Solving Equations with the Variable on Both Sides Worksheet for ... - Free Printable

Let’s solve each problem step by step. I’ll go through them one at a time, carefully checking my work.

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Problem 1: 8x + 10 = -13 + 9x

We want to get all x terms on one side and numbers on the other.

Subtract 8x from both sides:
→ 10 = -13 + x

Add 13 to both sides:
→ 23 = x

✔ Check: Plug x = 23 into original equation.
Left: 8(23) + 10 = 184 + 10 = 194
Right: -13 + 9(23) = -13 + 207 = 194 → ✔️ Correct.

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Problem 2: 4x + 15 = -13 + 6x

Subtract 4x from both sides:
→ 15 = -13 + 2x

Add 13 to both sides:
→ 28 = 2x

Divide by 2:
→ x = 14

✔ Check: Left: 4(14)+15=56+15=71; Right: -13+6(14)= -13+84=71 → ✔️

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Problem 3: (2/3)x = 1/6

Multiply both sides by 3 to eliminate denominator on left:
→ 2x = 3*(1/6) = 1/2

Now divide both sides by 2:
→ x = (1/2)/2 = 1/4

✔ Check: (2/3)(1/4) = 2/12 = 1/6 → ✔️

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Problem 4: (5/8)x = 3/4

Multiply both sides by 8:
→ 5x = 8*(3/4) = 6

Divide by 5:
→ x = 6/5

✔ Check: (5/8)(6/5) = 30/40 = 3/4 → ✔️

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Problem 5: 1.4y + 1.1 = 8.3 - y

Add y to both sides:
→ 2.4y + 1.1 = 8.3

Subtract 1.1:
→ 2.4y = 7.2

Divide by 2.4:
→ y = 7.2 / 2.4 = 3

✔ Check: Left: 1.4*3 + 1.1 = 4.2 + 1.1 = 5.3; Right: 8.3 - 3 = 5.3 → ✔️

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Problem 6: 0.4z + 1.2 = 0.2z + 0.6

Subtract 0.2z from both sides:
→ 0.2z + 1.2 = 0.6

Subtract 1.2:
→ 0.2z = -0.6

Divide by 0.2:
→ z = -3

✔ Check: Left: 0.4*(-3)+1.2 = -1.2+1.2=0; Right: 0.2*(-3)+0.6=-0.6+0.6=0 → ✔️

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Problem 7: (1/2)t - 2 = (1/3)t + 7

Get rid of fractions first. Multiply every term by 6 (LCM of 2 and 3):

6*(1/2 t) - 6*2 = 6*(1/3 t) + 6*7
→ 3t - 12 = 2t + 42

Subtract 2t:
→ t - 12 = 42

Add 12:
→ t = 54

✔ Check: Left: (1/2)*54 - 2 = 27 - 2 = 25; Right: (1/3)*54 + 7 = 18 + 7 = 25 → ✔️

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Problem 8: (1/5)m + 2 = (3/4)m - 10

Multiply every term by 20 (LCM of 5 and 4):

20*(1/5 m) + 20*2 = 20*(3/4 m) - 20*10
→ 4m + 40 = 15m - 200

Subtract 4m:
→ 40 = 11m - 200

Add 200:
→ 240 = 11m

Divide by 11:
→ m = 240/11 ≈ 21.818... but let’s keep as fraction: 240/11

Wait — let me double-check that arithmetic.

Original after multiplying by 20:

Left: 4m + 40
Right: 15m - 200

So: 4m + 40 = 15m - 200
Subtract 4m: 40 = 11m - 200
Add 200: 240 = 11m → m = 240/11 → yes, correct.

But maybe we can leave it as improper fraction or mixed number? The problem doesn’t specify, so 240/11 is fine.

✔ Check: Left: (1/5)(240/11) + 2 = 48/11 + 22/11 = 70/11
Right: (3/4)(240/11) - 10 = (720/44) - 10 = (180/11) - 110/11 = 70/11 → ✔️

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Problem 9: (1/2)n + 5 = (1/4)n - 3/2

Multiply every term by 4 (LCM of 2 and 4):

4*(1/2 n) + 4*5 = 4*(1/4 n) - 4*(3/2)
→ 2n + 20 = n - 6

Subtract n:
→ n + 20 = -6

Subtract 20:
→ n = -26

✔ Check: Left: (1/2)(-26) + 5 = -13 + 5 = -8
Right: (1/4)(-26) - 3/2 = -6.5 - 1.5 = -8 → ✔️

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Problem 10: (1/3)p + 4 = (1/6)p - 1/2

Multiply every term by 6:

6*(1/3 p) + 6*4 = 6*(1/6 p) - 6*(1/2)
→ 2p + 24 = p - 3

Subtract p:
→ p + 24 = -3

Subtract 24:
→ p = -27

✔ Check: Left: (1/3)(-27) + 4 = -9 + 4 = -5
Right: (1/6)(-27) - 0.5 = -4.5 - 0.5 = -5 → ✔️

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Problem 11: -0.8 - 0.1 = -0.1x + 1.1

Simplify left side:
→ -0.9 = -0.1x + 1.1

Subtract 1.1 from both sides:
→ -2.0 = -0.1x

Divide by -0.1:
→ x = (-2.0)/(-0.1) = 20

✔ Check: Right: -0.1*20 + 1.1 = -2 + 1.1 = -0.9 → matches left → ✔️

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Problem 12: -0.3y - 0.8 = 1.1y - 1.0

Bring like terms together. Add 0.3y to both sides:
→ -0.8 = 1.4y - 1.0

Add 1.0 to both sides:
→ 0.2 = 1.4y

Divide by 1.4:
→ y = 0.2 / 1.4 = 2/14 = 1/7

✔ Check: Left: -0.3*(1/7) - 0.8 = -3/70 - 56/70 = -59/70
Right: 1.4*(1/7) - 1.0 = 0.2 - 1.0 = -0.8 = -56/70 → Wait! That’s not matching.

Wait — mistake here.

Let me recalculate:

Left: -0.3y - 0.8 with y = 1/7
= -0.3*(1/7) - 0.8 = -3/10 * 1/7 - 4/5 = -3/70 - 56/70 = -59/70

Right: 1.4y - 1.0 = 14/10 * 1/7 - 1 = (14/70) - 1 = 1/5 - 1 = -4/5 = -56/70

Not equal → error in solving.

Go back:

Equation: -0.3y - 0.8 = 1.1y - 1.0

Add 0.3y to both sides:
→ -0.8 = 1.4y - 1.0

Add 1.0 to both sides:
→ 0.2 = 1.4y → same as before.

But 0.2 / 1.4 = 2/14 = 1/7 → seems right.

Wait — perhaps decimal precision?

Try using decimals:

y = 0.2 ÷ 1.4 = 0.142857...

Left: -0.3*(0.142857) - 0.8 ≈ -0.042857 - 0.8 = -0.842857
Right: 1.1*(0.142857) - 1.0 ≈ 0.157143 - 1.0 = -0.842857 → Oh! It does match.

Earlier I used fractions incorrectly.

Actually: 1.4 = 14/10 = 7/5, 0.2 = 1/5 → y = (1/5)/(7/5) = 1/7 → correct.

And -0.3 = -3/10, so:

Left: (-3/10)(1/7) - 4/5 = -3/70 - 56/70 = -59/70
Right: (11/10)(1/7) - 1 = 11/70 - 70/70 = -59/70 → YES! Matches.

I made a mistake earlier saying 1.4y = 14/10 * 1/7 = 14/70 = 1/5 — that’s wrong.

1.4 = 14/10 = 7/5 → (7/5)*(1/7) = 1/5 → then minus 1 → 1/5 - 1 = -4/5 = -56/70 — wait no:

Wait — 1.1y, not 1.4y on right!

Oh no! Big mistake.

Original equation: -0.3y - 0.8 = 1.1y - 1.0

When I added 0.3y to both sides, I got:

-0.8 = 1.4y - 1.0 → that’s correct because 1.1y + 0.3y = 1.4y.

Then adding 1.0: 0.2 = 1.4y → y = 0.2 / 1.4 = 1/7 → still correct.

Now check right side: 1.1y - 1.0 = 1.1*(1/7) - 1 = 11/10 * 1/7 - 1 = 11/70 - 70/70 = -59/70

Left side: -0.3*(1/7) - 0.8 = -3/10 * 1/7 - 4/5 = -3/70 - 56/70 = -59/70 → matches.

My earlier calculation for right side was wrong — I said 1.4y instead of 1.1y. But in the algebra, I correctly combined to 1.4y when moving terms. So answer is correct.

✔ Final: y = 1/7

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Problem 13: 1.2x + 3.7 = -2.1 - 0.3x

Add 0.3x to both sides:
→ 1.5x + 3.7 = -2.1

Subtract 3.7:
→ 1.5x = -5.8

Divide by 1.5:
→ x = -5.8 / 1.5

Convert to fractions: 5.8 = 58/10 = 29/5, 1.5 = 3/2 → better to use decimals.

-5.8 ÷ 1.5 = -58/15 ≈ -3.866..., but let's compute exactly.

Multiply numerator and denominator by 10: -58 / 15 → simplify? 58 and 15 share no common factors → x = -58/15

Or as decimal: -3.866... but probably keep as fraction.

Check:

Left: 1.2*(-58/15) + 3.7 = (6/5)*(-58/15) + 37/10 = (-348/75) + 37/10

Convert to common denominator 150:

-348/75 = -696/150
37/10 = 555/150
Sum: (-696 + 555)/150 = -141/150 = -47/50

Right: -2.1 - 0.3*(-58/15) = -21/10 + (3/10)*(58/15) = -21/10 + 174/150

Convert: -21/10 = -315/150, 174/150 → sum: (-315 + 174)/150 = -141/150 = -47/50 → matches.

So x = -58/15

But maybe they want decimal? Problem has decimals, so perhaps write as decimal.

-58 ÷ 15 = -3.8666... = -3.8̅6 or approximately -3.87, but exact is -58/15.

Looking at other problems, some have fractions, some decimals. Let’s see if we can write as mixed number: -3 and 13/15.

But I think -58/15 is acceptable.

Alternatively, re-solve with decimals:

1.5x = -5.8 → x = -5.8 / 1.5

Divide: 5.8 ÷ 1.5 = 58 ÷ 15 = 3.8666... so x = -3.8666... = -3.8̅6

But perhaps leave as fraction.

I'll go with -58/15

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Problem 14: 3p - 0.9 = 2p - 1.0

Subtract 2p from both sides:
→ p - 0.9 = -1.0

Add 0.9:
→ p = -0.1

✔ Check: Left: 3*(-0.1) - 0.9 = -0.3 - 0.9 = -1.2
Right: 2*(-0.1) - 1.0 = -0.2 - 1.0 = -1.2 → ✔️

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Problem 15: 0.4q + 1.2 = 0.2q + 0.6

This is identical to Problem 6! Which was z, now q.

Same steps:

Subtract 0.2q: 0.2q + 1.2 = 0.6
Subtract 1.2: 0.2q = -0.6
Divide: q = -3

✔ Same as Problem 6 → ✔️

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Problem 16: -1.3r + 1.7 = -2.1r - 1.3

Add 2.1r to both sides:
→ 0.8r + 1.7 = -1.3

Subtract 1.7:
→ 0.8r = -3.0

Divide by 0.8:
→ r = -3.0 / 0.8 = -30/8 = -15/4 = -3.75

✔ Check: Left: -1.3*(-3.75) + 1.7 = 4.875 + 1.7 = 6.575
Right: -2.1*(-3.75) - 1.3 = 7.875 - 1.3 = 6.575 → ✔️

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Problem 17: Twice a number increased by 12 is equal to 14 less than three times the number.

Let the number be x.

Twice the number: 2x
Increased by 12: 2x + 12

Three times the number: 3x
14 less than that: 3x - 14

Set equal:
2x + 12 = 3x - 14

Subtract 2x:
12 = x - 14

Add 14:
x = 26

✔ Check: Twice 26 + 12 = 52 + 12 = 64; Three times 26 - 14 = 78 - 14 = 64 → ✔️

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Problem 18: Eight minus two times a number is equal to the number plus 1. Find the number.

Let number be x.

Eight minus two times x: 8 - 2x
Equal to x + 1

So:
8 - 2x = x + 1

Add 2x to both sides:
8 = 3x + 1

Subtract 1:
7 = 3x

Divide by 3:
x = 7/3

✔ Check: Left: 8 - 2*(7/3) = 8 - 14/3 = 24/3 - 14/3 = 10/3
Right: 7/3 + 1 = 10/3 → ✔️

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Problem 19: The perimeter of a rectangle is 24 inches. Find the dimensions if its length is 2 inches greater than its width.

Let width = w
Length = w + 2

Perimeter = 2(length + width) = 2(w + 2 + w) = 2(2w + 2) = 4w + 4

Set equal to 24:
4w + 4 = 24

Subtract 4:
4w = 20

Divide by 4:
w = 5

Then length = 5 + 2 = 7

✔ Check: Perimeter = 2*(5+7) = 2*12 = 24 → ✔️

Dimensions: width 5 inches, length 7 inches.

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Problem 20: The greater of two consecutive odd integers is 10 less than twice the lesser. Find the integers.

Consecutive odd integers differ by 2.

Let lesser = x
Greater = x + 2

Given: greater = 2*lesser - 10
So: x + 2 = 2x - 10

Subtract x:
2 = x - 10

Add 10:
x = 12

But 12 is even! Contradiction.

Wait — consecutive odd integers: if x is odd, x+2 is next odd.

But we got x=12, which is even → mistake.

Equation: x + 2 = 2x - 10
→ 2 + 10 = 2x - x
→ 12 = x

But 12 is not odd. So what’s wrong?

Perhaps “consecutive odd integers” means like 3 and 5, 5 and 7, etc.

If x is the lesser odd integer, then greater is x+2.

Equation: x+2 = 2x - 10 → x=12 → not odd.

That suggests no solution? But that can’t be.

Wait — maybe I misread.

“The greater of two consecutive odd integers is 10 less than twice the lesser.”

So: greater = 2*lesser - 10

With greater = lesser + 2

So: lesser + 2 = 2*lesser - 10
→ 2 + 10 = 2*lesser - lesser
→ 12 = lesser

But 12 is even, so the integers would be 12 and 14 — but those are even, not odd.

Contradiction.

Unless... perhaps "consecutive odd integers" allows us to set up differently.

Maybe let the lesser be 2k+1, greater be 2k+3.

Then: 2k+3 = 2*(2k+1) - 10
→ 2k+3 = 4k + 2 - 10
→ 2k+3 = 4k - 8
→ 3 + 8 = 4k - 2k
→ 11 = 2k
→ k = 5.5 → not integer → impossible.

What if I try actual numbers.

Suppose lesser is 11, greater is 13. Is 13 = 2*11 - 10 = 22-10=12? No.

Lesser 13, greater 15: 15 vs 2*13-10=26-10=16? No.

Lesser 15, greater 17: 17 vs 30-10=20? No.

Lesser 9, greater 11: 11 vs 18-10=8? No.

Lesser 7, greater 9: 9 vs 14-10=4? No.

Lesser 5, greater 7: 7 vs 10-10=0? No.

Lesser 3, greater 5: 5 vs 6-10=-4? No.

Lesser 1, greater 3: 3 vs 2-10=-8? No.

Negative? Lesser -1, greater 1: 1 vs -2-10=-12? No.

Lesse -3, greater -1: -1 vs -6-10=-16? No.

Seems no solution? But that can't be for a homework problem.

Perhaps I misinterpreted "consecutive odd integers".

Another thought: sometimes "consecutive" might mean adjacent in value, but for odds, it's always +2.

Or perhaps the problem has a typo, but let's read again.

"The greater of two consecutive odd integers is 10 less than twice the lesser."

Mathematically: g = l + 2, and g = 2l - 10

So l + 2 = 2l - 10 → l = 12, g = 14 — but these are even.

So the only mathematical solution is 12 and 14, but they are not odd.

Perhaps the problem meant "consecutive integers" without specifying odd? But it says "odd".

Let me check the image text — but I can't, since I'm not describing.

Perhaps in context, we should report the solution even if not odd, but that doesn't make sense.

Another idea: "consecutive odd integers" could be interpreted as integers that are odd and consecutive, but perhaps starting from any point.

But mathematically, the equation gives l=12, which is not odd.

Unless the problem has a different interpretation.

Let me solve the equation as is.

From l + 2 = 2l - 10 → l = 12, g = 14

Even though they are even, perhaps that's the intended answer, or there's a mistake in the problem.

But let's see the next problem.

Perhaps I can assume that "consecutive odd integers" is a red herring, but unlikely.

Another approach: let the two integers be x and x+2, both odd, so x must be odd.

Then x+2 = 2x - 10 → x=12, which is not odd, so no solution in odd integers.

But that can't be for a standard homework.

Perhaps "10 less than twice the lesser" is misinterpreted.

"Twice the lesser" is 2l, "10 less than" that is 2l - 10, yes.

Perhaps "the greater is 10 less than twice the lesser" means g = 2l - 10, yes.

Or perhaps it's "twice the greater is 10 less than the lesser" — but no, the sentence is clear.

Let me try to plug in the answer.

Suppose the integers are 12 and 14. Greater is 14, twice lesser is 24, 10 less is 14 — yes, 14 = 24 - 10.

So mathematically it works, but they are not odd.

Perhaps the problem meant "consecutive integers" and forgot to remove "odd", or vice versa.

In many textbooks, such problems often have integer solutions, so likely the "odd" is a mistake, or we should ignore it.

Perhaps for the sake of completing, I'll go with 12 and 14.

But let's see problem 21.

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Problem 21: The perimeter of a rectangle is 24 inches. Find the dimensions if its length is 2 inches greater than its width.

This is identical to Problem 19!

In Problem 19, we had perimeter 24, length = width + 2, found width 5, length 7.

Here same thing.

So dimensions: width 5 inches, length 7 inches.

But why repeat? Perhaps a copy-paste error in the worksheet.

Anyway, same answer.

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Now, back to Problem 20.

Given that mathematically the only solution is 12 and 14, and they satisfy the equation, even though not odd, perhaps that's what is expected.

Maybe "consecutive odd integers" is a distractor, but unlikely.

Another thought: perhaps "consecutive" means something else, but no.

Or perhaps in some contexts, "consecutive odd integers" includes even, but that doesn't make sense.

I recall that sometimes problems have no solution, but rare.

Let me search online or think differently.

Suppose the lesser is x, greater is x+2, both odd, so x ≡ 1 mod 2.

Equation: x+2 = 2x - 10 → x=12 ≡ 0 mod 2, contradiction.

So no solution in odd integers.

But for the purpose of this assignment, perhaps the intended answer is 12 and 14, ignoring the "odd" part.

Maybe the problem is: "the greater of two consecutive integers is 10 less than twice the lesser" — then it makes sense.

In that case, integers 12 and 14.

I think that's likely a typo, and we should proceed with 12 and 14.

To confirm, let's see if there's another interpretation.

"Consecutive odd integers" might mean like 1,3 or 3,5, etc., but as above, no solution.

Perhaps "10 less than twice the lesser" is for the difference or something, but no.

I'll go with 12 and 14, noting that they are not odd, but satisfy the equation.

So integers are 12 and 14.

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Now, compiling all answers.

Final Answer:

1. x = 23
2. x = 14
3. x = 1/4
4. x = 6/5
5. y = 3
6. z = -3
7. t = 54
8. m = 240/11
9. n = -26
10. p = -27
11. x = 20
12. y = 1/7
13. x = -58/15
14. p = -0.1
15. q = -3
16. r = -15/4 or -3.75
17. x = 26
18. x = 7/3
19. Width = 5 inches, Length = 7 inches
20. Integers: 12 and 14 (note: not odd, but satisfy equation)
21. Width = 5 inches, Length = 7 inches

For Problem 20, since the problem specifies "odd integers", but no such integers satisfy, perhaps it's an error. However, based on the equation, 12 and 14 are the solution. If strict, no solution, but I think for homework, they expect 12 and 14.

Similarly, Problem 21 is duplicate of 19.

Now, for the final answer section, I need to list them clearly.

Since the user didn't specify format, I'll list by problem number.

Also, for Problem 13, I have x = -58/15, which is approximately -3.8667, but fraction is exact.

Similarly, others.

Let me write neatly.