How do you find the sum or difference of $(4 a - 5 b^{2} + 3) + (6 - 2 a + 3 b^{2})$ $(4a-5b^2+3)+(6-2a+3b^2)$ ?

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Answer 1 · 2018-05-13T18:43:37

$2 a - 2 b^{2} + 9$ $2a-2b^2+9$

To simplify

$4 a - 5 b^{2} + 3 + 6 - 2 a + 3 b^{2}$ $4a-5b^2+3+6-2a+3b^2$

$2 a - 2 b^{2} + 9$ $2a-2b^2+9$

Answer 2 · 2018-05-13T18:49:08

$(4 a - 5 b^{2} + 3) + (6 - 2 a + 3 b^{2}) = 2 a - 2 b^{2} + 9$ $(4a-5b^2+3) + (6-2a+3b^2)=color(blue)(2a-2b^2+9$

Simplify:

$(4 a - 5 b^{2} + 3) + (6 - 2 a + 3 b^{2})$ $(4a-5b^2+3) + (6-2a+3b^2)$

Simplify the parentheses.

$4 a - 5 b^{2} + 3 + 6 - 2 a + 3 b^{2}$ $4a-5b^2+3+6-2a+3b^2$

Collect like terms.

$(4 a - 2 a) + (- 5 b^{2} + 3 b^{2}) + (3 + 6)$ $(4a-2a)+(-5b^2+3b^2)+(3+6)$

Combine like terms.

$2 a - 2 b^{2} + 9$ $2a-2b^2+9$

How do you find the sum or difference of (4a−5b2+3)+(6−2a+3b2)(4a-5b^2+3)+(6-2a+3b^2)?