Solutions to Linear Equations | Worksheet - Free Printable

It looks like you've uploaded a worksheet titled "Solutions to Linear Equations" that asks students to determine whether each given equation has one solution, no solution, or infinitely many solutions. Then, there are challenge problems where students must create equations with specific types of solutions.

Let’s go through the problems step by step and solve them.

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Part 1: Determine the number of solutions

We analyze each equation by simplifying both sides and comparing:

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#### 1. $-4g + 8 = 20$

Solve:
$$
-4g = 20 - 8 = 12 \Rightarrow g = -3
$$
✔ One solution

---

#### 2. $-3h + 20 = 5h$

Bring variables to one side:
$$
20 = 5h + 3h = 8h \Rightarrow h = \frac{20}{8} = 2.5
$$
✔ One solution

---

#### 3. $-3f + 2 = 3f$

Bring variables together:
$$
2 = 3f + 3f = 6f \Rightarrow f = \frac{2}{6} = \frac{1}{3}
$$
✔ One solution

---

#### 4. $2n + 12 = -12 + 2n$

Subtract $2n$ from both sides:
$$
12 = -12
$$
✘ False — contradiction
➡️ No solution

---

#### 5. $3y - 4 = 3 + 7y$

Move variables to one side:
$$
-4 - 3 = 7y - 3y \Rightarrow -7 = 4y \Rightarrow y = -\frac{7}{4}
$$
✔ One solution

---

#### 6. $-5y - 2y = 24$

Combine like terms:
$$
-7y = 24 \Rightarrow y = -\frac{24}{7}
$$
✔ One solution

---

#### 7. $22 + n = 2(4n - 3)$

Distribute:
$$
22 + n = 8n - 6
$$
Bring all terms to one side:
$$
22 + 6 = 8n - n \Rightarrow 28 = 7n \Rightarrow n = 4
$$
✔ One solution

---

#### 8. $6h + 8 = 4h + 12 + 10h$

Simplify right side:
$$
6h + 8 = 14h + 12
$$
Bring variables to one side:
$$
8 - 12 = 14h - 6h \Rightarrow -4 = 8h \Rightarrow h = -\frac{1}{2}
$$
✔ One solution

---

#### 9. $12j = 4(3j - 8)$

Distribute:
$$
12j = 12j - 32
$$
Subtract $12j$ from both sides:
$$
0 = -32
$$
✘ False — contradiction
➡️ No solution

---

#### 10. $5(x - 9) = -2x - 45 + 7x$

Left: $5x - 45$

Right: $(-2x + 7x) - 45 = 5x - 45$

So:
$$
5x - 45 = 5x - 45
$$
Subtract $5x - 45$ from both sides:
$$
0 = 0
$$
✔ True for all $x$ → Infinitely many solutions

---

#### 11. $2(4m - 7) = 16 - 8m - 64$

Left: $8m - 14$

Right: $-8m - 48$

So:
$$
8m - 14 = -8m - 48
$$
Add $8m$ to both sides:
$$
16m - 14 = -48
\Rightarrow 16m = -34 \Rightarrow m = -\frac{17}{8}
$$
✔ One solution

---

#### 12. $3 - 7(d - 4d) = 3 - 5d$

Simplify inside parentheses:
$$
3 - 7(-3d) = 3 - 5d \Rightarrow 3 + 21d = 3 - 5d
$$
Subtract 3:
$$
21d = -5d \Rightarrow 26d = 0 \Rightarrow d = 0
$$
✔ One solution

---

Challenge Problems: Create equations

Now we need to write a number in each blank to make the equation have the specified type of solution.

---

#### 13. No solution:
$$
-4x + 12 = 2x + \boxed{?} = 9
$$

Wait — this seems poorly formatted. Let's interpret it carefully.

It says:
$$
-4x + 12 = 2x + \boxed{\quad} = 9
$$

This appears to be a typo or formatting issue. Likely intended as:
$$
-4x + 12 = 2x + \boxed{?}
$$
and then "equals 9" is separate? Or perhaps it's meant to be:
$$
-4x + 12 = 2x + \boxed{?} \quad \text{with no solution}
$$

But the instruction says: “Write a number in each blank” — so probably two blanks?

Looking at your image description, maybe it's:

> 13. No solution: $-4x + 12 = 2x + \boxed{?} = 9$

That doesn't make sense. Perhaps it's:
$$
-4x + 12 = 2x + \boxed{?}
$$
And we want no solution.

To have no solution, the equation should simplify to a contradiction (e.g., $a = b$, $a \ne b$).

Let’s suppose:
$$
-4x + 12 = 2x + c
$$
Solve:
$$
12 - c = 6x \Rightarrow x = \frac{12 - c}{6}
$$
This will always have one solution unless we manipulate it differently.

But if we want no solution, we need something like:
$$
-4x + 12 = 2x + 12 + 6x \quad \text{(but that would be } -4x + 12 = 8x + 12)
$$

Wait — let's reframe.

Suppose the equation is:
$$
-4x + 12 = 2x + \boxed{c}
$$
We want no solution.

Then we need the simplified form to be false.

Try:
$$
-4x + 12 = 2x + 12
\Rightarrow -6x = 0 \Rightarrow x = 0 \quad \text{(one solution)}
$$

Not good.

Try:
$$
-4x + 12 = 2x + 18
\Rightarrow 12 - 18 = 6x \Rightarrow -6 = 6x \Rightarrow x = -1 \quad \text{still one solution}
$$

To get no solution, we need the coefficients of $x$ to be equal but constants different.

So set:
$$
-4x + 12 = 2x + k
\Rightarrow -4x - 2x = k - 12 \Rightarrow -6x = k - 12
$$

For no solution, we need inconsistent: same variable coefficient, different constant.

But here, left side has $-4x$, right has $2x$. So unless we change the equation...

Ah! Maybe the problem is:

> $-4x + 12 = 2x + \boxed{?}$, and we want no solution

But since the $x$-coefficients are different (-4 vs 2), it will always have one solution unless we adjust.

Wait — perhaps the intended structure is:

$$
-4x + 12 = 2x + \boxed{?} \quad \text{and also equals 9?}
$$

Maybe it's a typo and meant to be:

> $-4x + 12 = 2x + \boxed{?}$, such that it has no solution

But again, unless we make the $x$-terms match and constants not, it can’t happen.

Alternatively, maybe the blank is part of the expression on the right, and we want to fill in a number so that the equation has no solution.

Let’s assume:
$$
-4x + 12 = 2x + c
$$
We want no solution → impossible unless we change the structure.

Wait — perhaps the equation is:
$$
-4x + 12 = 2x + \boxed{?} \quad \text{and this is supposed to be equivalent to } 9?
$$

Alternatively, maybe the full equation is:
$$
-4x + 12 = 2x + \boxed{?} \quad \text{and we want it to have no solution}
$$

But again, since $-4x$ and $2x$ are different, it will always have one solution.

So to get no solution, we need:
$$
\text{Same } x\text{-coefficient, different constant}
$$

Example:
$$
-4x + 12 = -4x + 15
\Rightarrow 12 = 15 \quad \text{→ no solution}
$$

So if the equation were:
$$
-4x + 12 = -4x + \boxed{15}
$$
then no solution.

But the problem says: $-4x + 12 = 2x + \boxed{?}$

So unless the blank is meant to be a number that makes the $x$-coefficient match, it won't work.

Wait — maybe the equation is:
$$
-4x + 12 = 2x + \boxed{?} \quad \text{and we want no solution}
$$

But that’s impossible because the $x$-coefficients differ.

Unless the blank is not just a number, but the whole right-hand side?

Alternatively, perhaps the format is:
$$
-4x + 12 = 2x + \boxed{?} \quad \text{and we want it to have no solution}
$$

But as written, it will always have one solution.

Wait — maybe it's a typo, and it's supposed to be:
$$
-4x + 12 = 2x + \boxed{?} \quad \text{and the equation is } -4x + 12 = 2x + \boxed{?} \text{ with no solution}
$$

But again, only possible if the $x$-coefficients are equal.

So perhaps the equation is:
$$
-4x + 12 = -4x + \boxed{?}
$$

Then to have no solution, set $ ? \ne 12 $

So $\boxed{15}$ → $-4x + 12 = -4x + 15$ → $12 = 15$ → no solution.

But the problem says $2x$, not $-4x$.

So unless the problem has a typo, we might need to reinterpret.

Wait — look at #14:

> 14. Infinitely many solutions: $-3(2 - 4p) = -6 + \boxed{?}$

That one makes sense.

Let’s do #14 first.

---

#### 14. Infinitely many solutions:
$$
-3(2 - 4p) = -6 + \boxed{?}
$$

Left side:
$$
-6 + 12p
$$

So:
$$
-6 + 12p = -6 + \boxed{?}
$$

To have infinitely many solutions, both sides must be identical.

So:
$$
-6 + 12p = -6 + 12p
\Rightarrow \boxed{?} = 12p
$$

But we are to write a number in the blank.

Wait — the blank is likely meant to be a term involving $p$, but the instruction says “write a number”.

Hmm.

Wait — maybe the blank is meant to be a number, and we’re supposed to write $12p$? But it says “number”.

Alternatively, perhaps the equation is:
$$
-3(2 - 4p) = -6 + \boxed{?}p
$$

But as written: $-6 + \boxed{?}$ — just a constant.

So:
$$
-6 + 12p = -6 + c
\Rightarrow 12p = c
$$

Only true for all $p$ if $c = 12p$, which is not a number.

So unless the blank is $12p$, but it asks for a number, this is problematic.

Wait — maybe the equation is:
$$
-3(2 - 4p) = -6 + \boxed{?} \cdot p
$$

But it's written as $-6 + \boxed{?}$ — so likely a constant.

So:
$$
-6 + 12p = -6 + c
\Rightarrow 12p = c
$$

This can only be true for all $p$ if $c = 12p$, which varies.

So impossible to have infinitely many solutions if the right side is a constant.

Therefore, likely the blank is meant to be a term, e.g., $12p$, but the instruction says “number”.

Alternatively, perhaps the equation is:
$$
-3(2 - 4p) = -6 + \boxed{?} \quad \text{and we want infinitely many solutions}
$$

But again, LHS has $p$, RHS is constant → only possible if $p=0$, not all $p$.

So unless the blank includes $p$, it can’t work.

Perhaps it's:
$$
-3(2 - 4p) = -6 + \boxed{?}p
$$

Then:
$$
-6 + 12p = -6 + cp \Rightarrow 12p = cp \Rightarrow c = 12
$$

So $\boxed{12}$

Then:
$$
-3(2 - 4p) = -6 + 12p
\Rightarrow -6 + 12p = -6 + 12p \quad \text{→ always true}
$$

✔ Infinitely many solutions

So likely, the blank is meant to be a coefficient of $p$, and we write $12$

So answer: $\boxed{12}$

---

#### 15. Infinitely many solutions:
$$
6(2z + 3) = \boxed{?} + 4z + 10
$$

Left:
$$
12z + 18
$$

Right:
$$
? + 4z + 10 = 4z + (? + 10)
$$

Set equal:
$$
12z + 18 = 4z + c + 10
\Rightarrow 12z + 18 = 4z + (c + 10)
\Rightarrow 8z + 8 = c
$$

Wait — to have infinitely many solutions, both sides must be identical.

So:
$$
12z + 18 = 4z + c + 10
\Rightarrow 12z + 18 = 4z + (c + 10)
$$

Match coefficients:

- $z$: $12 = 4$ → not possible!

So unless the blank is not just a number, but includes $z$, we can't match.

Wait — the blank is likely a number, but maybe it's meant to be a term.

But the equation is:
$$
6(2z + 3) = \boxed{?} + 4z + 10
$$

So:
$$
12z + 18 = \boxed{?} + 4z + 10
\Rightarrow 12z + 18 = 4z + (\boxed{?} + 10)
\Rightarrow 8z + 8 = \boxed{?}
$$

So $\boxed{?} = 8z + 8$, which is not a number.

So unless the blank is meant to be an expression, we can't do it.

But the instruction says “write a number”.

So perhaps the equation is:
$$
6(2z + 3) = \boxed{?}z + 4z + 10
$$

Then:
$$
12z + 18 = (\boxed{?} + 4)z + 10
$$

Set:
- Coefficients: $12 = ? + 4 \Rightarrow ? = 8$
- Constants: $18 = 10$ → no!

Still not matching.

Wait — what if the blank is meant to be a number, and we want:
$$
12z + 18 = \boxed{?} + 4z + 10
\Rightarrow 12z + 18 = 4z + (\boxed{?} + 10)
\Rightarrow 8z + 8 = \boxed{?}
$$

So $\boxed{?} = 8z + 8$, not a number.

So impossible unless we allow expressions.

But perhaps the problem is:
$$
6(2z + 3) = 8z + 4z + 10 \quad \text{but } 8z + 4z = 12z, and 10 ≠ 18
$$

Wait — try:
$$
6(2z + 3) = 12z + 18
$$

We want:
$$
\boxed{?} + 4z + 10 = 12z + 18
\Rightarrow \boxed{?} = 12z + 18 - 4z - 10 = 8z + 8
$$

Again, not a number.

So unless the blank is $8z + 8$, but it asks for a number, it’s impossible.

Wait — maybe the equation is:
$$
6(2z + 3) = \boxed{?} + 4z + 10
$$
and we want infinitely many solutions, so both sides must be identical.

So:
$$
12z + 18 = \boxed{?} + 4z + 10
\Rightarrow 12z + 18 = 4z + (\boxed{?} + 10)
\Rightarrow 8z + 8 = \boxed{?}
$$

So $\boxed{?} = 8z + 8$ — not a number.

So unless the problem allows expressions, it's flawed.

But perhaps the blank is meant to be $8z$, and then $+10$, so total $8z + 10$, but we need $8z + 8$.

No.

Alternatively, maybe the equation is:
$$
6(2z + 3) = 8z + 4z + 10 \quad \text{but } 8z + 4z = 12z, 10 ≠ 18
$$

Wait — what if the blank is $8z$, and the equation becomes:
$$
6(2z + 3) = 8z + 4z + 10 = 12z + 10
$$

But $12z + 18 = 12z + 10$ → $18 = 10$ → no solution.

Not good.

What if the blank is $12z$, then:
$$
12z + 18 = 12z + 4z + 10 = 16z + 10
\Rightarrow 12z + 18 = 16z + 10 \Rightarrow -4z = -8 \Rightarrow z = 2
$$

One solution.

Not infinite.

So to have infinitely many solutions, the two sides must be identical.

So:
$$
12z + 18 = A + 4z + 10
\Rightarrow 12z + 18 = 4z + (A + 10)
\Rightarrow 8z + 8 = A
$$

So $A = 8z + 8$

But $A$ is a blank — likely meant to be a number, so this is impossible.

Unless the blank is $8z + 8$, but it says “number”.

So perhaps the problem is misprinted.

Alternatively, maybe the equation is:
$$
6(2z + 3) = \boxed{?}z + 4z + 10
$$

Then:
$$
12z + 18 = (\boxed{?} + 4)z + 10
$$

Set:
- $12 = ? + 4 \Rightarrow ? = 8$
- $18 = 10$ → false

So still not.

Unless the constant is also adjusted.

But the blank is only one.

So perhaps the equation is:
$$
6(2z + 3) = 8z + 4z + 10 \quad \text{but } 8z + 4z = 12z, 10 ≠ 18
$$

Wait — what if the blank is $8z$, and the equation is:
$$
6(2z + 3) = 8z + 4z + 10 = 12z + 10
$$

Then $12z + 18 = 12z + 10$ → $18 = 10$ → no solution.

Not good.

So to have infinitely many solutions, we need:
$$
12z + 18 = 12z + 18
$$

So the right side must be $12z + 18$

But it's $\boxed{?} + 4z + 10$

So:
$$
\boxed{?} + 4z + 10 = 12z + 18 \Rightarrow \boxed{?} = 8z + 8
$$

So unless the blank is $8z + 8$, it won't work.

But the instruction says “write a number”, so likely it's a typo.

Perhaps the equation is:
$$
6(2z + 3) = 8z + 4z + 18
$$

Then $12z + 18 = 12z + 18$ → yes!

So $\boxed{?} = 8z$, but it's not a number.

Alternatively, maybe the blank is meant to be $8z$, and the equation is:
$$
6(2z + 3) = \boxed{8z} + 4z + 18
$$

Then $12z + 18 = 12z + 18$ → yes.

But the problem says “write a number”, so maybe it's $8$, and the blank is for the coefficient.

But it's written as $\boxed{?}$, so likely a number.

So perhaps the equation is:
$$
6(2z + 3) = \boxed{?}z + 4z + 18
$$

Then:
$$
12z + 18 = (? + 4)z + 18
\Rightarrow 12 = ? + 4 \Rightarrow ? = 8
$$

So $\boxed{8}$

Then:
$$
6(2z + 3) = 8z + 4z + 18 = 12z + 18
$$

Yes! ✔

So the blank is $8$, and it's the coefficient of $z$.

But in the original, it's written as $\boxed{?} + 4z + 10$, not $ \boxed{?}z + 4z + 10 $

So unless the $10$ is a typo and should be $18$, it won't work.

Given the ambiguity, I think there may be typos in the challenge problems.

Let’s move to #16.

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#### 16. No solution:
$$
-4 = \boxed{?} + 4(7 - 2) + 3
$$

First, simplify known parts:
$$
4(7 - 2) = 4 \times 5 = 20
$$

So:
$$
-4 = \boxed{?} + 20 + 3 = \boxed{?} + 23
\Rightarrow \boxed{?} = -4 - 23 = -27
$$

But this gives one solution.

We want no solution.

But this is an equation with only numbers — unless the blank is a variable.

But it's $\boxed{?}$, likely a number.

So:
$$
-4 = \boxed{?} + 23 \Rightarrow \boxed{?} = -27
$$

Always has one solution.

So how can it have no solution?

Unless the blank is a variable, like $x$, but the instruction says “write a number”.

So perhaps the equation is:
$$
-4 = \boxed{?}x + 4(7 - 2) + 3
$$

Then:
$$
-4 = \boxed{?}x + 20 + 3 = \boxed{?}x + 23
\Rightarrow \boxed{?}x = -27
$$

To have no solution, we need $\boxed{?} = 0$, then:
$$
0x = -27 \Rightarrow 0 = -27 \quad \text{→ no solution}
$$

So $\boxed{0}$

Then:
$$
-4 = 0 + 20 + 3 = 23 \quad \text{→ false, so no solution}
$$

Yes!

So if the blank is $0$, then:
$$
-4 = 0 + 20 + 3 = 23 \quad \text{→ false}
$$

So no solution

Thus, answer: $\boxed{0}$

But only if the blank is a coefficient of a variable.

If it's just a number, then it will always have one solution.

So likely, the equation is:
$$
-4 = \boxed{?}x + 4(7 - 2) + 3
$$

Then $\boxed{0}$ makes it have no solution.

Otherwise, if it's just a number, it's impossible.

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Final Answers Summary

#### Part 1:
| Problem | Answer |
|--------|--------|
| 1 | One solution |
| 2 | One solution |
| 3 | One solution |
| 4 | No solution |
| 5 | One solution |
| 6 | One solution |
| 7 | One solution |
| 8 | One solution |
| 9 | No solution |
| 10 | Infinitely many solutions |
| 11 | One solution |
| 12 | One solution |

#### Challenge Problems (best interpretation):

13. No solution: $-4x + 12 = 2x + \boxed{?}$ → Impossible with current form. If it were $-4x + 12 = -4x + \boxed{15}$, then no solution. But as is, likely typo.

14. Infinitely many solutions: $-3(2 - 4p) = -6 + \boxed{?}p$ → $\boxed{12}$

15. Infinitely many solutions: $6(2z + 3) = \boxed{?}z + 4z + 18$ → $\boxed{8}$ (if constant is 18)

16. No solution: $-4 = \boxed{?}x + 4(7 - 2) + 3$ → $\boxed{0}$

---

Conclusion

The first 12 problems are straightforward. The challenge problems appear to have formatting issues, but based on common patterns:

- 14: $\boxed{12}$
- 16: $\boxed{0}$
- 15: Likely $\boxed{8}$ if the equation is $6(2z+3) = 8z + 4z + 18$
- 13: Needs clarification — likely typo

Please double-check the challenge problems for correct formatting.