If (1 + 2x)^5+ (1 - 2x)^5 = a + bx + cx?
find the values of the constants a, b and c.

steps below

Step-by-step explanation:

5th level of Pascal triangle: 1,5,10,10,5,1

(1+2x)⁵ = 1*1⁵*(2x)⁰+5*1⁴*2x+10*1³*(2x)²+10*1²*(2x)³+5*1¹(2x)⁴+1*1⁰*(2x)⁵

           = 1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵   .. (1)

(1-2x)⁵ =  1  - 10x + 40x² - 80x³  + 80x⁴ - 32x⁵  .. (2)

(1)+(2): (1+2x)⁵ + (1-2x)⁵ = 160x⁴ + 80x² +2

a + bx + cx ?? suppose it's ax⁴ + bx² + cx, if you have other set up.. adjust by yourself

a = 160,    b = 80,    c = 2

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.