Solve the quadratic linear below :6x²-x-2=0
2 answers:
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>= means greater than or equal to
<= means less than or equal to
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Part A
The graph of y >= -3x+3 will have a solid boundary line and the shading will be above the boundary line.
The boundary line y = -3x+3 has a negative slope so it moves down as you read it from left to right. It goes through the points (0,3) and (1,0)
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The graph of y < (3/2)x - 6 will have a dashed or dotted boundary line. The shading is below the boundary.
The graph y = (3/2)x-6 goes through the two points (0,-6) and (2,-3)
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If you graph both y >= -3x+3 and y < (3/2)x - 6 together, you get what you see in the attached image. This solution shaded region is the result of the overlapping prior shaded regions.
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Part B
Plug (x,y) = (-6,3) into each inequality to see if we get a true inequality or not
For the first inequality we have
y >= -3x+3
3 >= -3*(-6)+3
3 >= 18+3
3 >= 21
which is false. The value 3 is not larger or equal to 21. So right off the bat we know that (-6,3) is NOT a solution. It is NOT in the solution region.
Let's check the other inequality just for the sake of completeness
y < (3/2)x - 6
3 < (3/2)*(-6) - 6
3 < -9 - 6
3 < -15
this is also false. The value -15 is smaller than 3, since it is to the left of 3
We're given more evidence that (-6,3) is NOT in the solution area. It is outside of both shaded areas.
<= means less than or equal to
---------------------------------------------
Part A
The graph of y >= -3x+3 will have a solid boundary line and the shading will be above the boundary line.
The boundary line y = -3x+3 has a negative slope so it moves down as you read it from left to right. It goes through the points (0,3) and (1,0)
--------------
The graph of y < (3/2)x - 6 will have a dashed or dotted boundary line. The shading is below the boundary.
The graph y = (3/2)x-6 goes through the two points (0,-6) and (2,-3)
--------------
If you graph both y >= -3x+3 and y < (3/2)x - 6 together, you get what you see in the attached image. This solution shaded region is the result of the overlapping prior shaded regions.
---------------------------------------------
Part B
Plug (x,y) = (-6,3) into each inequality to see if we get a true inequality or not
For the first inequality we have
y >= -3x+3
3 >= -3*(-6)+3
3 >= 18+3
3 >= 21
which is false. The value 3 is not larger or equal to 21. So right off the bat we know that (-6,3) is NOT a solution. It is NOT in the solution region.
Let's check the other inequality just for the sake of completeness
y < (3/2)x - 6
3 < (3/2)*(-6) - 6
3 < -9 - 6
3 < -15
this is also false. The value -15 is smaller than 3, since it is to the left of 3
We're given more evidence that (-6,3) is NOT in the solution area. It is outside of both shaded areas.
Answer:
{x,y} = {6/5,23/10}
Step-by-step explanation:
[1] 7x + 2y = 13
[2] 4x + 4y = 14 <---------- linear equations given
Graphic Representation of the Equations : PICTURE
2y + 7x = 13 4y + 4x = 14
Solve by Substitution :
// Solve equation [2] for the variable y
[2] 4y = -4x + 14
[2] y = -x + 7/2
// Plug this in for variable y in equation [1]
[1] 7x + 2•(-x +7/2) = 13
[1] 5x = 6
// Solve equation [1] for the variable x
[1] 5x = 6
[1] x = 6/5
// By now we know this much :
x = 6/5
y = -x+7/2
// Use the x value to solve for y
y = -(6/5)+7/2 = 23/10
// Plug this in for variable y in equation [1]
[1] 7x + 2•(-x +7/2) = 13
[1] 5x = 6
// Solve equation [1] for the variable x
[1] 5x = 6
[1] x = 6/5
// By now we know this much :
x = 6/5
y = -x+7/2
// Use the x value to solve for y
y = -(6/5)+7/2 = 23/10
