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4.(a) Subtract \(4a – 7ab + 3b + 12\) from \(12a – 9ab + 5b – 3\)
(b) Subtract \(3xy + 5yz – 7zx\) from \(5xy – 2yz – 2zx + 10xyz\)

(c) Subtract \(4p^2q – 3pq + 5pq^2 – 8p + 7q – 10 \) from \(18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q\)


Answer :

(a)We have the expressions to subtract:\(4a – 7ab + 3b + 12\) from \(12a – 9ab + 5b – 3\)
Subtracting these we get: \((12a – 9ab + 5b – 3)-(4a – 7ab + 3b + 12 )\)

\((12a – 9ab + 5b – 3)+(-4a +7ab - 3b - 12 )\quad\)[Taking (-) inside the second bracket]

\(\Rightarrow (12a-4a)+ (– 9ab+7ab)+ (+ 5b- 3b )+( – 3 - 12 )\quad\)[Placing like terms together]

\(\Rightarrow (8a)+ (-2ab)+ (2b)+(-15)\)

\(\Rightarrow (8a-2ab+ 2b-15)\)

(b)We have the expressions to subtract:\(3xy + 5yz – 7zx\) from \(5xy – 2yz – 2zx + 10xyz\)
Subtracting these we get: \((5xy – 2yz – 2zx + 10xyz)-(3xy + 5yz – 7zx)\)

\((5xy – 2yz – 2zx + 10xyz)+(-3xy - 5yz + 7zx )\quad\)[Taking (-) inside the second bracket]

\(\Rightarrow (5xy-3xy )+ (– 2yz - 5yz)+ (– 2zx+ 7zx )+( + 10xyz )\quad\)[Placing like terms together]

\(\Rightarrow (2xy)+ (-7yz)+ (5zx)+(10xyz)\)

\(\Rightarrow (2xy-7yz+5zx+10xyz)\)

(c)We have the expressions to subtract:\(4p^2q – 3pq + 5pq^2 – 8p + 7q – 10 \) from \(18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q\)
Subtracting these we get: \((18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q )-(4p^2q – 3pq + 5pq^2 – 8p + 7q – 10)\)

\(\Rightarrow(18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q)+(-4p^2q +3pq-5pq^2 +8p-7q+ 10)\quad\)[Taking (-) inside the second bracket]

\(\Rightarrow (+18+10)+(-3p+8p)+(-11q-7q)+(5pq+3pq)+(-2pq^2-5pq^2)+(+ 5p^2q-4p^2q) \quad\)[Placing like terms together]

\(\Rightarrow (+28)+ (+5p)+(-18q)+(8pq)+(-7pq^2)+(+p^2q)\)

\(\Rightarrow (+28+5p-18q+8pq-7pq^2+p^2q)\)

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