Algebra solver step by step, with a reason for every move
Need to see why each line of an algebra problem follows from the one before? Type a problem or snap a worksheet, and your agent solves it one move at a time, says which operation it applied to both sides, plugs the answer back in to check it, and can draw a graph so the solution makes sense visually.
The prompt
- Solve this algebra problem step by step: [PROBLEM OR PHOTO]. I am in [CLASS OR GRADE]. Show one move per line and write what was done to both sides (add, subtract, multiply, divide, factor, square root). If there is more than one method, use [PREFERRED METHOD] and mention the other in one sentence. Check every solution by substituting it back into the original, and throw out any extraneous solutions with the reason. If it helps, add a graph as a PNG showing where the solution is. Then give me [NUMBER] practice problems of the same type, getting a little harder, with answers in a separate message I can open later.
What to change
- [PROBLEM OR PHOTO]: E.g. "2x² − 7x + 3 = 0" or "solve the system 3x + 2y = 12, x − y = 1". Or attach a photo.
- [CLASS OR GRADE]: E.g. "Pre-algebra", "Algebra 1", "Algebra 2", "Year 10".
- [PREFERRED METHOD]: E.g. "factoring", "quadratic formula", "elimination", "substitution", or "the easiest one".
- [NUMBER]: How many practice problems, e.g. 3 or 5.
Example result
- Problem
- Solve 2x² − 7x + 3 = 0 (Algebra 1, by factoring)
- Step by step
- • Line 1: 2x² − 7x + 3 = 0. It is already in standard form ax² + bx + c = 0, with a = 2, b = −7, c = 3.
- • Line 2: Find two numbers that multiply to a × c = 6 and add to b = −7. They are −6 and −1.
- • Line 3: Split the middle term: 2x² − 6x − x + 3 = 0
- • Line 4: Group and factor each pair: 2x(x − 3) − 1(x − 3) = 0
- • Line 5: Factor out the common bracket: (2x − 1)(x − 3) = 0
- • Line 6: Zero product rule. Either 2x − 1 = 0, so x = 1/2, or x − 3 = 0, so x = 3.
- Check both answers
- • x = 1/2: 2(1/4) − 7(1/2) + 3 = 0.5 − 3.5 + 3 = 0. Correct
- • x = 3: 2(9) − 7(3) + 3 = 18 − 21 + 3 = 0. Correct
- Other method
- The quadratic formula gives the same result: x = (7 ± sqrt(49 − 24)) / 4 = (7 ± 5) / 4, so 3 or 1/2. Use it when factoring is not obvious.
- What the graph shows
- The parabola y = 2x² − 7x + 3 opens upward and crosses the x-axis at x = 0.5 and x = 3. Those crossing points are the solutions.
- Practice (answers sent separately)
- • 1. x² + 5x + 6 = 0
- • 2. 3x² − 10x + 8 = 0
- • 3. 4x² − 4x − 15 = 0
How to do it with todo.is
- Copy the prompt, then enter the problem, your class, the method you are using and how many practice problems you want.
- Paste it into todo.is on the Today screen, or send it with a photo to your agent on WhatsApp or Telegram.
- You get the solution one move per line, a check of every answer, a graph if useful and practice problems.
- Do the practice problems, then ask for the answers. Say "slower, explain line 2" whenever a step jumps too far.
Tips for a better result
- Write the reason next to each line, like "subtract 2x from both sides". Teachers give marks for it, and it stops careless errors.
- Always substitute answers back in, especially with square roots, fractions or variables in the denominator, where extraneous solutions appear.
- For systems, ask for both substitution and elimination once, then pick the one you find faster.
- Ask your agent for a daily set of 5 algebra problems on Telegram before a test. It becomes a short recurring warm-up.
algebra solver step by step: FAQ
- What kinds of algebra problems can it solve? Linear equations and inequalities, systems, quadratics, factoring, exponents and radicals, rational equations, absolute value, functions and logarithms.
- Does it show steps or just the answer? This prompt asks for one move per line with the reason, plus a check by substitution, so you can follow and copy the method by hand.
- Can it solve word problems? Yes. It first defines the variable, turns the words into an equation, then solves and answers the question in a full sentence with units.
- What is an extraneous solution? An answer that comes out of the algebra but does not work in the original equation, often after squaring both sides or multiplying by a variable. Checking by substitution catches it.
JavaScript is required to use the todo.is app.