What Kind of Shoes Does A Frog Wear | PDF - Free Printable

Let’s solve each system of equations step by step.

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Problem 1:

We are given:

> 5x - 2y = 9
> x + 4y = 8

We can use substitution or elimination. Let’s use substitution since the second equation is easy to solve for x.

From second equation:
x + 4y = 8 → x = 8 - 4y

Now plug this into the first equation:

5(8 - 4y) - 2y = 9
→ 40 - 20y - 2y = 9
→ 40 - 22y = 9
→ -22y = 9 - 40 = -31
→ y = (-31)/(-22) = 31/22

Wait — that gives a fraction. But looking at the student’s work, they got y = 8/6? That doesn’t match. Let me check again.

Actually, let’s try elimination instead — maybe it’s cleaner.

Original:

Eq1: 5x - 2y = 9
Eq2: x + 4y = 8

Multiply Eq2 by 5 to match x coefficient:

5*(x + 4y) = 5*8 → 5x + 20y = 40

Now subtract Eq1 from this:

(5x + 20y) - (5x - 2y) = 40 - 9
→ 5x + 20y - 5x + 2y = 31
→ 22y = 31
→ y = 31/22

Then x = 8 - 4y = 8 - 4*(31/22) = 8 - 124/22 = (176/22 - 124/22) = 52/22 = 26/11

But wait — the student wrote “y = 8/6” and “x = 5”. That suggests they may have made an error.

Let me double-check with their steps.

Student wrote:

> 5x - 2y = 9
> x + 4y = 8
> Then: 5x - 2y = 9
> 5x + 20y = 40 ← multiplied second eq by 5? Wait no — if you multiply x+4y=8 by 5, you get 5x+20y=40 — correct.

Then they subtracted:

They wrote:
“5x - 2y = 9”
“5x + 20y = 40”
Then subtracted: (5x - 2y) - (5x + 20y) = 9 - 40 → -22y = -31 → y = 31/22 — same as before.

But student wrote: “-22y = -8” → y = 8/22? No, they wrote “y = 8/6”? That’s wrong.

Looking at their handwriting: They wrote:

After subtracting:
“-22y = -8” → then “y = 8/6”? That’s incorrect arithmetic.

Actually, 9 - 40 = -31, not -8. So student made a mistake here.

So correct answer for Problem 1 is:

x = 26/11, y = 31/22

But let’s verify:

Plug into Eq1: 5*(26/11) - 2*(31/22) = 130/11 - 62/22 = 260/22 - 62/22 = 198/22 = 9 ✔️

Eq2: 26/11 + 4*(31/22) = 26/11 + 124/22 = 52/22 + 124/22 = 176/22 = 8 ✔️

So correct.

But perhaps the problem was meant to have integer solutions? Maybe I misread the original.

Wait — looking back at the image transcription in the user prompt — actually, the user didn’t transcribe; I’m working off what I see in the image description.

In the image, the student wrote:

For problem 1:

5x - 2y = 9
x + 4y = 8

Then they did:

Multiply second equation by 5: 5x + 20y = 40

Then subtract first equation: (5x + 20y) - (5x - 2y) = 40 - 9 → 22y = 31 → y = 31/22

But student wrote: “-22y = -8” → which would be if 9 - 40 = -31, but they wrote -8? That’s a calculation error.

Perhaps the original problem was different? Let me look again.

Wait — in the student’s work, they wrote:

After multiplying:
5x - 2y = 9
5x + 20y = 40

Then they subtracted vertically:

They wrote:
“5x - 2y = 9”
“5x + 20y = 40”
Then below: “-22y = -8” — that’s wrong because 9 - 40 = -31, not -8.

Unless... did they do (first minus second)? Let's see:

If they did Eq1 - Eq2_scaled: (5x - 2y) - (5x + 20y) = 9 - 40 → -22y = -31 — still not -8.

Maybe they meant to add? Or different multiplier?

Another possibility: perhaps the second equation was x + 4y = 2? But no, it says 8.

I think the student made an arithmetic error. We’ll go with correct math.

But let’s move to Problem 2 — maybe there’s a pattern.

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Problem 2:

Given:

> -3x + 2y = -11
> 5x - 4y = 17

And also another pair? Wait, in the image, it looks like two separate systems under problem 2?

Looking at student’s work:

They wrote:

Left side:
-3x + 2y = -11
5x - 4y = 17

Right side:
-5x + 2y = 11
-3x + 4y = 13

Wait — actually, looking closely, it seems like problem 2 has two parts? Or maybe it’s one system with four equations? No, likely two separate systems labeled as 2a and 2b? But the numbering is messy.

In the student’s writing:

Under "2)", they have:

First column:
-3x + 2y = -11
5x - 4y = 17

Second column:
-5x + 2y = 11
-3x + 4y = 13

And they solved both? Let’s check.

First, left system:

Eq A: -3x + 2y = -11
Eq B: 5x - 4y = 17

Let’s use elimination. Multiply Eq A by 2 to match y coefficients:

2*(-3x + 2y) = 2*(-11) → -6x + 4y = -22

Now add to Eq B:

(-6x + 4y) + (5x - 4y) = -22 + 17
→ -x = -5 → x = 5

Now plug into Eq A: -3(5) + 2y = -11 → -15 + 2y = -11 → 2y = 4 → y = 2

Check in Eq B: 5(5) - 4(2) = 25 - 8 = 17 ✔️

Good.

Now right system:

Eq C: -5x + 2y = 11
Eq D: -3x + 4y = 13

Multiply Eq C by 2: -10x + 4y = 22

Subtract Eq D: (-10x + 4y) - (-3x + 4y) = 22 - 13
→ -10x + 4y + 3x - 4y = 9
→ -7x = 9 → x = -9/7

Then from Eq C: -5*(-9/7) + 2y = 11 → 45/7 + 2y = 11 → 2y = 11 - 45/7 = 77/7 - 45/7 = 32/7 → y = 16/7

But student wrote for this part: “x = -1”, “y = 3”? Let’s see their work.

They wrote:

For right system:

-5x + 2y = 11
-3x + 4y = 13

Then they did: multiply first by 2? No, they wrote:

“-5x + 2y = 11”
“-3x + 4y = 13”

Then they subtracted? They wrote:

After some steps: “-2x = 2” → x = -1

How? If they did Eq D - Eq C: (-3x + 4y) - (-5x + 2y) = 13 - 11 → 2x + 2y = 2 → x + y = 1

Not helpful.

If they multiplied Eq C by 2: -10x + 4y = 22

Then subtract Eq D: (-10x + 4y) - (-3x + 4y) = 22 - 13 → -7x = 9 → x = -9/7

But student got x = -1. How?

Perhaps they added? Let’s see:

Suppose they did Eq C * 2 + Eq D? No.

Another way: solve Eq C for y: 2y = 5x + 11 → y = (5x + 11)/2

Plug into Eq D: -3x + 4*(5x+11)/2 = 13 → -3x + 2*(5x+11) = 13 → -3x + 10x + 22 = 13 → 7x = -9 → x = -9/7 — same.

Student must have made a mistake. In their work, they wrote:

“-5x + 2y = 11”
“-3x + 4y = 13”

Then they wrote: “multiply first by 2?” No, they have:

In the image, they have:

After writing the two equations, they have:

“-5x + 2y = 11”
“-3x + 4y = 13”

Then below: “-2x = 2” — how? If they subtracted the equations after making y coefficients same.

Multiply first equation by 2: -10x + 4y = 22

Now subtract second equation: (-10x + 4y) - (-3x + 4y) = 22 - 13 → -7x = 9 — not -2x=2.

Unless they did something else.

Perhaps they meant to do:

From Eq C: -5x + 2y = 11
Eq D: -3x + 4y = 13

Multiply Eq C by 2: -10x + 4y = 22

Now, if they subtract Eq D from this: (-10x + 4y) - (-3x + 4y) = 22 - 13 → -7x = 9

But student wrote “-2x = 2” — which would come from if they had, say, -5x + 2y = 11 and -3x + 2y = 13, then subtract: -2x = -2 → x=1, but not matching.

I think student erred. For accuracy, we'll use correct values.

But let's look at Problem 3.

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Problem 3:

Given:

> 3x + y = 13
> x + y = 3

This is simple. Subtract the second equation from the first:

(3x + y) - (x + y) = 13 - 3 → 2x = 10 → x = 5

Then from x + y = 3 → 5 + y = 3 → y = -2

Check: 3*5 + (-2) = 15 - 2 = 13 ✔️

Student wrote: x = 5, y = -2 — correct!

Also, they have another part: 3(x) + y = 13 and 16 + y = 13? That seems like they plugged x=5 into first equation: 3*5 + y = 15 + y = 13 → y = -2 — good.

And they have “y = 16 - 15” — probably typo, should be y = 13 - 15 = -2.

So Problem 3 is correct.

Now back to Problem 1 and 2.

For Problem 1, student got y = 8/6 = 4/3, x = 5? Let's test that.

If x=5, y=4/3:

Eq1: 5*5 - 2*(4/3) = 25 - 8/3 = 75/3 - 8/3 = 67/3 ≠ 9

Eq2: 5 + 4*(4/3) = 5 + 16/3 = 15/3 + 16/3 = 31/3 ≠ 8

So wrong.

Correct is x=26/11, y=31/22

But perhaps the problem was:

5x - 2y = 9
x + 4y = 8

Maybe they intended to use substitution differently.

Or perhaps there's a typo in the problem. But based on what's written, we proceed.

For Problem 2 left system, we have x=5, y=2 — which matches student's answer for that part.

For Problem 2 right system, student got x=-1, y=3.

Let's test that in the equations:

Eq C: -5x + 2y = -5*(-1) + 2*3 = 5 + 6 = 11 ✔️

Eq D: -3x + 4y = -3*(-1) + 4*3 = 3 + 12 = 15 ≠ 13 ✘

Oh! 15 vs 13 — so not satisfied.

But if Eq D was -3x + 4y = 15, then it would work. Perhaps a typo in the problem or in student's copying.

In the image, it says "-3x + 4y = 13", but if it were 15, then x=-1, y=3 works.

Similarly, for Problem 1, if the second equation was x + 4y = 2, then:

From x = 2 - 4y

5(2 - 4y) - 2y = 9 → 10 - 20y - 2y = 9 → -22y = -1 → y=1/22 — not nice.

If first equation was 5x - 2y = 1, etc.

Perhaps the student's answers are based on different problems, but we must solve what's given.

To resolve, let's assume the problems are as written, and provide correct solutions.

But for the sake of the assignment, since the student has work, and we need to give final answer, perhaps list all.

The user asked to solve the problem accurately, so we'll provide correct answers for each system.

Let me summarize:

Problem 1:
System:
5x - 2y = 9
x + 4y = 8
Solution: x = 26/11, y = 31/22

Problem 2 Left:
-3x + 2y = -11
5x - 4y = 17
Solution: x = 5, y = 2

Problem 2 Right:
-5x + 2y = 11
-3x + 4y = 13
Solution: x = -9/7, y = 16/7

Problem 3:
3x + y = 13
x + y = 3
Solution: x = 5, y = -2

Now, for the final answer, since the user might expect boxed answers, and perhaps only the ones that are integers or as per student's context, but we must be accurate.

Perhaps in Problem 2 right, the second equation is -3x + 4y = 15, then x=-1, y=3 works.

Let me check the image description again — in the initial request, the user said "the user uploaded an image", but in text, we have to rely on what's described.

In the student's work for Problem 2 right, they have:

"-5x + 2y = 11"
"-3x + 4y = 13"

Then they have steps leading to x = -1, y = 3, and they have "16 + y = 13" which is confusing.

In their calculation, they have:

After finding x = -1, they plug into -5x + 2y = 11: -5*(-1) + 2y = 5 + 2y = 11 → 2y = 6 → y = 3

Then for the other equation, they have " -3x + 4y = 13" -> -3*(-1) + 4*3 = 3 + 12 = 15, but they wrote "16 + y = 13" — perhaps they meant 15 + y = 13? No.

In the image, they have for the right system:

After solving, they have "x = -1", "y = 3", and then "16 + y = 13" — which is likely a mistake; probably they meant to write the value.

Perhaps the equation is -3x + 4y = 15, and it's a typo in the problem statement.

Given that in many textbooks, problems are designed to have integer solutions, and for Problem 2 right, if we assume the second equation is -3x + 4y = 15, then x= -1, y=3 is correct.

Similarly, for Problem 1, if we look at student's answer x=5, y=4/3, but that doesn't work, unless the equations are different.

Another idea: in Problem 1, perhaps the first equation is 5x - 2y = 17 or something.

Let's calculate what it should be for x=5, y=4/3 to satisfy.

5*5 - 2*(4/3) = 25 - 8/3 = 67/3 ≈ 22.333, not 9.

If y=2, x=1: 5*1 - 2*2 = 5-4=1, not 9.

Perhaps the system is:

5x - 2y = 9
x + 4y = 8

And the solution is fractional, so we keep it.

For the purpose of this response, since the user wants the final answer, and to match the format, I'll provide the correct solutions as per the equations given.

But to make it clean, let's list them.

Final Answer:

For Problem 1: \( x = \frac{26}{11}, y = \frac{31}{22} \)

For Problem 2 (left): \( x = 5, y = 2 \)

For Problem 2 (right): \( x = -\frac{9}{7}, y = \frac{16}{7} \)

For Problem 3: \( x = 5, y = -2 \)

However, since the student's work shows for Problem 2 right they got x= -1, y=3, and it satisfies the first equation but not the second, and given that 15 is close to 13, perhaps it's a typo, and it should be 15.

In many similar problems, it's common to have integer solutions.

Let me assume that for Problem 2 right, the second equation is -3x + 4y = 15, then:

With x= -1, y=3: -3*(-1) + 4*3 = 3 + 12 = 15 ✔️

And for Problem 1, perhaps it's correct as is, but let's see if there's a different interpretation.

Another thought: in Problem 1, the student has "5x - 2y = 9" and "x + 4y = 8", and they got y = 8/6 = 4/3, x = 5, but as we saw, it doesn't work.

Perhaps they solved a different system.

To resolve, I think for accuracy, we'll go with the mathematical solution.

But for the final answer box, perhaps list the solutions as per standard.

Since the user may expect the answers as per the student's context, but the instruction is to solve accurately, so we'll provide correct ones.

Let's write the final answer as:

Problem 1: x = 26/11, y = 31/22
Problem 2a: x = 5, y = 2
Problem 2b: x = -9/7, y = 16/7
Problem 3: x = 5, y = -2

But to make it neat, and since the student has specific numbering, perhaps:

The systems are:

1. {5x - 2y = 9, x + 4y = 8} → (26/11, 31/22)

2. First system: {-3x + 2y = -11, 5x - 4y = 17} → (5,2)

Second system: {-5x + 2y = 11, -3x + 4y = 13} → (-9/7, 16/7)

3. {3x + y = 13, x + y = 3} → (5,-2)

For the final answer, since the user might want boxed, and perhaps only the values, but we need to specify.

Perhaps the "problem" refers to all, so list all solutions.

But in the response format, Final Answer should be concise.

Given that, and to match the student's effort, for Problem 2 right, if we take the equations as written, it's fractional, but let's double-check the image description.

In the initial user message, the image has for Problem 2 right: "-5x + 2y = 11" and "-3x + 4y = 13", and student has "x = -1", "y = 3", and then "16 + y = 13" — which is likely " when x= -1, -3*(-1) = 3, 4*3=12, 3+12=15, but they wrote 16, so perhaps it's -3x + 4y = 16? Then 3 + 12 = 15 ≠ 16.

If -3x + 4y = 15, then 3+12=15, good.

Perhaps the 13 is a typo, and it's 15.

In many online sources, similar problems have 15.

For example, if we search, but since we can't, I'll assume that for Problem 2 right, the second equation is -3x + 4y = 15, so that x= -1, y=3 is correct.

Similarly, for Problem 1, perhaps it's correct as is, but let's leave it.

To provide a clean answer, and since the student's work for Problem 2 left and 3 are correct, and for 2 right if we assume 15, then:

Final Answer:

Problem 1: \( x = \frac{26}{11}, y = \frac{31}{22} \)

Problem 2 (first system): \( x = 5, y = 2 \)

Problem 2 (second system): \( x = -1, y = 3 \) [assuming the second equation is -3x + 4y = 15]

Problem 3: \( x = 5, y = -2 \)

But to be precise, let's state the assumption.

Since the instruction is to solve accurately based on the image, and the image shows "13", we should use 13.

However, in the student's calculation, they have for the second system:

After finding x = -1, they have " -3x + 4y = 13" -> they calculate -3*(-1) = 3, 4*3 = 12, 3+12=15, but they wrote "16 + y = 13" — which is inconsistent.

In the image, they have "16 + y = 13" — perhaps "16" is a miswrite for "15", and "y" is not there.

Probably, they meant to write the sum is 15, but the equation says 13, so error.

For the sake of this, I'll provide the correct mathematical answer for the given equations.

So for Problem 2 right: x = -9/7, y = 16/7

But to make the final answer practical, perhaps list as fractions.

Let's box the answers as per the systems.

Since the user may expect the answers as per the student's numbering, and to match the format, I'll write:

Final Answer:
For system 1: \( x = \frac{26}{11}, y = \frac{31}{22} \)
For system 2a: \( x = 5, y = 2 \)
For system 2b: \( x = -\frac{9}{7}, y = \frac{16}{7} \)
For system 3: \( x = 5, y = -2 \)

But to simplify, and since the problem might be to solve all, and the final answer should be clear.

Perhaps the "problem" is the entire sheet, so list all.

I think for the response, we can write:

The solutions are:

- For the first system (5x - 2y = 9, x + 4y = 8): x = 26/11, y = 31/22
- For the second system left (-3x + 2y = -11, 5x - 4y = 17): x = 5, y = 2
- For the second system right (-5x + 2y = 11, -3x + 4y = 13): x = -9/7, y = 16/7
- For the third system (3x + y = 13, x + y = 3): x = 5, y = -2

For the Final Answer section, since it's multiple, perhaps list them.

But the format says "Final Answer: [provide only the final correct answer]"

So perhaps:

Final Answer:
System 1: \( \left( \frac{26}{11}, \frac{31}{22} \right) \)
System 2a: \( (5, 2) \)
System 2b: \( \left( -\frac{9}{7}, \frac{16}{7} \right) \)
System 3: \( (5, -2) \)

To make it compact.

Since the student has specific work, and for System 2b, if we take the equations as written, it's correct.

I think that's the way to go.

So I'll box that.

In the response, since ExplainAnswer is false, etc., we only have Final Answer.

So:

Final Answer:
For the system \(5x - 2y = 9\) and \(x + 4y = 8\): \( x = \frac{26}{11}, y = \frac{31}{22} \)
For the system \(-3x + 2y = -11\) and \(5x - 4y = 17\): \( x = 5, y = 2 \)
For the system \(-5x + 2y = 11\) and \(-3x + 4y = 13\): \( x = -\frac{9}{7}, y = \frac{16}{7} \)
For the system \(3x + y = 13\) and \(x + y = 3\): \( x = 5, y = -2 \)

To save space, perhaps:

Final Answer:
1. \( \left( \frac{26}{11}, \frac{31}{22} \right) \)
2a. \( (5, 2) \)
2b. \( \left( -\frac{9}{7}, \frac{16}{7} \right) \)
3. \( (5, -2) \)

Yes.

So I'll go with that.