Grade 10 Algebra worksheet focusing on solving simultaneous equations by elimination and substitution.

Let’s solve each pair of simultaneous equations step by step. We’ll use either elimination or substitution, whichever is easier for each problem.

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1.

3x + 3y = 21
2x - 2y = 6

We can simplify both equations first:

Divide first equation by 3:
→ x + y = 7 → (Equation A)

Divide second equation by 2:
→ x - y = 3 → (Equation B)

Now add Equation A and B:
(x + y) + (x - y) = 7 + 3
→ 2x = 10 → x = 5

Plug x = 5 into Equation A:
5 + y = 7 → y = 2

✔ Check:
3(5)+3(2)=15+6=21 ✔️
2(5)-2(2)=10-4=6 ✔️

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2.

5x + y = 8
2x + 3y = 13

Use substitution. From first equation:
y = 8 - 5x

Plug into second equation:
2x + 3(8 - 5x) = 13
2x + 24 - 15x = 13
-13x + 24 = 13
-13x = -11 → x = 11/13? Wait — that seems messy. Let me check again.

Wait — maybe I made a mistake. Let’s try elimination instead.

Multiply first equation by 3:
15x + 3y = 24
Second equation: 2x + 3y = 13

Subtract second from first:
(15x + 3y) - (2x + 3y) = 24 - 13
13x = 11 → x = 11/13? Still fraction. But let’s keep going.

Then y = 8 - 5*(11/13) = 8 - 55/13 = (104 - 55)/13 = 49/13

But this feels off — maybe the numbers are meant to be integers? Let me double-check original problem.

Original:
5x + y = 8
2x + 3y = 13

Try plugging in small integer values.

Try x=1: then y=3 → 2(1)+3(3)=2+9=11 ≠13
x=2: y=8-10=-2 → 2(2)+3(-2)=4-6=-2≠13
x=0: y=8 → 0+24=24≠13
x= -1: y=13 → 2(-1)+3(13)= -2+39=37≠13

Hmm. Maybe it’s correct as fractions? But let’s re-solve carefully.

From 5x + y = 8 → y = 8 - 5x

Plug into 2x + 3y = 13:

2x + 3(8 - 5x) = 13
2x + 24 - 15x = 13
-13x = 13 - 24 = -11
x = 11/13

y = 8 - 5*(11/13) = 104/13 - 55/13 = 49/13

So answer is x=11/13, y=49/13 — but that’s unusual for Grade 10. Maybe typo? Or perhaps we proceed.

Wait — let me check if I copied right. Original says:

“2. 5x + y = 8
2x + 3y = 13”

Yes. So unless there’s a typo, we go with fractions.

But let’s hold on — maybe I should do all problems and see pattern. Perhaps some have nice answers.

Actually, let’s assume it’s correct and move on. We’ll write fractional answers where needed.

But wait — let me try solving #2 again with elimination properly.

Eq1: 5x + y = 8
Eq2: 2x + 3y = 13

Multiply Eq1 by 3: 15x + 3y = 24
Keep Eq2: 2x + 3y = 13

Subtract Eq2 from modified Eq1:
(15x + 3y) - (2x + 3y) = 24 - 13
13x = 11 → x = 11/13

Same result. So yes, it’s correct.

But since this is Grade 10, maybe they expect decimal? Or perhaps I misread.

Wait — looking back at image description — user said “image contains task” — but I don’t have image, only text provided. The text says:

“2. 5x + y = 8
2x + 3y = 13”

So we must solve as given.

Proceeding.

x = 11/13, y = 49/13

But let’s write as mixed numbers or decimals? Better as fractions.

Alternatively, maybe the problem was meant to be 5x + y = 9? Then x=1, y=4 → 2+12=14 no. Or 5x+y=7? x=1,y=2 → 2+6=8 no.

I think we have to accept fractions here.

But let’s continue — maybe other problems are nicer.

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3.

4x - y = 17
2x - 2y = 10

Simplify second equation: divide by 2 → x - y = 5 → y = x - 5

Plug into first:
4x - (x - 5) = 17
4x - x + 5 = 17
3x = 12 → x = 4

Then y = 4 - 5 = -1

Check:
4(4) - (-1) = 16 +1 =17 ✔️
2(4) -2(-1)=8+2=10 ✔️

Good.

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4.

4x + 3y = 42
3x - y = 12

From second equation: y = 3x - 12

Plug into first:
4x + 3(3x - 12) = 42
4x + 9x - 36 = 42
13x = 78 → x = 6

Then y = 3(6) - 12 = 18 - 12 = 6

Check:
4(6)+3(6)=24+18=42 ✔️
3(6)-6=18-6=12 ✔️

Perfect.

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5.

3x - 2y = 5
4x + 5y = 45

Use elimination. Make coefficients of y same.

Multiply first by 5: 15x - 10y = 25
Multiply second by 2: 8x + 10y = 90

Add them:
15x -10y +8x +10y = 25+90
23x = 115 → x = 5

Plug into first: 3(5) -2y =5 → 15 -2y=5 → -2y= -10 → y=5

Check:
3(5)-2(5)=15-10=5 ✔️
4(5)+5(5)=20+25=45 ✔️

Great.

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6.

2x - 3y = 10
4x + 5y = 38

Eliminate x. Multiply first by 2: 4x - 6y = 20
Second: 4x + 5y = 38

Subtract first from second:
(4x +5y) - (4x -6y) = 38 - 20
4x +5y -4x +6y = 18
11y = 18 → y = 18/11

Then from first: 2x -3*(18/11) =10
2x - 54/11 = 10
2x = 10 + 54/11 = 110/11 + 54/11 = 164/11
x = 82/11

Check:
2*(82/11) -3*(18/11) = 164/11 - 54/11 = 110/11 =10 ✔️
4*(82/11) +5*(18/11) = 328/11 + 90/11 = 418/11 = 38 ✔️

So x=82/11, y=18/11

Again fractions — but mathematically correct.

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7.

x + 5y = 27
3x + 5y = 21

Subtract first from second:
(3x +5y) - (x +5y) = 21 - 27
2x = -6 → x = -3

Plug into first: -3 +5y=27 → 5y=30 → y=6

Check:
-3 +5*6= -3+30=27 ✔️
3*(-3)+5*6= -9+30=21 ✔️

Good.

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8.

2x - 2y = 8
5x - 3y = 22

Simplify first: divide by 2 → x - y = 4 → x = y + 4

Plug into second:
5(y+4) -3y =22
5y +20 -3y =22
2y =2 → y=1

Then x=1+4=5

Check:
2(5)-2(1)=10-2=8 ✔️
5(5)-3(1)=25-3=22 ✔️

Perfect.

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9.

2x + 5y = 26
x + 4y = 22

From second: x = 22 - 4y

Plug into first:
2(22 - 4y) +5y =26
44 -8y +5y =26
44 -3y =26
-3y = -18 → y=6

Then x=22 -4*6=22-24= -2

Check:
2(-2)+5(6)= -4+30=26 ✔️
-2 +4*6= -2+24=22 ✔️

Good.

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Now, let’s list all answers clearly.

Note: Problems 2 and 6 gave fractional answers. Since the worksheet didn’t specify integer solutions, we present exact fractions.

Final Answers:

1. x = 5, y = 2
2. x = 11/13, y = 49/13
3. x = 4, y = -1
4. x = 6, y = 6
5. x = 5, y = 5
6. x = 82/11, y = 18/11
7. x = -3, y = 6
8. x = 5, y = 1
9. x = -2, y = 6

For problem 2 and 6, if decimals are preferred:

2. x ≈ 0.846, y ≈ 3.769
6. x ≈ 7.455, y ≈ 1.636

But since it's math, fractions are better.

However, let me double-check problem 2 — maybe I misread the original.

User wrote:
"2. 5x + y = 8
2x + 3y = 13"

What if it was 5x + y = 9? Then x=1, y=4 → 2+12=14 no.
Or 5x + y = 7? x=1,y=2 → 2+6=8 no.
Or 2x + 3y = 14? Then with x=1,y=3 → 2+9=11 no.

Perhaps it's correct as is.

Similarly for 6: 2x-3y=10, 4x+5y=38 — solution is indeed fractional.

So we'll go with fractions.

Final Answer:

1. x = 5, y = 2
2. x = \frac{11}{13}, y = \frac{49}{13}
3. x = 4, y = -1
4. x = 6, y = 6
5. x = 5, y = 5
6. x = \frac{82}{11}, y = \frac{18}{11}
7. x = -3, y = 6
8. x = 5, y = 1
9. x = -2, y = 6