Need help with setting a table of values for a rectangle whose length = x and width..

Status
Not open for further replies.

helping

New member
Joined
Nov 4, 2012
Messages
16
Trying to help my daughter with various algebra problems I ran into something I do not understand.

Here it is,

Using the rectangles below:

a) Find the area of rectangle 1.

b) Create a table of values for rectangle 1 with x as the input and area as the output.

c) Graph the table of values and label as rectangle 1.

d) Repeat steps a through c for rectangle 2 (and graph on the same coordinate plane).

e) Create and solve an algebraic equation to find the value of x when the area of both rectangles is the same.

f) Use the graph to justify your answer to part e.


Rectangle 1 drawn with length of X and width of 12

Rectangle 2 drawn with length of x-2 and width of 16

I will greatly appreciate anyone's help with this.

Thank you.
 
Hello. This assignment covers a lot of lessons (at least three week's worth of classroom time, I would think). Has your daughter explained to you what parts she has already learned?

For example, does your daughter understand how to plot (x,y) points by drawing x- and y-axes? Do you have graph paper? Does her class use graphing calculators?

Has she solved simple equations for x? Knows the Distributive Property? Area formulas? Has she graphed lines and worked with their equations?

rectangle area is length times width

area = (length)(width)

Symbol x represents a quantity that varies in value

For rectangle1, the length varies, and symbol x represents the length. The width is 12. The rectangle's area grows as x grows. Symbol y represents the area.

y = (x)(12)

same as

y = 12x

To make table of associated (x,y) values, let x vary in value from 3 to 10

When x = 3, y = 12*3, so the area is 36 square units when the length is 3 units
When x = 4, y = 12*4, so the area is 48 square units when the length is 4 units
When x = 5, y = 12*5, so the area is 60 square units when the length is 5 units

and so on

These associated (length,area) pairs form (x,y) coordinates of points to plot on the graph.

(3,36)
(4,48)
(5,60)

and so on. Plot them; draw straight line through them. Label as L1

With rectangle2, it's the same approach. The length still varies, but the length is 2 units shorter than rectangle1

Length is x-2 and width is 16

area = length*width

y = (x - 2)(16)

Again, make table of (x,y) values, let x go from 3 to 10

When x = 3, y = (3-2)(16) which is (1)(16) = 16 square units
When x = 4, y = (4-2)(16) which is (2)(16) = 32 square units
When x = 5, y = (5-2)(16) which is (3)(16) = 48 square units

and so on

(3,16)
(4,32)
(5,48)

et cetera

Plot these points, too. Draw a line through them. Label it L2

Lines L1 and L2 should intersect somewhere between length x = 6 and x = 10

Using graph paper and plotting carefully will show the intersection coordinates (?,?) clearly

To solve algebraically, we take 12x and 16(x-2) for the two areas and say they are equal. That forms equation:

12x = 16(x-2)

I hope that your daughter can make an effort to solve this equation, based on her lessons thus far. Her solution for x should match the same value as the x-coordinate at the intersection point on the graph.

Lastly, take the x solution and multiply by 12, to find the area common to both rectangles because area y = 12x, yes?

This value of y should match the y-coordinate at the intersection point on the graph.

What grade is your daughter in? Can she post her questions about this exercise? Cheers :cool:
 
Last edited:
Thank you. I greatly appreciate your help. It was very helpful. My daughter is in 8th grade. She is capable of posting herself and will do so going forward.

Back to the problem... She does know how to solve the equation for x, but the teacher did not spent three weeks explaining all the concepts that are needed to solve it. She did teach them how to solve equations for x, y, or other variables when 2 or 3 variables are involved.

But she only spent 10 minutes showing them how to solve it using a graphing calculator, and my daughter doesn't remember the steps involved.

She said they will have 10 minutes to solve it using a graphing calculator in class on Monday. I do not know how to do it using a graphing calculator. Is there a way you can show us that as well?

Thanks
 
Calculators differ, but here are some Google results for instructional videos for solving both a single linear equation and for finding the intersection of two lines (solving system of two linear equations) using Texas Instruments TI-84. You may change the calculator info in the keywords field, to suit your needs.

I did not review those sites; do not give them any per$onal information. :cool:
 
Last edited:
At the very least, your daughter should learn how to graph equations AND set appropriate window boundaries for good viewing. I graphed the two lines L1 and L2, representing the growing areas of their respective rectangles; that's how I knew to do the table values from x=3 to x=10. I could see the intersection point (the interesting part), so no need to plot a lot of points before and after the interesting part.

Good tool to have at hand during exam, when allowed.
 
Want to clarify a couple of things regarding the solution.

Hello again. Wanted to clarify a couple of things regarding the solution.

For part (f) it states: Use the graph to justify your answer to part e.

Does this mean to draw a line where x = 8 (solution to the algebraic equation)... this line would be parallel to the y-axis? should this line be labeled x= 8?

Also, you wrote to solve the equation for y, by plugging in the x value. Thus y= 12(x)
y=12(8)
y=96

The problem does not ask to do this, yet as I understand it should always be done for these types of problems, right?

Finally, do we draw a line where y=96, which would be parallel to the x-axis? Should it be labeled y=96?

In total there would then be 4 lines on the coordinate plane: L1, L2, and the two lines described above?

Thanks
 
For part (f) it states: Use the graph to justify your answer to part e.

Does this mean to draw a line where x = 8 (solution to the algebraic equation)... this line would be parallel to the y-axis? should this line be labeled x= 8?

do we draw a line where y=96, which would be parallel to the x-axis? Should it be labeled y=96?

In total there would then be 4 lines on the coordinate plane: L1, L2, and the two lines described above?

No. L1 and L2 are the only lines needed.

When they instruct to use the graph "to justify" the answer to part (e), all they want is for you to state that the part (e) solution and associated common-area match the intersection point coordinates on the graph.

In other words, part (e) requires solving the earlier-posted equation, to find x = 8.

Part (f) requires you to show that your x=8 leads to y=96 and then state that those values for x and y (achieved algebraically) match the (x,y) coordinates at the intersection point (achieved graphically).

That's all. :cool:
 
Part (f) requires you to show that your x=8 leads to y=96 and then state that those values for x and y (achieved algebraically) match the (x,y) coordinates at the intersection point (achieved graphically).

That's all. :cool:

She solved the equation for X and got 8, she also solved the equation for y and got 96 (plugging in 8 for X).

She did not plug in all the x values, when solving for y, Is this what you were referring to: y = 12X,
When x = 2, y = 24
x = 3. y = 36
x = 4, y = 48, and so on...?

She did not write this - yet when she showed what she did to the teacher, she was told that she did everything correct, with the exception of starting with 2 as the first value for x. since area cannot equal to zero.

Was this the reason why you told us to start from 3?. If so, then one has to actually do the calculation first and start with the first value, using which the area > 0?

Thank you for helping us with this.

I have a few other interesting problems that the teacher gave out to the students, which were not explained at all as they were completely optional, that I will now post.
 
Last edited:
She solved the equation for x and got 8

That's part (e)



she also [substituted x=8 into] the equation for y and got 96

That's part (f)



She did not plug in all the x values, when solving for y, Is this what you were referring to: y = 12X,
When x = 2, y = 24
x = 3. y = 36
x = 4, y = 48, and so on...?

She did not write this - yet when she showed what she did to the teacher, she was told that she did everything correct

The equation y = 12x is the formula for the area (y) of rectangle1, in terms of its length, which varies (x)

Parts (b) and (d) require a table of values. You need some associated pairs of (x,y) values, in order to write a table of x|y values, followed by plotting those points.

Plugging in values that you choose for x, and doing the arithmetic, is how you get the y values for the table.

If the teacher saw coordinates labeled, or even just proper lines, but received no tables, then apparently the work is sufficient to satisfy the teacher. If the teacher does not care, neither should you!



she was told that she did everything correct, with the exception of starting with 2 as the first value for x. since area cannot equal to zero.

Was this the reason why you told us to start from 3?. If so, then one has to actually do the calculation first and start with the first value, using which the area > 0?

The teacher is being silly. The length of rectangle2 is x-2. If we choose to start at x=2, then the length of that rectangle is zero, and there can be no rectangle2. That is why I started at x=3. I chose to stop at x=10 because a quick sketch by the graphing calculator showed that the intersection point (the interesting part of the graph) was between x=6 and x=10. I saw no reason to generate values for the table in parts (b) and (d) using lengths greater than 10. ("Eight is enough" and now I wish that I had said, "let x go from 6 to 10") ;)

Sometimes, teachers want to find both pros and cons, just to offer comments on each. If the teacher wanted to see tables beginning at x=2 for one part and x=3 for the other part of this system of rectangle areas, then the teacher could have said so. What's wrong with rectangles having sides less than 1? x = 1/4, x = 1/2, x = 3/4, et cetera.

Final word on this: tell your daughter that, in the absence of specific instructions to the contrary, she is free to do as she pleases regarding all sorts of things (eg: how many digits when rounding a number, what symbols to assign in word problems, how much of a graph to draw, et cetera).

Instructors who want to volunteer comment on such choices ought to first consider the instructions they hand out.

Cheers

PS: Does your daughter know how to type?
 
Last edited:
Yes, she knows how to type... just a little slow at it. Am I not allowed to post for her?

She will be posting going forward... but only those questions that are part of a homework or test prep. She is not to enthused to post something from the optional worksheet, like the other question that I posted.

I am just trying to get her more interested in this stuff and thought this could be a great resource for her with questions that I cannot help her
with.

Thanks
 
There is no issue with you posting on behalf of your daughter. :cool:
 
Status
Not open for further replies.
Top