EQ.1: `y=sqrtx` EQ.2: `y= (5-x)/4` EQ.3: `y = (3x-8)/2`
Determine the intersection points of the three equations using substitution method.
Susbstitute `y=sqrtx` to EQ.2 .
`4sqrtx = 5-x`
Square both sides.
` 16x = 25-10x+x^2`
`0 = 25 - 26x + x^2`
`0 = (x-25)(x-1)`
So values of x are: x = 1 and x = 25
Substitute the values of x to EQ.1 and EQ.2.
`x=1` , `y = sqrt1 = 1 ` and `y =(5-1)/4 = 1`
`x=25` , `y = sqrt25 = 5` and `y = (5-25)/4 = -5`
Note that the value of x that satisfy both equation is x=1. Hence, EQ.1 and EQ.2 intersect at (1,1).
Next, substitute `y=sqrtx` to EQ.3.
`sqrtx = (3x-8)/2`
`2sqrtx = 3x-8`
Square both sides.
` 4x = 9x^2-48x+64`
` 0= 9x^2-52x+64`
Using the quadratic formula, the values of x are: x=4 and x=16/9. Then, substitute the values of x to EQ.1 and EQ.3.
`x=4` , `y =sqrt4 = 2 ` and `y=[(3*4)-8]/2 =2`
`x=16/9` , `y = sqrt(16/9) =4/3` and `y = [(3*16/9)-8]/2 = -4/3`
Base on this, EQ.1 and EQ.3 intersect at (4,2).
Next, substitute y = (5-x)/4 to EQ.3.
`(5-x)/4 = (3x-8)/2`
`2(5-x) = 4(3x-8)`
`10-2x = 12x - 32`
`42 = 14x`
`3 = x `
Substitute the value of x to EQ.2 and EQ.3 .
`x = 3` , `y =(5-3)/4 = 1/2 ` and `y = (3(3)-8)/2 = 1/2`
Hence, EQ.2 and EQ.3 intersect at (3,1/2).
Then, plot the three equations.
(Note:Graph of EQ.1- Red; EQ.2-Green ; and EQ.3 - Blue)
To find the area bounded by the three equations, use the formula:
`A = int_a^b( y_U-y_L) dx `
Base on the graph above, the upper function is `y_U=sqrtx` . And there are two lower functions which are `y_L=(5-x)/4` and y`_L=(3x-8)/2` . Also, the limits of the integral are the x-coordinates of the intersection points. So, the area is:
`A = int_1^3(sqrtx-(5-x)/4)dx + int_3^4(sqrtx-(3x-8)/2)dx`
`A = int_1^3 (sqrtx -5/4 +x/4 )dx + int_3^4(sqrtx-(3x)/2 +4)dx`
`A = (2/3xsqrtx-(5x)/4+x^2/8)|_1^3 + (2/3xsqrtx - (3x^2)/2 +4x)|_3^4`
`A = (12sqrt3-13)/6 + (49-24sqrt3)/12 = 23/12 = 1.9`
Hence, the area bounded by` y=sqrtx` , `y=(5-x)/4` and `y = (3x-8)/2` is 1.9 square units.