Answer:
ther will be exactly one value. the answer is b
Im a keep editing since this is timed.
#1 x-8a/6=3a-2x
Multiply each term by 6 to remove fraction .
6(x-8a/6)= 3a(6)-2x(6)
X-8a=18a-12x
Add 12x both sides
X+12x-8a=18a-12x+12x
13x-8a= 18a
Add8a both sides
13x-8a+8a=18a+8a
13x=26a
Divide each side by 13
13x/13=26a/13
X= 26a/13
Reduce and cancel common factor .
X= 13(2a)/13(1)
X=2a/1
X=2a
Problem #2
3x-2a=7a find a
Subtract 3x both sides
3x-3x-2a=7a-3x
-2a=7a-3x
Subtract 7a both sides
-2a-7a=7a-7a-3x
5a=-3x
Divide both sides by 5
5a/5=-3x/5
A= -0.6 or -3/5
Problem#3
3(bx-2ab)=b(x-7a)+3ab
Distributed property
3(bx)+3(-2ab)=b(x)+b(-7a)+3ab
Simplify
3bx-6ab=bx-7ab+3ab
Add -7a plus 3ab on right side
3bx-6ab= bx-4ab
Subtract bx both sides
3bx-bx-6ab= bx -bx -4ab
Simplify
2bx-6ab= -4ab
Add 6ab both sides
2bx-6ab+6ab=6ab-4ab
2bx=2ab
Divide 2b both sides
X= 2ab/2b
Cancel 2 common
X= ab/b
X=a
1/6 since six people are eating the leftovers
Answer:
For x(θ) = 3cosθ + 2
y² = [9x²/(x - 2)²] - x²
For x(θ) = 3cosθ + 2
x² = [4y²/(y - 1)²] - y²
Step-by-step explanation:
Given the following equivalence:
x² + y² = r²
r = √(x² + y²)
x = rcosθ
cosθ = x/r
y = rsinθ
sinθ = y/r
Applying these to the given equations,
x(θ) = 3cosθ + 2
x = 3(x/r) + 2
xr = 3x + 2r
(x - 2)r = 3x
r = 3x/(x - 2)
Square both sides
r² = 9x²/(x - 2)²
(x - 2)²r² = 9x²
(x - 2)²(x² + y²) = 9x²
(x² + y²) = 9x²/(x - 2)²
y² = [9x²/(x - 2)²] - x²
y(θ) = 2sinθ - 1
y = 2y/r - 1
yr = 2y - r
(y - 1)r = 2y
r = 2y/(y - 1)
Square both sides
r² = 4y²/(y - 1)²
x² + y² = 4y²/(y - 1)²
x² = [4y²/(y - 1)²] - y²
Combine like terms:
7x-2x= 5x
-10+24= +14
Answer: 5x+14
