Substitution Calculator - Solve Systems by Substitution, Step by Step

Put both equations in standard form first: $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$. Then fill the six boxes of the substitution calculator and click Calculate.

• $a_1$, $b_1$, $c_1$ — Equation 1: the x-coefficient, the y-coefficient, and the right side. For x + y = 7 they are 1, 1, 7 — the substitution calculator's default system.
• $a_2$, $b_2$, $c_2$ — the same three numbers for Equation 2. Keep the sign as it stands: in x − y = 1, the y-coefficient $b_2$ is −1.
• Check — the substitution calculator's guard row: it tests whether the determinant $a_1b_2 - b_1a_2$ is nonzero. True means exactly one solution.
• x and y — the solution pair, rounded to four decimals. The substitution calculator prints one row per letter.

The substitution calculator starts pre-filled with x + y = 7 and x − y = 1; one click solves it: x = 4, y = 3. Swap in your own numbers and the substitution calculator re-solves instantly.

A false Check row means the determinant is 0: no solution or infinitely many. The special-cases table below separates them in seconds, so the substitution calculator never leaves you guessing.

This page is more than a static substitution calculator — it is also an AI tutor. The box under the form takes full questions, photos, and follow-ups:

• Ask in plain words. "My equations are 2x + 3y = 12 and x − y = 1 — solve the system by substitution." The AI fills the six boxes and runs the calculator on this page.
• Take a photo of your homework. The AI reads the system and fills the substitution calculator form for you.
• Ask about the ideas, not just the answer. "Why substitute into the other equation?" The AI explains step by step; every number still comes from the calculator on this page.
• Work backwards. Give a solution point and ask which coefficients produce it. The AI walks it backwards with you, then the substitution calculator confirms the pair.
• Keep asking follow-ups. Ask "why?" as often as you need; the substitution calculator page keeps teaching.

Substitution means replacing a letter with something equal to it. If x + y = 7, then x is 7 − y everywhere, so the x in the other equation can become 7 − y. One equation, one letter — and solving it is routine. That swap is the one move a substitution calculator makes, over and over. The substitution calculator just never gets tired of it.

A 1904 college algebra text defines the move: to eliminate x is to derive an equation in which x no longer occurs. That definition still drives every substitution method calculator, this one included. This substitution calculator runs it on whatever pair you type.

Read the picture: the blue line is every pair with x + y = 7, the orange line every pair with x − y = 1. The lines cross exactly once, at (4, 3). Every substitution calculator question ends at a crossing like this one. One crossing, one solution — that is the substitution calculator's Check row in graphical form.

Here is the full routine — you can solve the system by substitution by hand in six short steps:

1. Solve one equation for a letter with coefficient 1, to avoid fractions.
2. Substitute that expression into the other equation, parentheses and all.
3. Solve the result for the remaining letter.
4. Substitute back into either original equation for the second letter.
5. Write the answer as an ordered pair (x, y).
6. Check the pair in both original equations.

Every example below runs these six steps; the substitution calculator collapses them into one click.

The six steps always end at the same two fractions, so the substitution calculator can evaluate them directly:

$x = \dfrac{c_1b_2 - b_1c_2}{a_1b_2 - b_1a_2} \qquad y = \dfrac{a_1c_2 - c_1a_2}{a_1b_2 - b_1a_2}$

Both share one denominator — the determinant:

$D = a_1b_2 - b_1a_2$

The Check row of the substitution calculator computes exactly this number. A zero there is the substitution calculator's only stop sign.

Special cases — how the substitution calculator reports a false Check row:

Try 2x + 6y = 10 with x + 3y = 5: every ratio matches, one line drawn twice, infinitely many solutions. Change $c_1$ to 9 and no pair works: no solution. The substitution calculator flags both with a false Check row; the ratios tell you which.

Solve the anchor system by substitution, one line per step — the substitution calculator's own default:

1. Solve Equation 1 for x: $x = 7 - y$.
2. Substitute into Equation 2: $(7 - y) - y = 1$.
3. Combine like terms: $7 - 2y = 1$, so $y = 3$.
4. Substitute back: $x = 7 - 3 = 4$.
5. Check both equations: $4 + 3 = 7$ and $4 - 3 = 1$. The solution is $(4, 3)$.

The form above holds exactly this system: $a_1 = 1$, $b_1 = 1$, $c_1 = 7$, $a_2 = 1$, $b_2 = −1$, $c_2 = 1$. Click Calculate and the substitution calculator confirms x = 4, y = 3 in one run. That is the substitution calculator's whole promise: six coefficients in, two values out.

A 1904 college algebra text solves this pair, and it still teaches the right habits:

1. Solve Equation 1 for x: $x = 3 - 3y$.
2. Substitute into Equation 2, parentheses kept: $3(3 - 3y) + 5y = 1$.
3. Distribute the 3 over both terms: $9 - 9y + 5y = 1$.
4. Solve: $-4y = -8$, so $y = 2$.
5. Substitute back: $x = 3 - 3(2) = -3$.

The old book then verifies both: $−3 + 3 \cdot 2 = 3$ and $3(−3) + 5 \cdot 2 = 1$. Both hold, so $(x, y) = (−3, 2)$. The substitution calculator will agree.

Type 1, 3, 3 and 3, 5, 1 into the substitution calculator above. The substitution calculator's x and y rows return −3 and 2 without a single fraction. The substitution calculator and the 1904 text agree: the method is that stable.

Sometimes the solution is given and a coefficient is missing — work the substitution calculator's direction in reverse. Suppose x + y = 7 and 2x − by = 5 have the solution (4, 3). Find b.

1. A solution must satisfy both equations, so feed x = 4 and y = 3 to the second one.
2. That gives $2 \cdot 4 - 3b = 5$.
3. Solve: $8 - 3b = 5$, so $3b = 3$ and $b = 1$.
4. The full system is x + y = 7 with 2x − y = 5, and indeed $4 + 3 = 7$ while $8 - 3 = 5$.

With b = 1 the determinant is $1 \cdot (−1) - 1 \cdot 2 = −3$, not 0. One solution exists, and the substitution calculator reports it as x = 4, y = 3 — run it and see. Change b to 2 and the substitution calculator returns (4.75, 2.25) instead: the system itself changed.

Three slips the substitution calculator catches every day:

1. Dropping the parentheses. The 3 in 3(3 − 3y) must multiply both terms. Writing 9 − 3y instead of 9 − 9y wrecks the y value; the substitution calculator's x and y rows keep the algebra honest.
2. Losing a sign. A term that crosses the equals sign flips: −y becomes +y. That is why the anchor types $b_2 = −1$ exactly as x − y = 1 stands; the substitution calculator keeps every sign you type.
3. Solving for the wrong letter. When some coefficient equals 1, solve for that letter and skip the fractions. Forcing 3x + 5y = 1 to give x first invites fractions for nothing; the substitution calculator accepts any letter choice — the fractions are the only cost.