Question

solve this....................​

Answer 1

Step-by-step explanation:

18) 8c(-a+b+c)-[6a(a+b+c)-4b(a-b+c)]

= -8ac+8bc+8c^2-[6a^2+6ab+6ac-(4ab+4b^2-4bc)]

= -8ac+8bc+8c^2-[6a^2+6ab+6ac-4ab-4b^2+4bc]

= -8ac+8bc+8c^2- ( 6a^2 +2ab +6ac+4bc -4b^2)

= -8ac+8bc+8c^2- 6a^2 -2ab -6ac -4bc +4b^2

= 8c^2 +4b^2-6a^2 -14ac + 4bc-2ab

2) 4x(x-4)+13x

4x^2-16x+13x

4x^2-3x

when x= -1

4(-1)^2-3*-1

= 4+3

= 7

when x= 1/2

4x^2-3x

=4(1/2)^2-3*1/2

= 4*1/4 -3/2

= 1-3/2

= 2-3/2

= -1/2

Answer 2

Questions

(a) Subtract 6a (a + b + c) - 4b (a - b + c) from 8c (-a + b + c)?

(b) Simplify 4x (x - 4) + 13x and find its value when

(i) x = -1

(ii) x = ¹/₂

__________________________

Answers

(a) 8c (-a + b + c) - [6a (a + b + c) - 4b (a - b + c)]

First we will solve inside the box brackets

⇒ 8c (-a + b + c) - [6a (a + b + c) - 4b (a - b + c)]

⇒ 8c (-a + b + c) - [6a (a) + 6a (b) + 6a (c) - 4b (a) + 4b (b) - 4b (c)]

⇒ 8c (-a + b + c) - [6a² + 6ab + 6ac - 4ab  + 4b² - 4ac]

Now simplify and remove the box bracket

⇒ 8c (-a + b + c) - [6a² + 6ab + 6ac - 4ab  + 4b² - 4bc]

⇒ 8c (-a + b + c) - 6a² - 6ab - 6ac + 4ab  - 4b² + 4bc

Simplify the brackets

⇒ 8c (-a + b + c) - 6a² - 6ab - 6ac + 4ab  - 4b² + 4bc

⇒ 8c (-a) + 8c (b) + 8c (c) - 6a² - 6ab - 6ac + 4ab  - 4b² + 4bc

⇒ -8ac + 8bc + 8c² - 6a² - 6ab - 6ac + 4ab  - 4b² + 4bc

Combine Like Terms

⇒ -8ac + 8bc + 8c² - 6a² - 6ab - 6ac + 4ab  - 4b² + 4bc

⇒ -8ac - 6ac + 8bc + 4bc + 8c² - 6a² - 6ab + 4ab  - 4b²

⇒ -14ac + 12bc + 8c² - 6a² - 2ab - 4b²

∴ 8c (-a + b + c) - [6a (a + b + c) - 4b (a - b + c)] = -14ac + 12bc + 8c² - 6a² - 2ab - 4b²

______________________________

(b) First we will simplify then we will substitute the values of 'x' in both cases.

⇒ 4x (x - 4) + 13x

⇒ 4x (x) - 4x (4) + 13x

⇒ 4x² - 16x + 13x

⇒ 4x² - 3x

(i) x = -1

⇒ 4x² - 3x

⇒ 4 (-1)² - 3(-1)

⇒ 4(1) - (-3)

⇒ 4 + 3

⇒ 7

∴ 4x² - 3x = 7 when 'x' = -1.

(ii) x = ¹/₂

⇒ 4x² - 3x

⇒ $4(\frac{1}{2})^{2} - 3(\frac{1}{2})$

⇒ $4(\frac{1}{4}) - \frac{3}{2}$

⇒ $1 - \frac{3}{2}$

⇒ $\frac{2}{2}-\frac{3}{2}$

⇒ $\frac{-1}{2}$

∴ 4x² - 3x = ⁻¹/₂ when 'x' = ¹/₂

______________________________