solve the modulus equations below. (a) |2x+3|=6. (b) |x-2|=x+4. (c) |2x-7|=|x+3|​

(a) x=1.5 or  x=-4.5

(b) x=-1

(c) x=10  or  

Step-by-step explanation:

(a) |2x+3|=6

⇒ 2x+3=6  or  2x+3=-6

⇒ 2x=3 or  2x=-9

⇒ x=1.5 or  x=-4.5

(b) |x-2|=x+4

⇒ x-2=x+4  or  x-2=-(x+4)

x-2=x+4  

-2≠4 x∈Ф

or  x-2=-(x+4)

x-2=-x-4

2x=-2

x=-1

(c) |2x-7|=|x+3|​

⇒ 2x-7=x+3  or  2x-7=-(x+3)

2x-7=x+3

x=10

or  2x-7=-(x+3)

2x-7=-x-3

3x=4

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!