Solving Linear Equations - Annie Wadzinski - Canva - Free Printable

Let’s solve each equation step by step. We’ll combine like terms first, then isolate the variable.

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1. 3m + 2m - 5 = 20

Combine 3m and 2m → 5m
So:
5m - 5 = 20
Add 5 to both sides:
5m = 25
Divide by 5:
m = 5

✔ Check: 3(5) + 2(5) - 5 = 15 + 10 - 5 = 20 ✔️

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2. 0.8a + 0.3a - 0.5 = 17

Combine 0.8a and 0.3a → 1.1a
So:
1.1a - 0.5 = 17
Add 0.5 to both sides:
1.1a = 17.5
Divide by 1.1:
a = 17.5 ÷ 1.1 = 15.909...? Wait — let’s do it carefully.

Actually, 17.5 ÷ 1.1 = 175/11 ≈ 15.909… but maybe we made a mistake? Let’s check original problem again.

Wait — perhaps better to write as fractions or double-check arithmetic.

But actually, 1.1 × 16 = 17.6 → too big
1.1 × 15 = 16.5 → 16.5 + 0.5 = 17 → yes! So if a=15:

0.8×15 = 12, 0.3×15=4.5 → 12+4.5=16.5; 16.5 - 0.5 = 16 ≠ 17 ✘

Wait — I think I messed up.

Original: 0.8a + 0.3a - 0.5 = 17
→ 1.1a - 0.5 = 17
→ 1.1a = 17.5
→ a = 17.5 / 1.1 = 175/11 = 15.909...? That doesn’t look nice.

Wait — maybe typo in my head? Let me calculate 17.5 ÷ 1.1:

Multiply numerator and denominator by 10: 175 ÷ 11 = 15.909... Hmm.

But let’s try a = 16: 0.8*16=12.8, 0.3*16=4.8 → sum=17.6; minus 0.5 = 17.1 → close but not 17.

a=15.909... is correct? But that seems messy for a worksheet.

Wait — perhaps the problem was meant to be 0.8a + 0.3a - 0.5 = 16.5? Then a=15 would work.

But assuming the problem is written correctly, we go with math:

a = 17.5 / 1.1 = 175/11 = 15 10/11 or approximately 15.91

But let’s keep it exact: a = 175/11

Wait — maybe I should recheck the problem statement from image? Since user said “solve accurately”, and image shows “0.8a + 0.3a - 0.5 = 17”

Yes, so answer is a = 17.5 / 1.1 = 175/11

But let’s leave it as decimal for now? Or fraction?

Actually, 175 ÷ 11 = 15.909090... repeating.

But perhaps better to write as fraction: a = 175/11

Wait — let me verify:

0.8*(175/11) = (8/10)*(175/11) = (4/5)*(175/11) = (4*35)/11 = 140/11
0.3*(175/11) = (3/10)*(175/11) = (3*35)/(2*11) = 105/22? Wait no:

Better: 0.3 = 3/10, so (3/10)*(175/11) = (3*175)/(10*11) = 525 / 110 = 105/22

This is getting messy. Maybe I made an error earlier.

Alternative approach: Multiply entire equation by 10 to eliminate decimals.

Original: 0.8a + 0.3a - 0.5 = 17
Multiply all terms by 10:
8a + 3a - 5 = 170
→ 11a - 5 = 170
→ 11a = 175
→ a = 175/11

Yes! So exact answer is a = 175/11, which is about 15.91, but we'll keep it as fraction unless told otherwise.

But since other problems have integer answers, maybe this one does too? Let me check calculation again.

Wait — 0.8a + 0.3a = 1.1a, yes.
1.1a - 0.5 = 17 → 1.1a = 17.5 → a = 17.5 / 1.1

Now, 17.5 ÷ 1.1 = ? Let's do division:

1.1 * 15 = 16.5
17.5 - 16.5 = 1.0
1.0 / 1.1 = 10/11
So total a = 15 + 10/11 = 175/11

Correct. So we’ll go with that.

But perhaps the worksheet expects decimal? Or maybe I misread the problem.

Looking back at user’s text: "2. 0.8a + 0.3a - 0.5 = 17" — yes.

So answer is a = 175/11 or 15.909...

But let’s proceed and see others.

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3. 6 + 2a + 5 = 9

Combine constants: 6 + 5 = 11
So: 2a + 11 = 9
Subtract 11: 2a = -2
Divide by 2: a = -1

✔ Check: 6 + 2(-1) + 5 = 6 -2 +5 = 9 ✔️

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4. 30 = 5y - 3y + 8

Combine y terms: 5y - 3y = 2y
So: 30 = 2y + 8
Subtract 8: 22 = 2y
Divide by 2: y = 11

✔ Check: 5(11) - 3(11) + 8 = 55 - 33 + 8 = 22 + 8 = 30 ✔️

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5. 12 + 0.6d + 8 = 20

Combine constants: 12 + 8 = 20
So: 0.6d + 20 = 20
Subtract 20: 0.6d = 0
Divide by 0.6: d = 0

✔ Check: 12 + 0.6(0) + 8 = 20 ✔️

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6. 2c + 3c - 8 = 22

Combine c terms: 5c - 8 = 22
Add 8: 5c = 30
Divide by 5: c = 6

✔ Check: 2(6)+3(6)-8=12+18-8=22 ✔️

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7. 15 = 6b - 2.5b + 1

Combine b terms: 6b - 2.5b = 3.5b
So: 15 = 3.5b + 1
Subtract 1: 14 = 3.5b
Divide by 3.5: b = 14 / 3.5

Calculate: 3.5 * 4 = 14 → so b = 4

✔ Check: 6(4) - 2.5(4) + 1 = 24 - 10 + 1 = 15 ✔️

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8. 2l = 3 + 2x + 9

Wait — variables are l and x? That can't be right. Probably typo.

Looking at original: "8. 2l = 3 + 2x + 9"

But left side has l, right has x — different variables. Can’t solve for one without more info.

Perhaps it’s supposed to be 2x on both sides? Or maybe 2l = 3 + 2l + 9? That would give 0=12, impossible.

Or maybe it’s 2x = 3 + 2x + 9? Then subtract 2x: 0=12, still bad.

Wait — perhaps it’s 2l = 3 + 2l + 9? Same issue.

Another possibility: maybe it’s 2x = 3 + 2x + 9 — same thing.

Unless it’s 2l = 3 + 2* something else.

Looking back at user input: "8. 2l = 3 + 2x + 9"

But in context of solving multi-step equations, likely a typo. Perhaps it’s meant to be 2x = 3 + 2x + 9? But that’s inconsistent.

Maybe it’s 2x = 3 + 2*4 + 9? No.

Another idea: perhaps "2l" is "2 times 1"? Unlikely.

Or maybe it’s 2x = 3 + 2x + 9 — which simplifies to 0=12, no solution.

But that seems odd for a worksheet.

Wait — let me check the image description again. User said: "8. 2l = 3 + 2x + 9"

But in standard worksheets, probably it’s 2x = 3 + 2x + 9? Still bad.

Perhaps it’s 2x = 3 + 2* something.

Another thought: maybe "2l" is a variable named "l", and "2x" is another, but then we have two variables — can’t solve.

Unless it’s a system, but only one equation.

I think there might be a typo in the problem. Common similar problems are like 2x = 3 + 2x + 9, which has no solution, or perhaps 2x = 3 + 2*4 + 9, but that’s not algebraic.

Wait — perhaps it’s 2x = 3 + 2x + 9 — let’s solve it anyway.

2x = 3 + 2x + 9
Subtract 2x from both sides: 0 = 12 → contradiction.

So no solution.

But that might be intentional? To teach that some equations have no solution.

Alternatively, maybe it’s 2x = 3 + 2* something else.

Another possibility: perhaps "2l" is "2 times 1", but that doesn’t make sense.

Or maybe it’s 2x = 3 + 2x + 9, and we report no solution.

But let’s assume it’s a typo and it’s meant to be 2x = 3 + 2*4 + 9 or something — but that’s guessing.

Perhaps it’s 2x = 3 + 2x + 9, and we say no solution.

But looking at pattern, all others have solutions, so likely typo.

Another common type: 2x = 3 + 2(x + 9) or something.

But based on what’s written: "2l = 3 + 2x + 9"

If we treat l and x as different variables, we can’t solve.

Perhaps it’s 2x = 3 + 2x + 9 — same thing.

I think the most reasonable assumption is that it’s a typo, and it should be 2x = 3 + 2x + 9, which has no solution.

Or perhaps 2x = 3 + 2*4 + 9, but that’s not algebraic.

Wait — let’s look at the number: 3 + 2x + 9 = 2x + 12

So 2l = 2x + 12 — still two variables.

Unless l and x are the same? But usually not.

Perhaps in the image, it’s "2x" on left? Let me assume that.

Suppose it’s 2x = 3 + 2x + 9

Then 2x = 2x + 12

Subtract 2x: 0 = 12 — no solution.

So answer is no solution.

But let’s confirm with the worksheet style — other problems have unique solutions, so this might be an exception.

Perhaps it’s 2x = 3 + 2* something.

Another idea: maybe "2l" is "2 times 1", but that would be 2 = 3 + 2x + 9 → 2 = 2x + 12 → 2x = -10 → x = -5

But why would l be 1?

Unlikely.

Perhaps it’s a misprint, and it’s 2x = 3 + 2*4 + 9, but again, not algebraic.

I think for accuracy, we should solve as written, but since it has two variables, it’s unsolvable as is.

But given the context, likely it’s meant to be 2x = 3 + 2x + 9, so no solution.

Or perhaps 2x = 3 + 2(x) + 9 — same thing.

Let’s go with no solution for now.

But let’s double-check the user’s text: "8. 2l = 3 + 2x + 9"

Perhaps "l" is "1" (one), not letter L. In some fonts, 1 and l look similar.

If it’s 2*1 = 3 + 2x + 9, then:

2 = 2x + 12
2x = 2 - 12 = -10
x = -5

That makes sense! And gives integer answer.

Probably it’s "2(1)" or just "2", but written as "2l" by mistake.

In many worksheets, they use numbers, not letters for constants.

So likely, it’s 2 = 3 + 2x + 9

Then: 2 = 2x + 12
2x = 2 - 12 = -10
x = -5

✔ Check: 3 + 2(-5) + 9 = 3 -10 +9 = 2 ✔️

Perfect.

So I’ll assume that "2l" is a typo and should be "2" (the number two).

Thus, x = -5

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Now, compiling all answers:

1. m = 5
2. a = 175/11 or 15.909... but let's write as fraction or decimal? Since others are integers, but this one isn't, we'll keep exact.

But 175/11 is fine, or mixed number 15 10/11.

But perhaps decimal is acceptable. However, for accuracy, fraction is better.

Since the problem has decimals, decimal answer is ok.

a = 17.5 / 1.1 = 15.909090... but we can write as 15.91 if rounded, but better exact.

Note that 175/11 = 15.909090... with 90 repeating.

But for school level, perhaps they expect decimal approximation, but I think fraction is precise.

However, looking at problem 5 and 7, they have decimals but answers are integers.

For consistency, let's present exact values.

But for problem 2, a = 175/11

We can leave it as improper fraction.

Now, final answers:

1. m = 5
2. a = 175/11
3. a = -1
4. y = 11
5. d = 0
6. c = 6
7. b = 4
8. x = -5 (assuming "2l" was meant to be "2")

For problem 8, if we take it literally, it's unsolvable, but with correction, x=-5.

Given that, I'll go with that.

So Final Answer:

1. m = 5
2. a = \frac{175}{11}
3. a = -1
4. y = 11
5. d = 0
6. c = 6
7. b = 4
8. x = -5

But to box them properly.

Since the instruction is to provide final answer after reasoning, and list all.

Perhaps write as:

Final Answer:
1. \( m = 5 \)
2. \( a = \frac{175}{11} \)
3. \( a = -1 \)
4. \( y = 11 \)
5. \( d = 0 \)
6. \( c = 6 \)
7. \( b = 4 \)
8. \( x = -5 \)

For problem 2, if they want decimal, but I think fraction is fine.

To match the format, perhaps write all clearly.

Another thought: in problem 2, when I multiplied by 10, got 11a = 175, so a=175/11, which is correct.

So I'll keep it.

Now for the response.