Maths Year 6 algebra, KS3 or KS4 revision. Four games to try ... - Free Printable

Let's solve each equation in the hexagonal grid step by step. Each hexagon contains a linear equation with one variable. We'll solve for each variable individually.

---

Top Row

1. $ 2a + 2 = 14 $
Subtract 2:
$ 2a = 12 $
Divide by 2:
$ a = 6 $

2. $ 4b + 1 = 17 $
Subtract 1:
$ 4b = 16 $
Divide by 4:
$ b = 4 $

3. $ 2c - 3 = 7 $
Add 3:
$ 2c = 10 $
Divide by 2:
$ c = 5 $

---

Second Row

4. $ 2d = 8 $
Divide by 2:
$ d = 4 $

5. $ 7 - e = 4 $
Subtract 7:
$ -e = -3 $
Multiply by -1:
$ e = 3 $

6. $ 3f + 3 = 9 $
Subtract 3:
$ 3f = 6 $
Divide by 3:
$ f = 2 $

7. $ 10g + 1 = 11 $
Subtract 1:
$ 10g = 10 $
Divide by 10:
$ g = 1 $

---

Third Row

8. $ 2h + 6 = 10 $
Subtract 6:
$ 2h = 4 $
Divide by 2:
$ h = 2 $

9. $ 2j - 1 = 5 $
Add 1:
$ 2j = 6 $
Divide by 2:
$ j = 3 $

10. $ 5m - 7 = 3 $
Add 7:
$ 5m = 10 $
Divide by 5:
$ m = 2 $

11. $ 6n + 1 = 13 $
Subtract 1:
$ 6n = 12 $
Divide by 6:
$ n = 2 $

12. $ 4p + 3 = 19 $
Subtract 3:
$ 4p = 16 $
Divide by 4:
$ p = 4 $

---

Fourth Row

13. $ 3q = 15 $
Divide by 3:
$ q = 5 $

14. $ 4r - 9 = 3 $
Add 9:
$ 4r = 12 $
Divide by 4:
$ r = 3 $

15. $ 3s - 3 = 12 $
Add 3:
$ 3s = 15 $
Divide by 3:
$ s = 5 $

16. $ 6t = 18 $
Divide by 6:
$ t = 3 $

---

Bottom Row

17. $ 5x - 9 = 1 $
Add 9:
$ 5x = 10 $
Divide by 5:
$ x = 2 $

18. $ 3x + 1 = 16 $
Subtract 1:
$ 3x = 15 $
Divide by 3:
$ x = 5 $

Wait! Here we have a problem:
We just found $ x = 2 $ from the previous equation, but now $ x = 5 $? That’s a contradiction.

Let’s check both equations:

- $ 5x - 9 = 1 $ → $ 5x = 10 $ → $ x = 2 $
- $ 3x + 1 = 16 $ → $ 3x = 15 $ → $ x = 5 $

But both can't be true at the same time. So either:
- There's a typo in the puzzle,
- Or the variables are not meant to be consistent across the whole grid (but that would be unusual),
- Or perhaps there's a mislabeling.

But looking at the layout, these two equations are adjacent and both use $ x $. So unless it's a different variable, this is inconsistent.

Wait — maybe it's not the same variable? But they're both labeled as $ x $. Let's double-check.

Actually, let's look closely:

- The first one is: $ 5x - 9 = 1 $
- The second one is: $ 3x + 1 = 16 $

They both use $ x $, so unless it's a mistake, they must be consistent.

But:
- From $ 5x - 9 = 1 $: $ x = 2 $
- From $ 3x + 1 = 16 $: $ x = 5 $

Contradiction.

Now check the third one in the bottom row:

19. $ 2(z - 3) = 6 $
Divide both sides by 2:
$ z - 3 = 3 $
Add 3:
$ z = 6 $

So that one is fine.

But the two $ x $ equations conflict.

Let’s re-express:

- $ 5x - 9 = 1 $ → $ x = 2 $
- $ 3x + 1 = 16 $ → $ x = 5 $

No solution satisfies both.

Possibility: Maybe the middle one is not $ x $? Wait — no, it says $ 3x + 1 = 16 $

Alternatively, perhaps it's a typo, or maybe the puzzle has a trick.

Wait — could it be that the same letter appears in multiple equations, and we’re supposed to find values such that all are satisfied simultaneously?

But in that case, if $ x $ appears in two equations with conflicting values, it’s impossible.

Unless... wait — maybe I misread the variables.

Let me carefully recheck the image description.

From your image:

Bottom row (left to right):

- $ 5x - 9 = 1 $
- $ 3x + 1 = 16 $
- $ 2(z - 3) = 6 $

So yes — both equations involve $ x $, and give different values.

This suggests either:
- It's a mistake in the puzzle,
- Or perhaps the variables are not meant to be shared — but that doesn’t make sense because they’re labeled the same.

Alternatively, could it be that the same variable is used in multiple equations, so they must be consistent?

Then this system is inconsistent.

But let's see if any other variable repeats.

Let’s list all variables:

- a, b, c, d, e, f, g, h, j, m, n, p, q, r, s, t, x, z

Only x appears twice.

All others appear once.

So only $ x $ is repeated.

Therefore, unless it's a typo, this system has no solution due to inconsistency.

But likely, it's a typo.

Let’s test which equation might be wrong.

Suppose $ 5x - 9 = 1 $ → $ x = 2 $

Then plug into second: $ 3(2) + 1 = 6 + 1 = 7 \neq 16 $

Too small.

If $ 3x + 1 = 16 $ → $ x = 5 $

Then $ 5(5) - 9 = 25 - 9 = 16 \neq 1 $

So neither works.

Wait — what if the second equation was meant to be $ 3x + 1 = 7 $? Then $ x = 2 $, and it would work.

Or if the first was $ 5x - 9 = 16 $? Then $ 5x = 25 $, $ x = 5 $, and then $ 3(5)+1=16 $, which matches.

But as written, both are inconsistent.

Alternatively, maybe it's not $ x $ in both? But the labels are the same.

Another idea: Perhaps the middle one is not $ x $, but another variable?

Wait — looking at the image again:

The bottom row has:

- Left: $ 5x - 9 = 1 $
- Middle: $ 3x + 1 = 16 $
- Right: $ 2(z - 3) = 6 $

Yes — both use $ x $.

So unless it's a typo, the puzzle is flawed.

But let’s suppose the middle one is actually $ 3y + 1 = 16 $? But it says $ x $.

Alternatively, maybe it's a trick where you solve each independently, ignoring consistency?

But that would be strange.

Perhaps the goal is just to solve each equation separately, even if variables repeat.

In that case, we can still report the solutions, noting the inconsistency.

But since the task is likely to find values for all variables, and $ x $ cannot satisfy both, we must conclude:

> There is an inconsistency in the equations involving $ x $.

But let's assume it's a typo and try to fix it.

Suppose the left equation is correct: $ 5x - 9 = 1 $ → $ x = 2 $

Then the middle should be $ 3x + 1 = 3(2) + 1 = 7 $, so it should be $ 3x + 1 = 7 $, not 16.

Alternatively, if the middle is correct: $ 3x + 1 = 16 $ → $ x = 5 $, then $ 5x - 9 = 25 - 9 = 16 $, so the left should be $ 5x - 9 = 16 $

So likely, one of them is miswritten.

But based on what’s given, we can’t resolve both.

So perhaps the intended answer is to solve each equation as if independent, and just report values.

In that case, we’ll proceed with solving each, and note the issue.

But since the problem likely expects consistent values, and only $ x $ conflicts, let’s see if we can find a way.

Wait — maybe the middle one is not $ x $? Let's look at the position.

In the bottom row:

- First: $ 5x - 9 = 1 $
- Second: $ 3x + 1 = 16 $
- Third: $ 2(z - 3) = 6 $

It's possible that the second one is meant to be $ 3y + 1 = 16 $? But it's labeled $ x $.

Alternatively, maybe the first one is $ 5y - 9 = 1 $? But it says $ x $.

Without more context, we must go by what’s written.

So here’s the best approach:

---

✔ Solve Each Equation Individually

We’ll solve all equations as if they are separate, and list the values.

Even though $ x $ appears twice with conflicting values, we’ll report both.

But that means no single value of $ x $ satisfies both.

So final answer will include:

- $ a = 6 $
- $ b = 4 $
- $ c = 5 $
- $ d = 4 $
- $ e = 3 $
- $ f = 2 $
- $ g = 1 $
- $ h = 2 $
- $ j = 3 $
- $ m = 2 $
- $ n = 2 $
- $ p = 4 $
- $ q = 5 $
- $ r = 3 $
- $ s = 5 $
- $ t = 3 $
- $ x = 2 $ (from $ 5x - 9 = 1 $)
- $ x = 5 $ (from $ 3x + 1 = 16 $) ← Conflict
- $ z = 6 $

So the system is inconsistent due to the two equations involving $ x $.

---

🔍 Likely Resolution

Probably, one of the equations has a typo.

Let’s suppose the second equation in the bottom row is meant to be:

$ 3x + 1 = 7 $ instead of 16.

Then $ x = 2 $, and $ 5x - 9 = 10 - 9 = 1 $ → correct.

Alternatively, if the first equation is $ 5x - 9 = 16 $, then $ x = 5 $, and $ 3x + 1 = 16 $ → $ 15 + 1 = 16 $ → correct.

So likely, the intended equations were:

- $ 5x - 9 = 1 $ → $ x = 2 $
- $ 3x + 1 = 7 $ → $ x = 2 $

or

- $ 5x - 9 = 16 $ → $ x = 5 $
- $ 3x + 1 = 16 $ → $ x = 5 $

But as written, they are inconsistent.

Given that $ 3x + 1 = 16 $ gives $ x = 5 $, and $ 5x - 9 = 1 $ gives $ x = 2 $, and both are present, the puzzle is flawed.

However, if we must provide a solution, we can assume the bottom-middle equation is incorrect, and trust the left one.

Or vice versa.

But without further info, we can only report:

---

✔ Final Answer (with note)

Here are the solutions to each equation:

| Equation | Solution |
|--------|---------|
| $ 2a + 2 = 14 $ | $ a = 6 $ |
| $ 4b + 1 = 17 $ | $ b = 4 $ |
| $ 2c - 3 = 7 $ | $ c = 5 $ |
| $ 2d = 8 $ | $ d = 4 $ |
| $ 7 - e = 4 $ | $ e = 3 $ |
| $ 3f + 3 = 9 $ | $ f = 2 $ |
| $ 10g + 1 = 11 $ | $ g = 1 $ |
| $ 2h + 6 = 10 $ | $ h = 2 $ |
| $ 2j - 1 = 5 $ | $ j = 3 $ |
| $ 5m - 7 = 3 $ | $ m = 2 $ |
| $ 6n + 1 = 13 $ | $ n = 2 $ |
| $ 4p + 3 = 19 $ | $ p = 4 $ |
| $ 3q = 15 $ | $ q = 5 $ |
| $ 4r - 9 = 3 $ | $ r = 3 $ |
| $ 3s - 3 = 12 $ | $ s = 5 $ |
| $ 6t = 18 $ | $ t = 3 $ |
| $ 5x - 9 = 1 $ | $ x = 2 $ |
| $ 3x + 1 = 16 $ | $ x = 5 $ ← Conflict |
| $ 2(z - 3) = 6 $ | $ z = 6 $ |

⚠️ Note: The two equations involving $ x $ are inconsistent. No single value of $ x $ satisfies both. This suggests a typo in the puzzle. If we assume $ 3x + 1 = 7 $, then $ x = 2 $, and both equations are consistent. Alternatively, if $ 5x - 9 = 16 $, then $ x = 5 $, and the second equation holds.

✔ Most likely, the intended value is $ \boxed{x = 2} $, assuming the second equation should be $ 3x + 1 = 7 $, or $ \boxed{x = 5} $ if the first is $ 5x - 9 = 16 $.

But as written, the system has no solution.

---

🟩 Recommended Fix

Change either:
- $ 3x + 1 = 16 $ → $ 3x + 1 = 7 $, or
- $ 5x - 9 = 1 $ → $ 5x - 9 = 16 $

Then all equations can be solved consistently.

With that correction, the complete solution set is:

$$
\boxed{
\begin{aligned}
&a = 6,\ b = 4,\ c = 5,\ d = 4,\ e = 3,\ f = 2,\ g = 1,\\
&h = 2,\ j = 3,\ m = 2,\ n = 2,\ p = 4,\ q = 5,\ r = 3,\\
&s = 5,\ t = 3,\ x = 2\ \text{(or 5, depending on fix)},\ z = 6
\end{aligned}
}
$$

But due to inconsistency, no consistent solution exists unless one equation is corrected.

Let me know if you'd like to assume a specific fix.