Answer

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Hint: Take the quantity that should be taken away as P then write the equation as $3{{x}^{2}}-4{{y}^{2}}+5xy+20-P=-{{x}^{2}}-{{y}^{2}}+6xy+20,$ then, do necessary operations to get P which is the answer.

Complete step-by-step answer:

Let the quantity be that should be taken as P.

So, we can write it as,

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

Now taking P from right hand side of equation to left hand side we can write it as,

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

Now reversing the equation we get,

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

Now for solving an equation we will make use of facts which are,

i) If a = b then a+c = b+c for any value c. This means that we can add a number ‘c’ to both the sides of the equation and the value of the equation does not change.

ii) If a = b then a-c = b-c for any value c. This means that we can subtract a number ‘c’ from both the sides of the equation and the value of the equation does not change.

iii) If a = b then ac = bc for any non-zero value of c, so that the value of the equation remains unaltered.

iv) If a = b then $\dfrac{a}{b}=\dfrac{b}{c}$ for any non-zero value of c, so that the value of the equation remains unaltered.

These points are very important and help very much while solving any liner type of equation.

They should be kept in mind.

Now adding ${{x}^{2}}+{{y}^{2}}$ on both sides in equation (i), we get

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

Now subtracting 6xy from both the sides we get,

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

Now subtracting ‘20’ from both sides we get,

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

Hence the quantity which should be subtracted is $4{{x}^{2}}-3{{y}^{2}}-xy$

Note: Be careful about the calculation needed in the problem as they are a bit complex. There is also another way is to just subtract the 2nd expression from the 1st expression to get the answer.

Another way of solving this problem is just transferring the unknowns on the left hand side and known values as well as the same terms on the right hand side. Then solve accordingly.

Complete step-by-step answer:

Let the quantity be that should be taken as P.

So, we can write it as,

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

Now taking P from right hand side of equation to left hand side we can write it as,

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

Now reversing the equation we get,

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

Now for solving an equation we will make use of facts which are,

i) If a = b then a+c = b+c for any value c. This means that we can add a number ‘c’ to both the sides of the equation and the value of the equation does not change.

ii) If a = b then a-c = b-c for any value c. This means that we can subtract a number ‘c’ from both the sides of the equation and the value of the equation does not change.

iii) If a = b then ac = bc for any non-zero value of c, so that the value of the equation remains unaltered.

iv) If a = b then $\dfrac{a}{b}=\dfrac{b}{c}$ for any non-zero value of c, so that the value of the equation remains unaltered.

These points are very important and help very much while solving any liner type of equation.

They should be kept in mind.

Now adding ${{x}^{2}}+{{y}^{2}}$ on both sides in equation (i), we get

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

Now subtracting 6xy from both the sides we get,

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

Now subtracting ‘20’ from both sides we get,

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

Hence the quantity which should be subtracted is $4{{x}^{2}}-3{{y}^{2}}-xy$

Note: Be careful about the calculation needed in the problem as they are a bit complex. There is also another way is to just subtract the 2nd expression from the 1st expression to get the answer.

Another way of solving this problem is just transferring the unknowns on the left hand side and known values as well as the same terms on the right hand side. Then solve accordingly.

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