Math Buddyv1.0

System of equations calculator with steps

Type or scan a system of equations. MathBuddy solves it by substitution or elimination and shows each step.

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What this page covers

  • Two linear equations in two unknowns, solved by substitution or by elimination.
  • Three equations in three unknowns, reduced to two equations and then to one.
  • Systems with no solution or with infinitely many, which the steps identify rather than force a single answer.
  • Not covered here: one equation in one unknown (use Solve an equation for x) and the matrix method (use Gauss-Jordan elimination calculator).

How to enter the problem

  • Type each equation on its own line, or separate the equations with a comma.
  • Scan it. Fit the one problem in the frame.
  • Choose a photo. Several problems in one photo are solved as a list.
  • Then tap Solve.

Worked examples

Example 1

Problem

Solve by substitution: y=2x−1\displaystyle y = 2x - 1 and 3x+2y=12\displaystyle 3x + 2y = 12.

Answer

Verified

The answer is

x=2,y=3x = 2, y = 3

Explanation

  1. Use the equation already solved for yy
    y=2x−1y = 2x - 1 can go straight into the other equation.
  2. Substitute
    3x+2(2x−1)=123x + 2(2x - 1) = 12
  3. Solve for xx
    3x+4x−2=123x + 4x - 2 = 12, so 7x=147x = 14 and x=2x = 2.
  4. Find yy
    y=2(2)−1=3y = 2(2) - 1 = 3

Example 2

Problem

Solve by elimination: 4x+3y=10\displaystyle 4x + 3y = 10 and 2x−5y=18\displaystyle 2x - 5y = 18.

Answer

Verified

The answer is

x=4,y=−2x = 4, y = -2

Explanation

  1. Match the xx coefficients
    Multiply both sides of the second equation by 2
    4x−10y=364x - 10y = 36
  2. Subtract to remove xx
    (4x+3y)−(4x−10y)=10−36(4x + 3y) - (4x - 10y) = 10 - 36, so 13y=−2613y = -26.
  3. Solve for yy
    y=−2y = -2
  4. Back-substitute
    4x+3(−2)=104x + 3(-2) = 10, so 4x=164x = 16 and x=4x = 4.

Example 3

Problem

Solve x+y+z=6\displaystyle x + y + z = 6, 2x−y+z=3\displaystyle 2x - y + z = 3 and x+2y−z=2\displaystyle x + 2y - z = 2.

Answer

Verified

The answer is

x=1,y=2,z=3x = 1, y = 2, z = 3

Explanation

  1. Remove zz using the first two equations
    Second minus first
    x−2y=−3x - 2y = -3
  2. Remove zz using the first and third
    First plus third
    2x+3y=82x + 3y = 8
  3. Solve the two-unknown system
    From x=2y−3x = 2y - 3: 2(2y−3)+3y=82(2y - 3) + 3y = 8, so 7y=147y = 14, y=2y = 2 and x=1x = 1.
  4. Find zz
    z=6−x−y=6−1−2=3z = 6 - x - y = 6 - 1 - 2 = 3

Common mistakes

  • Scaling only one side of an equation for elimination. Twice 2x−5y=182x - 5y = 18 is 4x−10y=364x - 10y = 36, right-hand side included.
  • Losing a sign when subtracting equations: 3y−(−10y)=13y3y - (-10y) = 13y, not −7y-7y.
  • Stopping after one unknown. The system is solved only when every unknown has a value.
  • Treating 0=50 = 5 as an arithmetic slip. It means the system has no solution, while 0=00 = 0 means infinitely many.

Checks, assumptions and limits

  • Substitute into every original equation, not just one: x=3,y=1x = 3, y = 1 gives 2(3)+1=72(3) + 1 = 7 and 3−1=23 - 1 = 2.
  • A linear system in two unknowns has exactly one solution, none, or infinitely many.
  • Values stay exact: a solution such as x=73x = \frac{7}{3} is kept as a fraction unless you ask for a decimal.
  • MathBuddy can make mistakes. Double check important steps.

Related tasks

Frequently asked questions

Yes. Write "by substitution" or "by elimination" before the system. To see the other method afterwards, ask a follow-up in the same thread; a typed follow-up needs sign-in.

Yes. The steps remove one unknown at a time until a single equation in one unknown is left, then work back to find the others.

The steps reach a false statement such as 0 = 5 and say there is no solution. A true statement such as 0 = 0 means there are infinitely many.

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