Question

subtract 3a²-4+6a from 4a²-7a-3​

Answer 1

To solve  : according to question we are asked to subtract  $3a^{2} - 4 + 6a \ from \ 4a^{2} - 7a - 3$

given expression : $3a^{2} - 4 + 6a$  and $4a^{2} - 7a - 3$

• To perform subtraction of given algebraic expressions we need to arrange them in standard form i.e   $Ax + By + C = 0$
•     $3a^{2} - 4 + 6a$                               (I)

             $4a^{2} - 7a - 3$                               (II)

• Subtracting eq (I) from (II)

           $= ( 4a^{2} - 7a - 3) - ( 3a^{2} - 4 + 6a)$

          $= 4a^{2} - 7a - 3 - 3a^{2} + 4 - 6a$

       arranging like terms together

         $= 4a^{2} - 3a^{2} - 7a - 6a + 4 - 3$

          $= a^{2} - 13a + 1$

Hence , $a^{2} - 13a + 1$ is the correct answer.

Answer 2

Given equations are,

$3a^{2} -4+6a...(1)$

$4a^{2}-7a-3...(2)$

Now we have to subtract the equation $(1)$ from equation $(2)$.

We get,

$4a^{2}-7a-3-( 3a^{2} -4+6a)$

By using mathematical operations,

$(+)\times(-)=(-)\\(-)\times(+)=(-)\\(+)\times(+)=(+)\\(-)\times(-)=(+)$

$4a^{2}-7a-3-3a^{2} +4-6a$

Write all like terms together.

$4a^{2}-3a^{2} -7a-6a-3+4$

$a^{2} -13a+1$

Therefore, the final answer after subtracting two equations is $a^{2} -13a+1$