Question

what should be added to 3x² - 4y² + 5xy + 20 to obtain -x² - y² +6xy +20​

Answer 1

Answer:

Solution

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Given  

A =7X  

2

+5xy−9y  

2

 

B =4x  

2

+xy+5y  

2

 

C =4y  

2

−3x  

2

−6xy

To show A + B + C = 0

Substitute the value of A, B and C in A + B + C

A + B + C = (7x  

2

+5xy−9y  

2

)+(−4x  

2

+xy+5y  

2

)+(4y  

2

−3x  

2

−6xy)

=7x  

2

+5xy−9y  

2

−4x  

2

+xy+5y  

2

+4y  

2

−3x  

2

−6xy

Rearranging and collecting the like terms, we get:

=7x  

2

−4x  

2

−3x  

2

+5xy+xy−6xy−9y  

2

+5y  

2

+4y  

2

 

=(7−4−3)x  

2

+(5+1−6)xy−(9−5−4)y  

2

 

=0x  

2

+0xy−0y  

2

 

=0

Step-by-step explanation:

Answer 2

Given equations are,

$3x^{2} - 4y^{2} + 5xy+20$

$-x^{2} - y^{2} +6xy +20$

We have find an equation which is added to the first equation to get the second equation.

Let the equation be $A$

$3x^{2} - 4y^{2} + 5xy+20+A=-x^{2} - y^{2} +6xy +20$

Except A transpose the complete equation to RHS.

We get,

$A=-x^{2} - y^{2} +6xy +20-(3x^{2} - 4y^{2} + 5xy+20)$

$A=-x^{2} - y^{2} +6xy +20-3x^{2} +4y^{2} -5xy-20$

$A=-4x^{2} +3y^2+ xy$

The required equation is $-4x^{2} +3y^2+ xy$

Therefore, $-4x^{2} +3y^2+ xy$ is added to $3x^{2} - 4y^{2} + 5xy+20$ to get $-x^{2} - y^{2} +6xy +20$