(I am not sure how to get the y value for part C and the whole part D)
(I answered a couple of them but am not confident on my answers, so can you please tell me if my answers are not correct to the ones I answered and for the ones I haven not answered that is because I do not know how to do them. PLEASE HELP!!! This assignment is already late).
a) The sales representative informs you that there are two floor plans still available, and that there are a total of 56 houses available. Use x to represent floor plan #1 and y to represent floor plan #2. Write an equation that illustrates the situation. x+y=56
b) The sales representative later indicates that there are 3 times as many homes available with the second floor plan than the first. Write an equation that illustrates this situation. Use the same variables you used in part a. x+y(3)=56
c) Use the equations from part a and b of this exercise as a system of equations. Use substitution to determine how many of each type of floor plan is available. Describe the steps you used to solve the problem.
d) What are the intercepts of the equation from part a of this problem? What are the intercepts from part b of this problem? Where would the lines intersect if you solved the system by graphing?
algebra - Quidditch, Monday, November 16, 2009 at 10:31pm
a. is correct
b. is not correct You have the right idea. y = 3x. Now substitute 3x in for y in the equation and solve it.
algebra - LeAnn/URGENT, Monday, November 16, 2009 at 11:01pm
I am not very confident with algebra and am failing my class. Can you please help with parts c and d?
algebra - Quidditch, Monday, November 16, 2009 at 11:39pm
X + y = 56
y = 3x
Substituting that into the first equation...
x + 3x = 56
combining...
4x = 56
Divide both sides by 4 to find the value for x.
Once you have the value for x, use the second equation of y = 3x to find the value for y.
algebra - LeAnn/URGENT, Tuesday, November 17, 2009 at 12:04am
I'm so sorry that I'm asking so many questions but like I said I do not understand all of this. How do you solve for y in those steps? Please give details so I can better understand it. The answers you gave me, were they just for part c or d too? I think just c right?
Algebra - LeAnn/Needs turned in within an hour, Tuesday, November 17, 2009 at 10:10pm
Sorry, doesn't need turned in within an hour but does need to be turned in asap cause it is late.
Algebra - LeAnn/Needs turned in within an hour, Tuesday, November 17, 2009 at 11:12pm
Once you have the value for x, use the second equation of y = 3x to find the value for y. (This is the part I don't understand)(I also need help with part d please).I'm working on getting an tutor here in mt town cause I need alot of help understanding all this
d) What are the intercepts of the equation from part a of this problem? What are the intercepts from part b of this problem? Where would the lines intersect if you solved the system by graphing?
c) To solve the system of equations using substitution, we can substitute the value of y from the second equation into the first equation.
From the second equation, we have y = 3x.
Substituting this into the first equation: x + y = 56, we get x + 3x = 56.
Combining like terms, we have 4x = 56.
Dividing both sides by 4, we find x = 14.
Now, we can substitute this value of x back into the second equation to find y.
From the second equation, y = 3x.
Substituting x = 14, we get y = 3(14) = 42.
Therefore, there are 14 houses of floor plan #1 and 42 houses of floor plan #2 available.
d) In part a, the equation x + y = 56 represents the situation where there are a total of 56 houses available. To find the x-intercept, we set y = 0 and solve for x:
x + 0 = 56
x = 56
Therefore, the x-intercept is 56.
To find the y-intercept, we set x = 0 and solve for y:
0 + y = 56
y = 56
Therefore, the y-intercept is also 56.
In part b, the equation y = 3x represents the situation where there are three times as many homes available with the second floor plan than the first. To find the x-intercept, we set y = 0 and solve for x:
0 = 3x
x = 0
Therefore, the x-intercept is 0.
To find the y-intercept, we set x = 0 and solve for y:
y = 3(0)
y = 0
Therefore, the y-intercept is 0.
If you were to graph these two equations, the lines would intersect at the point (14, 42), representing the solution to the system of equations.
To solve part c, we need to use substitution. We have the system of equations:
x + y = 56 (from part a)
y = 3x (from part b)
To use substitution, we can substitute the value of y from the second equation into the first equation.
Substituting y = 3x into the equation x + y = 56, we get:
x + 3x = 56
Combining like terms, we have:
4x = 56
To find the value for x, we divide both sides by 4:
x = 56/4
Simplifying, we get:
x = 14
Now that we have the value for x, we can substitute it back into the second equation y = 3x to find the value for y:
y = 3(14)
Solving, we get:
y = 42
Therefore, there are 14 houses with floor plan #1 and 42 houses with floor plan #2.
Regarding part d, we need to find the intercepts of the equations from parts a and b.
For part a, the equation is x + y = 56. To find the intercepts, we set x and y equal to 0 and solve for the other variable.
When x = 0:
0 + y = 56
y = 56
So the y-intercept is (0, 56).
When y = 0:
x + 0 = 56
x = 56
So the x-intercept is (56, 0).
For part b, the equation is y = 3x. To find the intercepts, we again set x and y equal to 0 and solve for the other variable.
When x = 0:
y = 3(0)
y = 0
So the y-intercept is (0, 0).
When y = 0:
0 = 3x
x = 0
So the x-intercept is (0, 0).
If you graph these lines, they will intersect at (14, 42), which is the solution to the system of equations.