question_answer

(a) Subtract \[4a\text{ }-\text{ }7ab\text{ }+\text{ }3b\text{ }+\text{ }12\]from

\[12a\text{ }-\text{ }9ab\text{ }+\text{ }5b\text{ }-\text{ }3\]

(b) Subtract \[3xy+\text{ }5yz-7zx\]from

\[5xy-\text{ }2yz-2zx+10xyz\]

(c) Subtract \[4{{p}^{2}}q-3pq+5p{{q}^{2}}-8p+7q-10\]

From \[18-3p-11q+5pq-2p{{q}^{2}}+5{{p}^{2}}q\]

Answer:

(a) \[12a9ab+5b3\]

\[4a7ab+3b+12\]

\[\frac{\left(  \right)\text{ }\left( + \right)\text{ }\left(  \right)\text{ }\left(  \right)\text{ }subtracting}{8a\text{ }\text{ }2ab\text{ }+\text{ }2b\text{ }\text{ }15}\]

(b)\[5xy\text{ }\text{ }2yz\text{ }\text{ }2zx\text{ }+\text{ }10xyz\]

\[3xy\text{ }+\text{ }5yz\text{ }\text{ }7zx\]

\[\frac{\left(  \right)\text{ }\left(  \right)\text{ }\left( + \right)\text{ }subtracting}{2xy\text{ }\text{ }7yz\text{ }+\text{ }5zx\text{ }+\text{ }10xyz}\]

(c) \[183p11q+5pq2p{{q}^{2}}+5{{p}^{2}}q\]

\[-108p+7q3pq+5p{{q}^{2}}+4{{p}^{2}}q\]

\[\frac{\left( + \right)\text{ }\left( + \right)\text{ }\left(  \right)\text{ }\left( + \right)\text{ }\left(  \right)\,\,\text{ }\left(  \right)}{28+5p18q+8pq7p{{q}^{2}}+{{p}^{2}}q}\]

Therefore, \[{{p}^{2}}q7p{{q}^{2}}+8pq18q+5p+28\]