19.1 should be your answer
X/2-y/3=3/2
(6×x/2)-(6×y/3)=6×3/2
3x-2y=9______(1)
x/3+y/2=16/3
(6×x/3)+(6×y/2)=6×16/3
2x+3y=32_____(2)
(1)×3____9x-6y=27____(3)
(2)×2____4x+6y=64____(4)
(3)+(4)___13x=91
x=7
3(7)-2y=9
-2y=-12
y=6
Answer:
Equation of the tangent to the curve
y = 240x - 215994
Equation of the normal
y = (-1/240)x + 9.75 = - 0.00417x + 9.75
Step-by-step explanation:
y = (6 + 4x)² = 36 + 48x + 16x² = 16x² + 48x + 36
dy/dx = 32x + 48
At the point (6,900),
dy/dx = 32(6) + 48 = 240
Equation of the tangent at point (a,b) is
(y - b) = m(x - a)
a = 6, b = 900, m = 240
y - 6 = 240(x - 900)
In the y = mx + b form,
y - 6 = 240x - 216000
y = 240x - 215994
The slope of the normal line = -(1/slope of the tangent line) (since they're both perpenducular to each other)
Slope of the normal line = -1/240
Equation of normal
y - 6 = (-1/240)(x - 900)
y - 6 = (-x/240) + 3.75
y = (-1/240)x + 9.75
y = - 0.00417x + 9.75
Minus 6 both sides
4>x/3
times3 both sides
12>x
