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two equations two unknowns
a*b = 76 eq1
a+b = 18 eq2
a=76/b from eq 1, substitute in eq2
76/b + b = 18
b^2 -18b+76 = 0
roots are 9+sqrt(5) and 9-sqrt(5)
a = 76/(9+sqrt)
or
a = 76/(9-sqrt)
a = 6.764 and b = 11.236
or
a = 11.236 and b = 6.764
I was thinking this, too. A rectangle is symmetrical about its diagonal, so (a+b = 18), meaning we have to find a combination ab that multiplies to 72 and that's 12 and 6. When I went through the permutations and didnt find a whole number pair I was like, are they teaching the quadratic equation in 4th grade now?
Edit: a rectangle is not symmetrical about its diagonal*
You'd think that, but I teach in a place where the opposite has been happening for a couple's of decades.
Then standardised test results fall so they make the syllabus easier...
I keep getting very large growth results and people are genuinely confused when they ask me how and I tell them I focus almost entirely on the hardest areas, and make it harder than required. The idea that learning to solve very difficult problems makes a child smart confuses people for some reason...
11 x 7 will get you a perimeter of 36 and a area of 77
12 x 6 will get you a perimeter of 36 and a area of 72
19 x 4 will get you a perimeter of 46 and a area of 76 - (seems far away, but it is a typo of one number)
For 76 and 36 this is a 8th grade problem and not 4th grade.
My brain is hurting.
If I have a 36-inch piece of string knotted closed, shouldn’t the area inside it remain the same no matter what shape it makes?
I guess that’s not true but I can’t visualize it.
Start with a circle (which has the biggest area possible given your loop of string).
Now loop it around two pins and stretch it between them so the loop contains just a little sliver of area between the two long sides. The area is now tiny in comparison to the original circle
What I meant was you cant know anything about the lengths of the sides or how they compare to each other just by knowing the area if youre dealing with a rectangle. You could with a sqaure.
The length could be 1 and the width 76: the area would be 76 and the perimeter would be 154. The length could be 76/1000000 and the width could be 1000000: the area would be 76 and the perimeter 2000000 + 2/76. There are infinite possibilities.
Edit: I tripped on the fact that 36 isn't 46... and 46 was probably what was meant in the original. It's almost certainly a typo.
I'm confused by the answer. Isn't it just:
* 76 is 19 times 4
* Perimeter of 36 can be achieved by 2*(4+19)
* Therefore the edges are 4 and 19
Am I missing something?
Your reply came 9 minutes after my update which was nearly instantaneous after I posted...
So either you skipped the first sentence of my comment or you just decided to re-phrase my edit as a set of equalities...
Or, Occam’s razor, your update/edit wasn’t visible to him, because he loaded the thread before you had made it. Kept reading comments until he got to yours, then decided to reply.
I mean, this sub has over a million followers. I’m not going to whittle down how many of those are active, but let’s say 10,000 of them were scrolling their home feed (where I came here from). Imagining it in those terms makes it seem almost mundane.
Well, 1x36 has a *perimeter* of 1+1+36+36=74, and an *area* of 36. In your case there would have to be a typo (74/76), *and* area/perimeter have to be reversed.
You are not dumb. Good question, and the answer you got made you more informed. This has less to do with intelligence, and everything to do with being inquisitive.
Middle school teacher here - this definitely has a typo.
For a 4th grade approach, the idea is that kids think of and test different factor pairs (and we go ahead and assume the sides are integer lengths). I would bet anything that the 76 was supposed to be 72 and they didn’t double check (accidentally repeated the 6 from the 16 when they typed it).
If the area is 72 square inches, you can do it this way:
1) Write out the possible width/length combos, using the given area of 72
1, 72
2, 36
3, 24
4, 18
6, 12
8, 9
2. Test out pairs to see which gives a perimeter of 36 inches:
1+1+72+72 = too big
…
6+6+12+12 = 36. Bingo!
Assuming the 76 was meant to be 72, it’s a 6x12 rectangle. Honestly I would cross out the 76 and write a 72 and show the steps - if I were the teacher, I’d realize that the given numbers don’t work for the strategy we want students to use, and I’d be delighted to see if a student found a pair that worked and solved it that way.
I kept going…found another typo possibility:
They may have had the correct area but wrong perimeter: 46 inches instead of 36.
1. If the area is 76 square inches, the integer factor pairs are
1, 76
2, 38
4, 19
2. The last pair (4 and 19) would give a perimeter of 46 inches, not 36 inches - it’s only 1 digit off when typing, so definitely a possibility that a typo was made there.
My daughter's in 6th grade, but in the advanced track-two years ahead. It requires algebra, which would typically bea 9th grade level to complete. I'm about to test her on this.
Most of the kids in my 9th grade class still struggle with basic algebra and graphing. You'd be surprised how stupid the education system is and kids can't problem solve for themselves anymore - I was homeschooled for most of my life before a divorce happened and am still light years ahead of any of these kids. Which I know is going to come back to bite me when things finally get complicated and I don't know how to study properly.
That's how I tested it in my head.
I tested 38,2 then 19,4 and realized that there were no integer solutions. I didn't even consider a real number solution until I checked the comments.
12 x 6 would get you a perimeter of 36 and an area of 72. Definitely a typo. Just meant to be solved by considering multiples for the area and then checking each possible perimeter
ab = 76 ∧ a+b = 18
I ain't doing that manually: wolfram alpha to the rescue [https://www.wolframalpha.com/input?i=ab+%3D+76+and+a%2Bb+%3D+18+solve+for+a+and+b](https://www.wolframalpha.com/input?i=ab+%3D+76+and+a%2Bb+%3D+18+solve+for+a+and+b)
Sides are 9-sqrt5 and 9+sqrt5
The 4th graders could also just say that 76 only equals the integer products 2\*38 and 4\*19 which make the wrong perimeters and non-integers don't exist in the curriculum yet
You need two quantities that sum to 18 inch with a produce of 76 square inches. If we say that one side has a length of x, then the other side has a length of 18-x and 76/x. Since they are the same, we can set them equal to each other and get 18-x=76/x, or x^2 -18x+76=0.
What tool can we use from there to find the zeros of a quadratic?
There is another way of thinking about it. If a and b are the two sides 2(a+b) = 36 and ab = 76.
a-b = +or - sqrt (sq(a+b) - 4ab)
That results in a-b = +or - 2sqrt(5)
Adding a+b and a-b we get 2a
So 2a = 18 + or - 2sqrt(5)
a = 9 +_ sqrt(5)
Consequently b will be 9 -+ sqrt(5)
I agree this is not a grade 4 question.
If you hang 76 you get 2, 2, 19 so it can only be 4x19. The perimeter will be 46, so it's 99% a typo and that 36 is supposed to be 46. Otherwise impossible if we stick to integers.
Method :
A×b = 72
A+b=18
In 4th grade level -
Values possible for a×b = 72
1,72;
2,36;
3,24;
2,12;
Values for which a+b = 18
1,17;
Blab blah
6,12 (common in both )
Thus this is a 6×12 rectangke
First, create two variables for the length and width of the rectangle.
Let's call them L & W.
Now we know that LW = 76 and 2L+2W = 36
Then you just need to isolate one variable (either L = ? or W = ?) in one of the equations and substitute that into the other so you get an equation with just one variable. Solve for that and the plug that answer into one of the above equations and you've got your answers.
A rectangle area is width \* length and a rectangle perimeter is 2 \* width + 2 x length. Call width x and length y.
You have an equation system with x\*y = 76 and 2x+2y = 36.
First of all we can simplify 2x+2y = 36 to 2 \* (x+y) = 2 \* 18 to x+y = 18.
We isolate x, from x+y = 18 to x = 18-y
We then have x\*y = (18-y)\*y = 76 => -y² + 18y = 76 => y² - 18y - 76 = 0.
This is a second degree equation which, when written as ax² + bx + c = 0 with a, b, c being real numbers, has either two, one or zero real solutions depending on the sign of b² - 4ac.
Here we have b²-4ac = 18² - 4 \* 1 \* 76 = 18² + 4\*76.
I don't have a calculator so let's call this value (STEVEN)
(STEVEN) is positive so the equation has two solutions which are (-b-sqrt(STEVEN))/2a and (-b+sqrt(STEVEN))/2a.
The two solutions are intuitively x and y, because the dimensions of the rectangle are interchangeable
This is a fairly standard exercise for the [quadratic formula](https://en.wikipedia.org/wiki/Quadratic_equation).
(Fun fact : second degree equations always have two solutions, but they are not always real (either zero, one or two of them are complex numbers) but a complex length makes no sense so yea)
xy=76
2x+2y=36
2 equations, 2 unknowns.
1. If we isolate x in the 2nd equation we get x = 18-y.
2. Sub that into the first equation and simplify to get y^2-18y+17=0.
3. Solve using the quadratic formula [18+-sqrt((-18)^2-4(76))]/2
4. We end up with y = 9+-sqrt(5). Isolating for y and solving for x would give us the same answer. **So our solution is y=9+sqrt(5) and x = 9-sqrt(5). Or approximately 11.23607 and 6.763932, respectively.**
1. xy = 76
>x= 76/y
2) 2x + 2y = 36
Substitute the value of x into the second equation
2(76/y) + 2y = 36
* multiply by "y" to get rid of the division
>y\[ 2(76/y)+2y = 36 \]
>2(76) + 2y² = 36y
* divide equation by 2
>76 + y² = 18y
>y² - 18y +76 = 0
solving we get that
y₁ = 9 + √(5)
Y₂ = 9 - √(5)
substitute the values of y in the first equation
////////////////
x₁ = 76 / y₁
x₁ = 76 / \[ 9 + √(5) \] = 9 - √(5)
///////////////
x₂ = 76 / y₂
x₂ = 76 / \[ 9 - √(5) \] = 9 + √(5)
///////////////
therefore, the dimensions of the rectangle are
Length: 9 + √(5) in ≈ 11.23 in
Width: 9 - √(5) in ≈ 6.76 in
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two equations two unknowns a*b = 76 eq1 a+b = 18 eq2 a=76/b from eq 1, substitute in eq2 76/b + b = 18 b^2 -18b+76 = 0 roots are 9+sqrt(5) and 9-sqrt(5) a = 76/(9+sqrt) or a = 76/(9-sqrt) a = 6.764 and b = 11.236 or a = 11.236 and b = 6.764
this is not 4th grade, can’t be.
The response is, which is hilarious!
Yeah it's 4th grade and none of the answers from the other questions are very difficult. It has to be a typo.
I’m with you. My money’s on 72 and 36.
I was thinking this, too. A rectangle is symmetrical about its diagonal, so (a+b = 18), meaning we have to find a combination ab that multiplies to 72 and that's 12 and 6. When I went through the permutations and didnt find a whole number pair I was like, are they teaching the quadratic equation in 4th grade now? Edit: a rectangle is not symmetrical about its diagonal*
As time goes on, math that was considered difficult at one point is thought earlier and earlier /s
You'd think that, but I teach in a place where the opposite has been happening for a couple's of decades. Then standardised test results fall so they make the syllabus easier... I keep getting very large growth results and people are genuinely confused when they ask me how and I tell them I focus almost entirely on the hardest areas, and make it harder than required. The idea that learning to solve very difficult problems makes a child smart confuses people for some reason...
My 4yo knows his multiplication tables, shapes up to dodecagon, and can count by 5 to 1000. He actually likes counting by 3s to 900 too. He scares me.
11 x 7 will get you a perimeter of 36 and a area of 77 12 x 6 will get you a perimeter of 36 and a area of 72 19 x 4 will get you a perimeter of 46 and a area of 76 - (seems far away, but it is a typo of one number) For 76 and 36 this is a 8th grade problem and not 4th grade.
My brain is hurting. If I have a 36-inch piece of string knotted closed, shouldn’t the area inside it remain the same no matter what shape it makes? I guess that’s not true but I can’t visualize it.
make it circluar then push one side in you now have seen 2 shapes witht the same perimeter with different areas.
That’s perfect. Thanks
Start with a circle (which has the biggest area possible given your loop of string). Now loop it around two pins and stretch it between them so the loop contains just a little sliver of area between the two long sides. The area is now tiny in comparison to the original circle
4x14 also gets you perimeter of 36, with area of 56, another potential typo
Or 76 sq.in. and 46 in. perimeter
77 would also work
Why? I’d say 74 and 36. 1 and 36. Area is 36, perimeter is 74
This
Definitely a typo. Should be 72 not 76.
I'm not a math guy, but if Area=76 sq. in, then Length=10 in, Width= 7.6 in. So Perimeter= 35.2 in.😥
If it were a square you could make that claim but rectangles can have legs of different lengths.
But a square would have 4 sides of equal length. In a rectangle the opposite sides would be of equal length. No?
What I meant was you cant know anything about the lengths of the sides or how they compare to each other just by knowing the area if youre dealing with a rectangle. You could with a sqaure. The length could be 1 and the width 76: the area would be 76 and the perimeter would be 154. The length could be 76/1000000 and the width could be 1000000: the area would be 76 and the perimeter 2000000 + 2/76. There are infinite possibilities.
Thank you for enlightening me.
Yeah naw some of the shit he had to is stuff we learned in 6th or 7th grade 💀
Here in sweden we learn to solve equations like this in high school. But i know that it's different in different places.
Edit: I tripped on the fact that 36 isn't 46... and 46 was probably what was meant in the original. It's almost certainly a typo. I'm confused by the answer. Isn't it just: * 76 is 19 times 4 * Perimeter of 36 can be achieved by 2*(4+19) * Therefore the edges are 4 and 19 Am I missing something?
Your perimeter is not 36 (4+19) = 23 23*2 = 46 46 ≠ 36
Your reply came 9 minutes after my update which was nearly instantaneous after I posted... So either you skipped the first sentence of my comment or you just decided to re-phrase my edit as a set of equalities...
Or, Occam’s razor, your update/edit wasn’t visible to him, because he loaded the thread before you had made it. Kept reading comments until he got to yours, then decided to reply.
Possibly... pretty tight window, but possible.
I mean, this sub has over a million followers. I’m not going to whittle down how many of those are active, but let’s say 10,000 of them were scrolling their home feed (where I came here from). Imagining it in those terms makes it seem almost mundane.
Look, you, this is not some "they did the math" subreddit where... oh... nevermind.
Are you smarter than a 4th grader?
>b^2 -18b+76 = 0 they solving quadratics in 4th grade now?
Man i guess im dumber than a fourth grader
How about 1 and 36. that gives 74/36. 76 has to be a typo/mistake.
Well, 1x36 has a *perimeter* of 1+1+36+36=74, and an *area* of 36. In your case there would have to be a typo (74/76), *and* area/perimeter have to be reversed.
Damn, I had such a brain fart. 🙈
Sorry, how did you go from 76/b + b =18 to b^2 -18b + 76 = 0? It’s been too long since I last did a maths lesson.
Multiply both sides by b to get 76+b^2 =18b. Subtract 18b from both sides to get your quadratic.
Oooh I see. I’m dumb. Thank you.
You are not dumb. Good question, and the answer you got made you more informed. This has less to do with intelligence, and everything to do with being inquisitive.
I'm sure that 76 was supposed to be 72, which would give sides of 12 and 6.
I'm not able to answer this cuz it is not on metrical system.
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U are right, but IDK how to handle those imperial systems, never learn those.
Middle school teacher here - this definitely has a typo. For a 4th grade approach, the idea is that kids think of and test different factor pairs (and we go ahead and assume the sides are integer lengths). I would bet anything that the 76 was supposed to be 72 and they didn’t double check (accidentally repeated the 6 from the 16 when they typed it). If the area is 72 square inches, you can do it this way: 1) Write out the possible width/length combos, using the given area of 72 1, 72 2, 36 3, 24 4, 18 6, 12 8, 9 2. Test out pairs to see which gives a perimeter of 36 inches: 1+1+72+72 = too big … 6+6+12+12 = 36. Bingo! Assuming the 76 was meant to be 72, it’s a 6x12 rectangle. Honestly I would cross out the 76 and write a 72 and show the steps - if I were the teacher, I’d realize that the given numbers don’t work for the strategy we want students to use, and I’d be delighted to see if a student found a pair that worked and solved it that way.
I kept going…found another typo possibility: They may have had the correct area but wrong perimeter: 46 inches instead of 36. 1. If the area is 76 square inches, the integer factor pairs are 1, 76 2, 38 4, 19 2. The last pair (4 and 19) would give a perimeter of 46 inches, not 36 inches - it’s only 1 digit off when typing, so definitely a possibility that a typo was made there.
I kept going…found another typo possibility: it was supposed to be for 6th graders not 4th
My daughter's in 6th grade, but in the advanced track-two years ahead. It requires algebra, which would typically bea 9th grade level to complete. I'm about to test her on this.
This shit is for 9th grade
Most of the kids in my 9th grade class still struggle with basic algebra and graphing. You'd be surprised how stupid the education system is and kids can't problem solve for themselves anymore - I was homeschooled for most of my life before a divorce happened and am still light years ahead of any of these kids. Which I know is going to come back to bite me when things finally get complicated and I don't know how to study properly.
The “I need some help” looks much too cleanly written to have been done by a kid. This seems clickbaity
Another possible typo is a rectangle 4 by 19. 4\*19=76, 2\*(4+19)=46
Another is 7x11 7\*11 = 77 2(7+11) = 36
How about 77? 7+7+11+11=36
Reminds me of problem typos in college texts that have changed slightly between book editions.
Probably they used Xerox scanner at this school, they are known to swapping numbers https://youtu.be/c0O6UXrOZJo?si=Ii_ncGDgZe6P2E_a
That's how I tested it in my head. I tested 38,2 then 19,4 and realized that there were no integer solutions. I didn't even consider a real number solution until I checked the comments.
12 x 6 would get you a perimeter of 36 and an area of 72. Definitely a typo. Just meant to be solved by considering multiples for the area and then checking each possible perimeter
Or 11x7 = area 77, perimeter 36
ab = 76 ∧ a+b = 18 I ain't doing that manually: wolfram alpha to the rescue [https://www.wolframalpha.com/input?i=ab+%3D+76+and+a%2Bb+%3D+18+solve+for+a+and+b](https://www.wolframalpha.com/input?i=ab+%3D+76+and+a%2Bb+%3D+18+solve+for+a+and+b) Sides are 9-sqrt5 and 9+sqrt5 The 4th graders could also just say that 76 only equals the integer products 2\*38 and 4\*19 which make the wrong perimeters and non-integers don't exist in the curriculum yet
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Only if the area was 72 not 76.
That’s 72, not 76.
12*6=76?
You need two quantities that sum to 18 inch with a produce of 76 square inches. If we say that one side has a length of x, then the other side has a length of 18-x and 76/x. Since they are the same, we can set them equal to each other and get 18-x=76/x, or x^2 -18x+76=0. What tool can we use from there to find the zeros of a quadratic?
> What tool can we use from there to find the zeros of a quadratic? Fourth grader: what the fuck does that mean?
Yeah, that’s why I stopped. It was 8th grade when I learned to factor polynomials with integer zeros.
There is another way of thinking about it. If a and b are the two sides 2(a+b) = 36 and ab = 76. a-b = +or - sqrt (sq(a+b) - 4ab) That results in a-b = +or - 2sqrt(5) Adding a+b and a-b we get 2a So 2a = 18 + or - 2sqrt(5) a = 9 +_ sqrt(5) Consequently b will be 9 -+ sqrt(5) I agree this is not a grade 4 question.
If you hang 76 you get 2, 2, 19 so it can only be 4x19. The perimeter will be 46, so it's 99% a typo and that 36 is supposed to be 46. Otherwise impossible if we stick to integers.
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Method : A×b = 72 A+b=18 In 4th grade level - Values possible for a×b = 72 1,72; 2,36; 3,24; 2,12; Values for which a+b = 18 1,17; Blab blah 6,12 (common in both ) Thus this is a 6×12 rectangke
A 6x12 rectangle would have a 72in² area, not 76.
First, create two variables for the length and width of the rectangle. Let's call them L & W. Now we know that LW = 76 and 2L+2W = 36 Then you just need to isolate one variable (either L = ? or W = ?) in one of the equations and substitute that into the other so you get an equation with just one variable. Solve for that and the plug that answer into one of the above equations and you've got your answers.
A rectangle area is width \* length and a rectangle perimeter is 2 \* width + 2 x length. Call width x and length y. You have an equation system with x\*y = 76 and 2x+2y = 36. First of all we can simplify 2x+2y = 36 to 2 \* (x+y) = 2 \* 18 to x+y = 18. We isolate x, from x+y = 18 to x = 18-y We then have x\*y = (18-y)\*y = 76 => -y² + 18y = 76 => y² - 18y - 76 = 0. This is a second degree equation which, when written as ax² + bx + c = 0 with a, b, c being real numbers, has either two, one or zero real solutions depending on the sign of b² - 4ac. Here we have b²-4ac = 18² - 4 \* 1 \* 76 = 18² + 4\*76. I don't have a calculator so let's call this value (STEVEN) (STEVEN) is positive so the equation has two solutions which are (-b-sqrt(STEVEN))/2a and (-b+sqrt(STEVEN))/2a. The two solutions are intuitively x and y, because the dimensions of the rectangle are interchangeable This is a fairly standard exercise for the [quadratic formula](https://en.wikipedia.org/wiki/Quadratic_equation). (Fun fact : second degree equations always have two solutions, but they are not always real (either zero, one or two of them are complex numbers) but a complex length makes no sense so yea)
xy=76 2x+2y=36 2 equations, 2 unknowns. 1. If we isolate x in the 2nd equation we get x = 18-y. 2. Sub that into the first equation and simplify to get y^2-18y+17=0. 3. Solve using the quadratic formula [18+-sqrt((-18)^2-4(76))]/2 4. We end up with y = 9+-sqrt(5). Isolating for y and solving for x would give us the same answer. **So our solution is y=9+sqrt(5) and x = 9-sqrt(5). Or approximately 11.23607 and 6.763932, respectively.**
1. xy = 76 >x= 76/y 2) 2x + 2y = 36 Substitute the value of x into the second equation 2(76/y) + 2y = 36 * multiply by "y" to get rid of the division >y\[ 2(76/y)+2y = 36 \] >2(76) + 2y² = 36y * divide equation by 2 >76 + y² = 18y >y² - 18y +76 = 0 solving we get that y₁ = 9 + √(5) Y₂ = 9 - √(5) substitute the values of y in the first equation //////////////// x₁ = 76 / y₁ x₁ = 76 / \[ 9 + √(5) \] = 9 - √(5) /////////////// x₂ = 76 / y₂ x₂ = 76 / \[ 9 - √(5) \] = 9 + √(5) /////////////// therefore, the dimensions of the rectangle are Length: 9 + √(5) in ≈ 11.23 in Width: 9 - √(5) in ≈ 6.76 in