Question

ii) Solve : (2x + y) (z - 2d) + (x - y)(2z + 3d) + 14.​

Answer 1

(2x + y)(z - 2d) + (x - y)(2z + 3d) + 14

Firstly, we have to multiply a binomial to another binomial.

⇒ (2x + y)(z - 2d) + (x - y)(2z + 3d) + 14

⇒ 2x (z - 2d) + y (z - 2d) + x (2z + 3d) - y (2z + 3d) + 14

⇒ 2x (z) - 2x (2d) + y(z) - y(2d) + x(2z) + x(3d) - y(2z) - y(3d) + 14

⇒ 2xz - 4dx + yz - 2dy + 2xz + 3dx - 2yz - 3dy + 14

Now we will combine like terms.

⇒ 2xz - 4dx + yz - 2dy + 2xz + 3dx - 2yz - 3dy + 14

⇒ 2xz + 2xz - 4dx + 3dx + yz - 2yz - 2dy - 3dy + 14

Now add and subtract accordingly.

⇒ 2xz + 2xz - 4dx + 3dx + yz - 2yz - 2dy - 3dy + 14

⇒ 4xz - dx - yz - 5dy + 14

∴ (2x + y)(z - 2d) + (x - y)(2z + 3d) + 14 = 4xz - dx - yz - 5dy + 14