Graph and solve the system Y=-3x+3 Y=2x-7
2 answers:
An equation of the form y=ax+b, where a and b are constants; y and x are variable, is called a linear equation.
They are called "linear" because if we draw all the solutions of such an equation, they form a line.
This also means that it is enough to:
1. find to solutions
2. plot them on the coordinate axis
3.join those 2 points
and all the solutions of the equation y=ax+b will be on that line.
Thus, to draw line y=-3x+3
let x=0, then y=3, so one solution is (0, 3)
let y=0, then x=1 so a second solution is (1, 0)
(remark we can pick x or y to be any value, then solve for the other)
to draw the second line y=2x-7
let x=0, then y=-7
let x=3, then y=-1
plot these points and join them as shown in the picture.
The solution of the system is the point where the lines intersect, which is
(2, -3)
They are called "linear" because if we draw all the solutions of such an equation, they form a line.
This also means that it is enough to:
1. find to solutions
2. plot them on the coordinate axis
3.join those 2 points
and all the solutions of the equation y=ax+b will be on that line.
Thus, to draw line y=-3x+3
let x=0, then y=3, so one solution is (0, 3)
let y=0, then x=1 so a second solution is (1, 0)
(remark we can pick x or y to be any value, then solve for the other)
to draw the second line y=2x-7
let x=0, then y=-7
let x=3, then y=-1
plot these points and join them as shown in the picture.
The solution of the system is the point where the lines intersect, which is
(2, -3)
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Refer to the figure shown below.
x = the width of the rectangle (meters)
y = the height of the rectangle (meters(
The fencing for the perimeter of the rectangle costs $30 per meter.
The two inner partitions cost $25 per meter.
The total cost of the fencing is
C = 2(x+y)*$30 + 2y*$25
= 60(x+y) + 50y
= 60x + 110y
Because the amount available to spend is $600, therefore
60x + 110y = 6000
or
6x + 11y = 600
That is,
y = (600 - 6x)/11 (1)
The area is
A = x*y (2)
Substitute (1) into (2).
A = (x/11)*(600 - 6x) = (1/11)*(600x - 6x²)
To maximize A, the derivative of A with respect to x is zero.
That is,
600 - 12x = 0
x = 600/12 = 50
From (1), obtain
y = (1/11)*(600 - 6*50) = 300/11 = 27.273
Because the second derivative of A with respect to x is negative, x=50, y = 27.273 will yield the maximum area.
The maximum area is
50*27.273 = 1363.64 m² = 1364 m² (nearest integer)
Answer: 1364 m² (nearest integer)
x = the width of the rectangle (meters)
y = the height of the rectangle (meters(
The fencing for the perimeter of the rectangle costs $30 per meter.
The two inner partitions cost $25 per meter.
The total cost of the fencing is
C = 2(x+y)*$30 + 2y*$25
= 60(x+y) + 50y
= 60x + 110y
Because the amount available to spend is $600, therefore
60x + 110y = 6000
or
6x + 11y = 600
That is,
y = (600 - 6x)/11 (1)
The area is
A = x*y (2)
Substitute (1) into (2).
A = (x/11)*(600 - 6x) = (1/11)*(600x - 6x²)
To maximize A, the derivative of A with respect to x is zero.
That is,
600 - 12x = 0
x = 600/12 = 50
From (1), obtain
y = (1/11)*(600 - 6*50) = 300/11 = 27.273
Because the second derivative of A with respect to x is negative, x=50, y = 27.273 will yield the maximum area.
The maximum area is
50*27.273 = 1363.64 m² = 1364 m² (nearest integer)
Answer: 1364 m² (nearest integer)