Question

Subtract p2 – 6pq + q2 from 2p2 + 7ab – 2q2​

Answer 1

As per data given in the question,

We have to subtract $(p^2-6pq+q^2)$ from $(2p^2-7ab-2q^2)$

As we know that,

While subtracting one number from other number, we must kept in mind that the addition and subtraction must be done on like terms.

Like the terms containing $p^2$ must be added or subtracted from the terms containing $p^2$ and same should be followed for all the variables.

So, applying the rules,

We will get,

$(2p^2-7ab-2q^2)-(p^2-6pq+q^2)$

Now separating the like terms,

$(2p^2-7ab-2q^2)-(p^2-6pq+q^2)\\=>(2p^2-7ab-2q^2-p^2+6pq-q^2)\\=>2p^2-p^2-2q^2-q^2-7ab+6pq\\=>p^2-3q^2-7ab+6pq$

Hence, we will get the answer as $p^2-3q^2-7ab+6pq$.

Answer 2

Given: The equation is$p^{2} -6pq + q^{2} from2p^{2} + 7ab -2q^{2}$.

We have to subtract$p^{2} -6pq + q^{2} from2p^{2} + 7ab -2q^{2}$.

As we know that the subtraction is an arithmetic operation that represents the operation of removing objects from a collection.

By using the properties of subtraction, we are solving the above equation.

We are solving in the following way:

We have,

$p^{2} -6pq + q^{2} from2p^{2} + 7ab -2q^{2}$

$=>2p^{2} + 7ab -2q^{2} -p^{2} -6pq + q^{2} \\=>2p^{2}-p^{2}-2q^{2}+ q^{2}+ 7ab-6pq\\=>p^{2} -q^{2} + 7ab-6pq$

Hence, after subtracting the above equation we get, the solution is$p^{2} -q^{2} + 7ab-6pq$.