Question

3x²+4xy-y²
ANSWER IS ☝️☝️☝️
Simplify and verify the result for x = 2 and y = 1
(x + y)(2x + y) + (x + 2y)(x - y)​
solve and tell how answer comes ​

Answer 1

$\huge\red{AnsWeR}$

$1)(x+y)(2x+ y) + (x+ 2y)(x— y)(x+y)(2x+y)+(x+2y)(x—y)$

$=>(x)(2x+ y)+y (2x+ y)+ (x) (x— y)+2y(x— y)$

$=>2x^2+xy+2xy+ y^2+ x^2-xy+2yx- 2y^2$

$=>3x^2+4xy- y^2$ ANSWER

•VERIFY

LHS

$(x+y)(2x+ y) + (x+ 2y)(x-y)$

x = 2 and y = 1

$=>(2+1)(2×2+1)+(2+2×1)(2-1)$

$=>(3)(4+1)+(4)(1)=>(3)(4+1)+(4)(1)$

$=>3×5+4=>15+4$

$=>19$

RHS

$3x^2+4xy-y^2$

x = 2 and y = 1

$=>(3×2^2)+(4×2×1)-1^2$

$=>(3×4)+(8×1)-1$

$=>12+8-1$

$=>20-1=>19$

HOPE SO IT IS HELPFUL..❣️✌️

Answer 2

Answer:

1)(x+y)(2x+y)+(x+2y)(x—y)(x+y)(2x+y)+(x+2y)(x—y)

=>(x)(2x+ y)+y (2x+ y)+ (x) (x— y)+2y(x— y)=>(x)(2x+y)+y(2x+y)+(x)(x—y)+2y(x—y)

=>2x^2+xy+2xy+ y^2+ x^2-xy+2yx- 2y^2=>2x2+xy+2xy+y2+x2−xy+2yx−2y2

=>3x^2+4xy- y^2=>3x2+4xy−y2 ANSWER

•VERIFY

LHS

(x+y)(2x+ y) + (x+ 2y)(x-y)(x+y)(2x+y)+(x+2y)(x−y)

x = 2 and y = 1

=>(2+1)(2×2+1)+(2+2×1)(2-1)=>(2+1)(2×2+1)+(2+2×1)(2−1)

=>(3)(4+1)+(4)(1)=>(3)(4+1)+(4)(1)=>(3)(4+1)+(4)(1)=>(3)(4+1)+(4)(1)

=>3×5+4=>15+4=>3×5+4=>15+4

=>19=>19

RHS

3x^2+4xy-y^23x2+4xy−y2

x = 2 and y = 1

=>(3×2^2)+(4×2×1)-1^2=>(3×22)+(4×2×1)−12

=>(3×4)+(8×1)-1=>(3×4)+(8×1)−1

=>12+8-1=>12+8−1

=>20-1=>19=>20−1=>19

HOPE SO IT IS HELPFUL..❣️✌️