What is (x^2+3xy+4y^2)-(2x^2-xy+3y^2-5) ?

The solution to (x^2+3xy+4y^2)-(2x^2-xy+3y^2-5) is -x^2+4xy+y^2+5

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simplify (x^2+3xy+4y^2)-(2x^2-xy+3y^2-5)

simplify

(x^{2} + 3 x y + 4 y^{2}) - (2 x^{2} - x y + 3 y^{2} - 5)

- x^{2} + 4 x y + y^{2} + 5

Solution steps

(x^{2} + 3 x y + 4 y^{2}) - (2 x^{2} - x y + 3 y^{2} - 5)

Apply rule:

(a) = a

(x^{2} + 3 x y + 4 y^{2}) = x^{2} + 3 x y + 4 y^{2}

= x^{2} + 3 x y + 4 y^{2} - (2 x^{2} - x y + 3 y^{2} - 5)

Apply the distributive law:

- (a - b) = - a + b

- (2 x^{2} - x y + 3 y^{2} - 5) = - 2 x^{2} + x y - 3 y^{2} + 5

= x^{2} + 3 x y + 4 y^{2} - 2 x^{2} + x y - 3 y^{2} + 5

Group like terms

= x^{2} - 2 x^{2} + 3 x y + x y + 4 y^{2} - 3 y^{2} + 5

Add similar elements:

x^{2} - 2 x^{2} = - x^{2}

= - x^{2} + 3 x y + x y + 4 y^{2} - 3 y^{2} + 5

Add similar elements:

3 x y + x y = 4 x y

= - x^{2} + 4 x y + 4 y^{2} - 3 y^{2} + 5

Add similar elements:

4 y^{2} - 3 y^{2} = y^{2}

= - x^{2} + 4 x y + y^{2} + 5

\frac{sin ( x ) + tan ( x )}{1 + cos ( x )}

11 \cdot 8

lo g_{10} (100) + lo g_{10} (1 0^{- 6})

(x + a) (x + b) (x + c)

(a + b)^{2} - (c + d)^{2}

What is (x^2+3xy+4y^2)-(2x^2-xy+3y^2-5) ?
The solution to (x^2+3xy+4y^2)-(2x^2-xy+3y^2-5) is -x^2+4xy+y^2+5