Question

By How much must be the sum of 6p^2q-4pq+2q^2p and 5q^2p + 5pq be increased to get 8p^2q+5pq÷7q^2p​

Answer 1

Answer:

$sum \: of \: {6p}^{2q} - 4pq + {2q}^{2p} \: and \: {5q}^{2p} + 5pq \: is$

${6p}^{2q} - 4pq + {2q}^{2p} + ({5q}^{2p} + 5pq)$

${6p}^{2q} - 4pq + {2q}^{2p} + {5q}^{2p} + 5pq$

${6p}^{2q} + {5q}^{2p} + {2q}^{2p} - 4pq + 5pq$

${6p}^{2q} + {7q}^{2p} + pq$

${6p}^{2q} + {7q}^{2p} + pq + (x) = \frac{{8p}^{2q} + 5pq}{ {7q}^{2p} }$

$x =\frac{{8p}^{2q} + 5pq}{ {7q}^{2p} } - ({6p}^{2q} + {7q}^{2p} + pq)$

$x = \frac{{8p}^{2q} + 5pq}{ {7q}^{2p} } - {6p}^{2q} - {7q}^{2p} - pq$