y=a+b,ax-by=a^2-b^2​

Answer 1

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

−(a

2

−b

2

)−(−b)×−(a+b)

x

=

−(a

2

−b

2

)−a×−(a+b)

−y

=

1×−b−a×1

1

⇒

−a

2

+b

2

−ab−b

2

x

=

−a

2

+b

2

+a

2

+ab

−y

=

−b−a

1

⇒

−a(a+b)

x

=

b(a+b)

−y

=

−(a+b)

1

⇒

−a(a+b)

x

=

−b(a+b)

y

=

−(a+b)

1

⇒x=

−(a+b)

−a(a+b)

=a and y=

−(a+b)

−b(a+b)

=b

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

Step-by-step explanation:

I hope it's helpful mark me as brainliest please

Answer 2

Step-by-step explanation:

Given:

$x + y = a + b \\ ax - by = {a}^{2} - {b}^{2}$

To find: Solve linear equations using cross multiplication method.

Solution:

If $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ are linear equations in two variables, then

using cross multiplication method

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{a_2c_1-a_1c_2}=\frac{1}{a_1b_2-a_2b_1}$

or

$\boxed{\bold{x=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}}}\\\\\boxed{\bold{y=\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}}}$

here

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

Put these values in formula

$x=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1} \\ \\ x = \frac{ - {a}^{2} + {b}^{2} + b( - a - b)}{ - b - a} \\ \\ x = \frac{ - {a}^{2} + {b}^{2} - ab - {b}^{2} }{ - (a + b)} \\ \\ x = \frac{ - a(a + b)}{- (a + b)} \\ \\ \bold{x = a}$

Same way, put the coefficients to find y

$y=\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}\\\\y = \frac{ a( - a - b) -(- {a}^{2} + {b}^{2}) }{- (a + b)} \\ \\ y = \frac{ - {a}^{2} - ab + {a}^{2} - {b}^{2} }{- (a + b)} \\ \\ y = \frac{ - b(a + b)}{ - (a + b)} \\ \\\bold{y = b}$

Final answer:

$\bold{\red{x = a}} \\ \bold{\green{y = b}}$

Hope it helps you.

To learn more:

If the system of equation 3×+2y-7=0 and k×+2y+11=o has unique solution then

https://brainly.in/question/45012709