SOLUTION: This question is one question with four parts! I asked the first part already bur I'm going to stick all four parts together on this question so that it makes sense. 1.Suppose

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: This question is one question with four parts! I asked the first part already bur I'm going to stick all four parts together on this question so that it makes sense. 1.Suppose       Log On


   

Question 127985: This question is one question with four parts!
I asked the first part already bur I'm going to stick all four parts together on this question so that it makes sense.
1.Suppose you are in the market for a new home and are interested in a new housing community under construction in a different city.
1a.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.
MY ANSWER but I don't think it is right-----
x+y=56
1b. 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.
MY ANSWER: 3y+x=56
1c. 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 used to solve the problem.
NOW for this problem I am completely lost but I'm going to give it a try-------
MY ANSWER: 3y^2+x^2=56*3 OR IS IT THIS: 3y^2+x^2=168?
1d. 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 systems by graphing?
THIS QUESTION i'M TOTALLY STUMPED ON!.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
1a.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+%2B+y+=+56+……………(a)
1b. 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.
+y=3x+…………(b)
1c. 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 used to solve the problem.
x+%2B+y+=+56+……………(a)
+y=3x+++…………(b)….substitute yfrom this equation from (a)
---------------------------------------
x+%2B+y+=+56+……………(a)
+x%2B+3x+=+56
+4x+=+56
+x+=+56%2F4
+x+=+14
+y=3x+++…………(b)….
+y=3%2A14++
+y=42++

1d. 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 systems by graphing?
the intercepts are:
x+%2B+y+=+56+……………(a)…if x=0, y-intercept will be:
0%2By=56
y=56……………………… so, y-intercept is (0,56)
if y=0, x-intercept will be:
x%2B0=56
x=56…………….so, x-intercept is (56,0)

Solved by pluggable solver: Graphing Linear Equations


1%2Ax%2B1%2Ay=56Start with the given equation



1%2Ay=56-1%2Ax Subtract 1%2Ax from both sides

y=%281%29%2856-1%2Ax%29 Multiply both sides by 1

y=%281%29%2856%29-%281%29%281%29x%29 Distribute 1

y=56-%281%29x Multiply

y=-1%2Ax%2B56 Rearrange the terms

y=-1%2Ax%2B56 Reduce any fractions

So the equation is now in slope-intercept form (y=mx%2Bb) where m=-1 (the slope) and b=56 (the y-intercept)

So to graph this equation lets plug in some points

Plug in x=0

y=-1%2A%280%29%2B56

y=0%2B56 Multiply

y=56 Add

So here's one point (0,56)





Now lets find another point

Plug in x=1

y=-1%2A%281%29%2B56

y=-1%2B56 Multiply

y=55 Add

So here's another point (1,55). Add this to our graph





Now draw a line through these points

So this is the graph of y=-1%2Ax%2B56 through the points (0,56) and (1,55)


So from the graph we can see that the slope is -1%2F1 (which tells us that in order to go from point to point we have to start at one point and go down -1 units and to the right 1 units to get to the next point), the y-intercept is (0,56)and the x-intercept is (56,0) . So all of this information verifies our graph.


We could graph this equation another way. Since b=56 this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,56).


So we have one point (0,56)






Now since the slope is -1%2F1, this means that in order to go from point to point we can use the slope to do so. So starting at (0,56), we can go down 1 units


and to the right 1 units to get to our next point



Now draw a line through those points to graph y=-1%2Ax%2B56


So this is the graph of y=-1%2Ax%2B56 through the points (0,56) and (1,55)