TenTors Math Teacher Resources: Solving Systems of Linear ... - Free Printable

Let’s solve each system of linear equations one by one, then match the solution (x, y) to the letter in the table.

We’ll go step by step for all 12 problems.

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Problem 1:
3x + 2y = 9
x + 2y = 7

Subtract second equation from first:
(3x + 2y) - (x + 2y) = 9 - 7 → 2x = 2 → x = 1
Plug into second equation: 1 + 2y = 7 → 2y = 6 → y = 3
→ Solution: (1, 3) → Look at table: M has x=6,y=3? No. Wait — check again.

Wait! Let me recheck:

Actually, let’s do substitution properly.

From second equation: x = 7 - 2y
Plug into first: 3(7 - 2y) + 2y = 9 → 21 - 6y + 2y = 9 → 21 - 4y = 9 → -4y = -12 → y = 3
Then x = 7 - 2*3 = 1 → (1,3)

Now look at table:

Check letters with x=1, y=3 → none? Wait, maybe I misread.

Look again:

A: (-2,4)
B: (9,1)
C: (2,7)
D: (2,4)
E: (1,9)
F: (-1,1)
G: (2,-7)
H: (1,-9)
I: (3,6)
J: (4,9)
K: (5,1)
L: (1,2) ← close but y=2
M: (6,3) ← x=6,y=3
N: (9,4)
O: (-1,0)
P: (-5,-2)
Q: (-4,9)
R: (-2,5)
S: (1,3) ← YES! S is x=1, y=3

So Problem 1 → S

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Problem 2:
5x + y = 14
3x + y = 10

Subtract second from first: (5x+y)-(3x+y)=14-10 → 2x=4 → x=2
Plug into second: 3*2 + y = 10 → 6 + y = 10 → y=4
→ (2,4) → Table: D has x=2,y=4 → D

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Problem 3:
9x + 5y = 50
12x + 5y = 65

Subtract first from second: (12x+5y)-(9x+5y)=65-50 → 3x=15 → x=5
Plug into first: 9*5 + 5y = 50 → 45 + 5y = 50 → 5y=5 → y=1
→ (5,1) → Table: K has x=5,y=1 → K

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Problem 4:
4x + y = 37
2x - y = 17

Add both equations: (4x+y)+(2x-y)=37+17 → 6x=54 → x=9
Plug into second: 2*9 - y = 17 → 18 - y = 17 → y=1
→ (9,1) → Table: B has x=9,y=1 → B

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Problem 5:
x + 2y = 22
2x + y = 20

Multiply first by 2: 2x + 4y = 44
Subtract second: (2x+4y) - (2x+y) = 44 - 20 → 3y=24 → y=8
Plug into first: x + 2*8 = 22 → x + 16 = 22 → x=6
→ (6,8) → Table: Z has x=6,y=8 → Z

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Problem 6:
x - y = -1
2x + y = 4

Add both: (x - y) + (2x + y) = -1 + 4 → 3x = 3 → x=1
Plug into first: 1 - y = -1 → -y = -2 → y=2
→ (1,2) → Table: L has x=1,y=2 → L

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Problem 7:
2x - y = 14
5x + y = 14

Add both: (2x - y) + (5x + y) = 14 + 14 → 7x = 28 → x=4
Plug into first: 2*4 - y = 14 → 8 - y = 14 → -y = 6 → y=-6
→ (4,-6) → Table: T has x=4,y=-6 → T

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Problem 8:
x + y = -7
x - y = -3

Add both: (x+y)+(x-y) = -7 + (-3) → 2x = -10 → x=-5
Plug into first: -5 + y = -7 → y = -2
→ (-5,-2) → Table: P has x=-5,y=-2 → P

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Problem 9:
2x + 3y = 1
5x + 4y = -1

Use elimination. Multiply first by 4, second by 3:

First *4: 8x + 12y = 4
Second *3: 15x + 12y = -3

Subtract: (15x+12y) - (8x+12y) = -3 - 4 → 7x = -7 → x = -1
Plug into first: 2*(-1) + 3y = 1 → -2 + 3y = 1 → 3y=3 → y=1
→ (-1,1) → Table: F has x=-1,y=1 → F

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Problem 10:
3x - 9 = -4y → rewrite as 3x + 4y = 9
2x - y = -16 → multiply by 4: 8x - 4y = -64

Now add to first equation:

3x + 4y = 9
8x - 4y = -64
Add: 11x = -55 → x = -5
Plug into 2x - y = -16: 2*(-5) - y = -16 → -10 - y = -16 → -y = -6 → y=6
→ (-5,6) → Table: V has x=-5,y=6 → V

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Problem 11:
6x - 5y = -32
5x + 4y = 6

Use elimination. Multiply first by 4, second by 5:

First *4: 24x - 20y = -128
Second *5: 25x + 20y = 30

Add: 49x = -98 → x = -2
Plug into second: 5*(-2) + 4y = 6 → -10 + 4y = 6 → 4y=16 → y=4
→ (-2,4) → Table: A has x=-2,y=4 → A

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Problem 12:
6x - 3y + 27 = 0 → 6x - 3y = -27 → divide by 3: 2x - y = -9
5x + 4y = 10

From first: y = 2x + 9
Plug into second: 5x + 4*(2x + 9) = 10 → 5x + 8x + 36 = 10 → 13x = -26 → x = -2
Then y = 2*(-2) + 9 = -4 + 9 = 5
→ (-2,5) → Table: R has x=-2,y=5 → R

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Now we have:

1 → S
2 → D
3 → K
4 → B
5 → Z
6 → L
7 → T
8 → P
9 → F
10 → V
11 → A
12 → R

The code boxes are labeled:

Box 1: 1 - 2 - 3 → S - D - K
Box 2: 4 - 5 - 6 → B - Z - L
Box 3: 7 - 8 - 9 → T - P - F
Box 4: 10 - 11 - 12 → V - A - R

So the final code is:

SDK BZL TPFA VR

But wait — the boxes are shown as four separate boxes, each with three letters. So we write them as groups:

Final Answer should be the four groups concatenated or separated? The image shows four boxes with labels “1-2-3”, “4-5-6”, etc., so likely we output the letters in order for each box.

So:

Box 1: S D K → SDK
Box 2: B Z L → BZL
Box 3: T P F → TPF
Box 4: V A R → VAR

Thus, the complete code is: SDK BZL TPF VAR

But since it's a "code breaker" and locks, probably just the string without spaces? Or with hyphens? Looking at the bottom, it says “complete the code in the four boxes”, and each box has three positions.

In the image, the boxes are drawn as:

[ _ - _ - _ ] [ _ - _ - _ ] [ _ - _ - _ ] [ _ - _ - _ ]

So likely, we fill each box with the three letters corresponding to the problem numbers.

Therefore, the answer is:

Box 1: S, D, K
Box 2: B, Z, L
Box 3: T, P, F
Box 4: V, A, R

So writing as requested: the code is formed by taking the letters for problems 1-2-3, then 4-5-6, etc.

Final Answer: SDK BZL TPF VAR

But to match format, perhaps without spaces? Or with hyphens? Since the boxes show hyphens between positions, but the letters are what matter.

Looking back at user instruction: “complete the code in the four boxes at the bottom”

And the boxes are labeled “1-2-3”, “4-5-6”, etc., meaning each box contains the letters for those three problems.

So the final code is the sequence of letters for problems 1 through 12, grouped as (1,2,3), (4,5,6), (7,8,9), (10,11,12)

Thus:

Group 1: S, D, K → SDK
Group 2: B, Z, L → BZL
Group 3: T, P, F → TPF
Group 4: V, A, R → VAR

So the full code is: SDK BZL TPF VAR

But since it’s a single code, maybe concatenate: SDKBZLTPFVAR? But that seems unlikely because there are four distinct boxes.

Given the visual layout, I think we present it as four groups.

However, the instruction says: “complete the code in the four boxes”, so likely each box gets three letters.

For the final answer, since it’s a text response, I’ll write the four groups separated by spaces.

But let me double-check all solutions once more for accuracy.

Rechecking Problem 1: got (1,3) → S ✔️
Problem 2: (2,4) → D ✔️
Problem 3: (5,1) → K ✔️
Problem 4: (9,1) → B ✔️
Problem 5: (6,8) → Z ✔️
Problem 6: (1,2) → L ✔️
Problem 7: (4,-6) → T ✔️
Problem 8: (-5,-2) → P ✔️
Problem 9: (-1,1) → F ✔️
Problem 10: (-5,6) → V ✔️
Problem 11: (-2,4) → A ✔️
Problem 12: (-2,5) → R ✔️

All correct.

So the code is:

Box 1 (problems 1,2,3): S, D, K → SDK
Box 2 (problems 4,5,6): B, Z, L → BZL
Box 3 (problems 7,8,9): T, P, F → TPF
Box 4 (problems 10,11,12): V, A, R → VAR

Final Answer: SDK BZL TPF VAR