1. Imagine you are in the market for a new home and are interested in a new housing community under construction in a different city.
a. The sales representative informs you that there are 56 houses for sale with two floor plans still available. Use x to represent floor plan one and y to represent floor plan two. Write an equation that illustrates the situation.
x+y=56 ?
b. The sales representative later indicates that there are three 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.
3x=y ?
c. Use the equations from Parts 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?
need help with C and d
c. Substitute 3x for y in the first equation to solve for x. Use that value in the second equation to find y. Check by putting both values in the first equation.
d. I cannot help you with the intercepts.
Floor plan one = x = 14 available
Floor plan two = y = 42 available
I substituted y = 3x into the equation x + y = 56 for x. Therefore I had (3x + x = 56) and I solved for x which gave me x = 14. To find y, I plugged x into the equation in (a) which gave me 14 + y = 56 and solved for y which gave me y = 42.
c. To solve the system of equations using substitution, we can substitute the value of y from equation b into equation a.
From equation b, we know that y = 3x. We can substitute this value into equation a:
x + y = 56
x + 3x = 56
4x = 56
x = 14
Now that we know the value of x, we can substitute it back into equation b to find y:
y = 3x
y = 3 * 14
y = 42
Therefore, there are 14 houses with floor plan one (x) and 42 houses with floor plan two (y) available in the new housing community.
d. The intercepts of an equation represent the points where the line intersects the x-axis and y-axis.
In Part a, the equation is x + y = 56. To find the x-intercept, we set y = 0 and solve for x:
x + 0 = 56
x = 56
So, the x-intercept is 56.
To find the y-intercept, we set x = 0 and solve for y:
0 + y = 56
y = 56
Thus, the y-intercept is also 56.
In Part b, the equation is 3x = y. To find the x-intercept, we set y = 0 and solve for x:
3x = 0
x = 0
Therefore, the x-intercept is 0.
To find the y-intercept, we set x = 0 and solve for y:
3 * 0 = y
y = 0
Hence, the y-intercept is also 0.
If we graph the lines corresponding to both equations, they will intersect at the point (14, 42) since that is the solution we obtained in part c.