How many problems do I need to solve to be confident with linear equations?

A useful rule of thumb: 5 to 10 problems per sitting, three sittings spaced across three different days before a test. That is roughly 15 to 30 problems you have actually worked through — enough to recognize the standard forms without burning out.

What if I get completely stuck on a problem?

Three steps. First: open the hint and give yourself 30 seconds to think it through. Second: read the linear-equations tutorial — it walks through the four-step method explicitly. Third: paste your problem into the step-by-step calculator and compare each transformation with your attempt. That triangulation pinpoints the exact step where things went sideways.

What is the boss question?

The boss question is the final and hardest problem in a set. It combines multiple steps — typically variable on both sides or brackets together with fractions. If you can crack the boss, you own the topic.

What types of equations does this set cover?

Three problems in standard form ax + b = c, one bracketed problem 3(x − 2) = 15, and one boss with x on both sides 4x + 9 = 2x + 21. That covers the three problem types that matter most for Grade 7 algebra tests.

Which grade level is this for?

Linear equations land in most curricula at Grade 7 (ages 12–13). Grade 8 typically introduces systems of equations, Grade 9 quadratics.

Do I need an account, and is there a paid tier?

No to both. The practice set is fully free, no signup, no email gate, no in-app upsells.

Home/Grade 7/Algebra/Linear equations/Practice

Practice · Grade 7 · Algebra

Practice solving linear equations

19 training problems in rising difficulty plus a boss question with variables on both sides. Solve directly below — hints available, no signup.

Q1 of 20

0 correct

Solve for x.

2x + 3 = 11

Your answer for x

Need backup?

Read tutorial Check with calculator

Quick answer

What's the most effective way to practice linear equations?

Solve at least five problems in rising difficulty: start with the standard form ax + b = c, then move on to bracketed problems like 3(x − 2) = 15, and finish with a problem that has the variable on both sides. Write each transformation explicitly in the margin (e.g. "| − 7"), always verify your answer by substituting it back into the original equation, and use a hint sooner rather than later — understanding beats rushing.

HowTo

A four-step solving strategy

This sequence works for any linear equation — standard form ax + b = c, with brackets, or with x on both sides.

1
Step 1 of 4
Read the equation and identify its form
Is it already standard form ax + b = c? Or does it have brackets, fractions, or x on both sides? This diagnosis decides whether you can transform immediately or have to clean up first.
2
Step 2 of 4
Clean up: expand brackets, collect x-terms
Brackets via the distributive property: 3(x − 2) becomes 3x − 6. With x on both sides: move all x-terms to one side (e.g. "− 2x" on both sides). Only after cleanup do you have the form ax + b = c.
3
Step 3 of 4
Isolate the constant: subtract or add
For 3x + 7 = 22 you subtract 7 from both sides → 3x = 15. Write "| − 7" in the margin so the step is traceable.
4
Step 4 of 4
Divide by the coefficient, then verify
Divide both sides by a (here 3): x = 5. Substitute that value back into the original equation — both sides should match. Only then is the problem actually solved.

Examples

Worked examples with full solving steps

Four typical problem types from Grade 7 algebra tests. Try each on your own first, then compare with the worked solution.

Easy

Solve: 2x + 3 = 11

2x + 3 = 11 | − 3

2x = 8 | ÷ 2

x = 4

Check: 2 · 4 + 3 = 8 + 3 = 11 ✓

Classic ax + b = c form. Peel off the constant first, then divide.

Easy

Solve: 5x − 7 = 18

5x − 7 = 18 | + 7

5x = 25 | ÷ 5

x = 5

Check: 5 · 5 − 7 = 25 − 7 = 18 ✓

Same shape as above but with a minus. Watch the sign flip: "− 7" becomes "+ 7" when moved across.

Medium

Solve: 3(x − 2) = 15

3(x − 2) = 15 | expand

3x − 6 = 15 | + 6

3x = 21 | ÷ 3

x = 7

Check: 3 · (7 − 2) = 3 · 5 = 15 ✓

Expand the bracket via the distributive property first. Faster trick: divide directly by 3 → x − 2 = 5 → x = 7.

Hard

Boss: 4x + 9 = 2x + 21

4x + 9 = 2x + 21 | − 2x

2x + 9 = 21 | − 9

2x = 12 | ÷ 2

x = 6

Check: 4 · 6 + 9 = 24 + 9 = 33; 2 · 6 + 21 = 12 + 21 = 33 ✓

Variable on both sides — first move all x-terms to one side (here, the left), then run the standard routine.

Pitfalls

Common mistakes — and how to avoid them

These five traps show up in almost every test on this topic.

Forgetting to flip the sign when moving terms

When you move "+ 7" to the other side, it becomes "− 7" — not "+ 7". Write "| − 7" in the margin and the step is hard to forget.

Misreading or mishandling brackets

In 3(x − 2), the minus also applies to 3 times 2. The result is 3x − 6, not 3x − 2. And in −(x + 4), both signs flip: −x − 4.

Verifying against an intermediate step instead of the original

Substitute x into the original problem, not into a transformed line — otherwise you miss arithmetic errors made early.

Treating a divide-by-negative wrong

For equations, nothing else changes. For inequalities the comparison flips — that confusion trips students up at exam time.

Avoiding fractions or decimals

For 0.5x + 1.5 = 4, the method is identical to whole-number problems. If you prefer, multiply the whole equation by 2 first and work with whole numbers.

Study

Practice with a plan — three short tips

15 minutes spread out, not 90 in one go

Three short sessions on three different days stick better than one long session the night before the test. The technical name is "spaced repetition".

Solve first, then check the worked answer

Tempting as it is, write down your steps before revealing the hint. Active recall is three to four times more effective than passive reading.

After every wrong answer, ask why

Was it a sign flip? A skipped step? A bracket? Note the cause — and on your next round you will spot the same trap immediately.

FAQ

Practice — frequently asked

Glossary

Definitions in one sentence

Linear equation: An equation of the form ax + b = c with a ≠ 0; the variable x appears only to the first power.
Coefficient: The number multiplying the variable — in 3x the coefficient is 3.
Constant: A standalone number — in 3x + 7 the constant is 7.
Equivalent transformation: An operation (e.g. "subtract 7 from both sides") that does not change the equation's solution set.
Distributive property: The rule a(b + c) = ab + ac — the basis for expanding brackets.
Verification: Substituting the computed solution back into the original equation as a sanity check.
Boss question: The final and hardest problem in a practice set, combining multiple steps or problem types.

Practice solving linear equations

Solve for x.

A four-step solving strategy

Read the equation and identify its form

Clean up: expand brackets, collect x-terms

Isolate the constant: subtract or add

Divide by the coefficient, then verify