System of Equations Calculator (Elimination and Substitution with Steps)

Enter the coefficients and the right-hand sides of equations (1) and (2). The system below is linked to the input fields, so you can also edit the numbers directly in it.

Method

(1) x coefficient a

(1) y coefficient b

(1) right side c

(2) x coefficient d

(2) y coefficient e

(2) right side f

Decimals, negative numbers and fractions such as 3/4 can be used. A blank coefficient of x or y is treated as 1 (x means 1x), and a blank right-hand side is treated as 0. Enter 0 for any term you do not use.

Result and graph

Enter the coefficients of the two equations in the fields on the left and press "Calculate". The steps and a graph will appear here.

[ CLOSE ]

What you can do on this page
What is this calculation used for?
How to Use
Formulas and graphs
Symbols and terms
- Symbols
- Terms
Good to know before you start
How to calculate it in Excel
How to calculate it in Google Sheets
How to calculate it in Python
How to write it in LaTeX and other math languages (copy and paste)
How to have ChatGPT do the calculation
DataChef Features
Related Features
NumberChef Calculators List

What you can do on this page

Solve a system of two linear equations, $ax + by = c,\ dx + ey = f$, on the spot just by entering the coefficients and the right-hand sides. The calculator shows the system as you would write it by hand, and you can edit the numbers right in it
Choose the elimination method (match the coefficients, then add or subtract the equations) or the substitution method (rewrite one equation as $y = $ and substitute). The steps follow the same flow as a school algebra textbook
Coefficients can be decimals, negative numbers or fractions such as 3/4 (for fractions and decimals, the steps start by multiplying both sides to make the coefficients whole numbers)
Besides a single solution, it detects no solution (parallel lines) and infinitely many solutions (the same line), and explains why. Answers are shown as fractions in lowest terms (exact values), with decimals added when they are fractions. You also get a graph of the two lines (the intersection is the solution) and a check that substitutes the solution into both equations
A plain-language explanation of the formulas and copy-and-paste formulas for Excel, Google Sheets and Python are all on this page

This page solves a system of two linear equations in two variables (x and y). Systems with three or more variables, or with squared terms, are not supported.

What is this calculation used for?

Finding the prices of two items (shopping and bookkeeping)

Suppose you have two receipts that show only totals: "1 coffee and 1 bagel for $5.50" and "2 coffees and 1 bagel for $9.00". Let a coffee be $x$ dollars and a bagel $y$ dollars. The system $x + y = 5.5$, $2x + y = 9$ gives a coffee at $3.50 and a bagel at $2.00.
"The single prices are not shown, but the totals of combinations are" is the most classic use of a system of equations, and it also comes up when breaking down bills and quotes in bookkeeping.

Finding where two price plans cross (phone, electricity, subscriptions)

Plan A costs $15 a month plus $5 per GB, and Plan B costs $30 a month plus $2 per GB. With the monthly bill $y$ dollars and the data $x$ GB, they are $y = 5x + 15$ and $y = 2x + 30$. Solving the system gives $x = 5$ GB, where both bills are the same, $40.
So "if you use more than 5 GB, Plan B is cheaper". Finding the intersection of two lines, the crossover point, works the same way for comparing electricity plans or subscriptions (assuming the price grows in proportion to how much you use).

Finding how far you walked and how far you ran (commuting)

It is 1.5 miles from home to the station. You walk at 3 mph, then run at 6 mph, and arrive in 20 minutes ($\dfrac{1}{3}$ hour). With $x$ miles walked and $y$ miles run, the distance equation is $x + y = 1.5$ and the time equation is $\dfrac{x}{3} + \dfrac{y}{6} = \dfrac{1}{3}$. Solving gives 0.5 mile walked and 1 mile run.
Whenever you have a total and a second amount decided by how the total is split (time, cost, and so on), you get a system of equations. This pattern is useful far beyond speed problems.

Mixing solutions of different strengths (cooking, manufacturing, farming)

To make 300 mL of an 8% solution by mixing a 5% solution and a 10% solution, let $x$ mL be the 5% solution and $y$ mL the 10% solution. The total amount gives $x + y = 300$, and the amount of the dissolved substance gives $0.05x + 0.1y = 24$ (8% of 300 mL is 24 mL). Solving gives 120 mL of the 5% solution and 180 mL of the 10% solution.
Diluting cleaning concentrate, mixing fertilizer, making metal alloys: this calculation is used wherever two ingredients are mixed to reach a target strength.

Finding the break-even point (shops and business)

A product sells for $8, its materials cost $5 per unit, and fixed costs such as rent are $900 a month. With $x$ units sold per month and $y$ dollars, the revenue equation is $y = 8x$ and the cost equation is $y = 5x + 900$. Their intersection is $x = 300$ units and $y = 2{,}400$ dollars, so 300 units a month is the line between profit and loss (the break-even point).
It is one of the first numbers to check in a business plan or when setting a price, and it is a system of equations at heart (assuming the price and the unit cost do not change with the quantity).

Formulas and graphs

A system of two linear equations and its solution

Graph

Standard notation (the usual math form)

$ax + by$ $=$ $c$

$dx + ey$ $=$ $f$

In words (symbols replaced with words)

① $ax + by$: left side of (1) (the $x$ term plus the $y$ term) $=$ ② $c$: right side of (1)

③ $dx + ey$: left side of (2) (the $x$ term plus the $y$ term) $=$ ④ $f$: right side of (2)

The formula in words

① The equation that $ax + by$: left side of (1) (the $x$ term plus the $y$ term)

② equals $c$: right side of (1) and the equation that

③ $dx + ey$: left side of (2) (the $x$ term plus the $y$ term)

④ equals $f$: right side of (2) must both be true at the same time. A pair $x,\ y$ that makes both true is called the "solution"

Quick example

For (1) $x + y = 5$ and (2) $2x - y = 1$, the only pair that makes both true is $x = 2,\ y = 3$

left side of (1), $x + y$ $=$ right side of (1), $5$

left side of (2), $2x - y$ $=$ right side of (2), $1$

$x = 2,\ y = 3$

$2 + 3 = 5$

$2 \times 2 - 3 = 1$

Key idea

With the single equation $x + y = 5$, there are infinitely many pairs that work: $(1,\ 4)$, $(2,\ 3)$, $(0.5,\ 4.5)$, and so on. Adding the condition that a second equation must also be true narrows the answer down to one pair. On a graph, the points that satisfy one equation form a line, and the solution of the system is the point where the two lines cross (the point of intersection). So if the two lines are parallel there is no solution, and if they lie exactly on top of each other there are infinitely many solutions.

Elimination method (match the coefficients and subtract the equations)

Standard notation (the usual math form)

$(ax + by)$ $-$ $(ax + ey)$ $=$ $c - f$

$(b - e)y$ $=$ $c - f$

In words (symbols replaced with words)

① $(ax + by)$: left side of (1) $-$ ② $(ax + ey)$: left side of (2) ($x$ coefficient matched to (1)) $=$ ③ $c - f$: difference of the right sides

④ $(b - e)y$: $x$ is gone and only $y$ is left $=$ $c - f$: difference of the right sides

The formula in words

① First make the $x$ coefficients of the two equations match. From $(ax + by)$: left side of (1)

② subtract $(ax + ey)$: left side of (2) with its $x$ coefficient matched to (1) (subtract the right sides in the same way)

③ The right side becomes $c - f$: difference of the right sides

④ and on the left side the $ax$ terms cancel, leaving $(b - e)y$: $x$ is gone and only $y$ is left (then divide both sides by the coefficient of $y$ to find $y$)

Quick example

Eliminate $y$ from (1) $x + y = 5$ and (2) $2x - y = 1$. The coefficients of $y$ are $+1$ and $-1$ (same absolute value, opposite signs), so you can add the equations as they are

left side of (1), $x + y$ $+$ left side of (2), $2x - y$ $=$ sum of the right sides, $5 + 1$

$(x + y) + (2x - y) = 5 + 1$

$3x = 6$

$x = 2$

$y = 5 - 2 = 3$

Key idea

Whether to subtract or add depends on the signs of the matched coefficients. If they have the same sign (such as $+2$ and $+2$), subtract the equations; if they have opposite signs (such as $+2$ and $-2$), add them. Either way, that variable is eliminated. If the coefficients do not match yet, multiply both sides of the equations to make them match. For example, if the $x$ coefficients are 2 and 3, match them at the least common multiple, 6, by doing (1) × 3 and (2) × 2. "Multiply both sides by the same number" and "add or subtract left side to left side and right side to right side" are both based on the properties of equality (doing the same thing to both sides keeps them equal), so the answer does not change. You may eliminate either $x$ or $y$. A good tip is to eliminate the one whose coefficients are easier to match.

Substitution method (rewrite as $y = $ and substitute)

Standard notation (the usual math form)

$y$ $=$ $px + q$

$dx + e(px + q)$ $=$ $f$

In words (symbols replaced with words)

① $y$: from rewriting (1) $=$ ② $px + q$: an expression in $x$ only

③ $dx + e(px + q)$: left side of (2) with $px + q$ substituted for $y$, now in $x$ only $=$ ④ $f$: right side of (2)

The formula in words

① Rewrite (1) by moving terms and dividing, so that $y$: from rewriting (1) is written as

② $px + q$: an expression in $x$ only

③ Put this in place of $y$ in (2) (substitute it), and the left side of (2) becomes $dx + e(px + q)$: left side of (2) in $x$ only

④ Set this equal to $f$: right side of (2) and solve this one-variable linear equation to find $x$. Then substitute $x$ into $px + q$ to find $y$

Quick example

For (1) $x + y = 5$ and (2) $2x - y = 1$, solving (1) for $y$ gives $y = 5 - x$. Substitute this for $y$ in (2)

left side of (2) with $5 - x$ for $y$, $2x - (5 - x)$ $=$ right side of (2), $1$

$2x - (5 - x) = 1$

$3x - 5 = 1$

$3x = 6$

$x = 2$

$y = 5 - 2 = 3$

Key idea

You are free to choose which equation to solve and for which variable. Choosing a variable whose coefficient is 1 or $-1$ avoids fractions and makes the work easier (this calculator chooses the same way). When you substitute, put the whole expression in parentheses. Getting the signs wrong when removing the parentheses in $2x - (5 - x)$ is the most common mistake with the substitution method. Elimination and substitution are two routes with the same plan: "get rid of one variable to make a one-variable equation". So they always give the same answer. When one equation is already in the form $y = $, such as $y = 2x + 1$, substitution is the better choice.

A formula for the solution in one step, and how to spot no solution or infinitely many

Graph

Standard notation (the usual math form)

$x$ $=$ $\dfrac{ce - bf}{ae - bd}$

$y$ $=$ $\dfrac{af - cd}{ae - bd}$

In words (symbols replaced with words)

② $x$: solution for $x$ $=$ ① $\dfrac{ce - bf}{ae - bd}$: fraction made from the coefficients

④ $y$: solution for $y$ $=$ ③ $\dfrac{af - cd}{ae - bd}$: fraction made from the coefficients

The formula in words

① For the denominator, cross-multiply $a$ with $e$ and $b$ with $d$ and subtract. For the numerator, do the same with $c$ and $e$, $b$ and $f$. Working out $\dfrac{ce - bf}{ae - bd}$: fraction made from the coefficients gives

② $x$: solution for $x$

③ Keeping the same denominator and changing only the numerator to $af - cd$, working out $\dfrac{af - cd}{ae - bd}$: fraction made from the coefficients gives

④ $y$: solution for $y$

Quick example

For (1) $x + y = 5$ and (2) $2x - y = 1$ ($a = 1,\ b = 1,\ c = 5,\ d = 2,\ e = -1,\ f = 1$)

$x$: solution for $x$ $=$ fraction made from the coefficients $\dfrac{5 \times (-1) - 1 \times 1}{1 \times (-1) - 1 \times 2}$

$ae - bd = 1 \times (-1) - 1 \times 2 = -3$

$x = \dfrac{ce - bf}{ae - bd} = \dfrac{5 \times (-1) - 1 \times 1}{-3} = \dfrac{-6}{-3} = 2$

$y = \dfrac{af - cd}{ae - bd} = \dfrac{1 \times 1 - 5 \times 2}{-3} = \dfrac{-9}{-3} = 3$

Key idea

If the denominator $ae - bd$ is not 0, there is exactly one solution. This formula is called Cramer's rule. It is the result of elimination, written with the coefficient letters instead of numbers. If the denominator $ae - bd = 0$, you cannot divide, and there is no single solution. This happens when the left sides of the two equations are proportional (the two lines have the same slope). If the right sides are in the same ratio too, the lines are the same (infinitely many solutions); if only the right sides are off, the lines are parallel (no solution). In formulas: if the numerators $ce - bf$ and $af - cd$ are both 0 as well, there are infinitely many solutions; otherwise there is no solution. The formula is handy for checking answers and for solving many systems at once in a spreadsheet. On a school test, though, you are usually expected to show the steps of elimination or substitution.

The basic plan for solving a system of equations is to "get rid of one variable to make a one-variable equation". The elimination method matches the coefficients and adds or subtracts the equations; the substitution method rewrites one equation as $y = $ and substitutes it. Both give the same answer. When there is no single solution, the two lines are either parallel (no solution) or the same line (infinitely many solutions).

Symbols and terms

Symbols

$x,\ y$	ex, why	The letters for the two numbers you want to find (the unknowns). By custom, letters near the end of the alphabet, $x,\ y,\ z$, are used for numbers we do not know yet.
$a,\ b,\ c,\ d,\ e,\ f$	a, b, c, d, e, f	Letters that stand for the coefficients and the right-hand constants (the known numbers). By custom, letters near the start of the alphabet are used for fixed numbers. On this page, (1) is $ax + by = c$ and (2) is $dx + ey = f$.
(1), (2)	equation one, equation two	Numbers given to the equations. They are used to write briefly which operation was done to which equation, as in "(1) × 2" or "(1) + (2)". Textbooks and worked solutions use them too.
$\{$	brace	The large brace on the left of a system of equations. It means "these two equations must be true at the same time", showing that they belong together as a set.
$ae - bd$	a e minus b d	The value in the denominator of the one-step formula, found by cross-multiplying the coefficients and subtracting. Whether it is 0 decides whether there is exactly one solution. In the language of matrices, it is the determinant of the matrix of coefficients.
$(x,\ y)$	the pair x, y	A way of writing two values as a set. The solution of a system is a pair of $x$ and $y$, so it is written as an ordered pair such as $(2,\ 3)$, the same way as the coordinates of a point on a graph.

Terms

system of equations	Two or more equations taken as a set, where you look for the values of the variables that make all of them true at the same time. Taught in Grade 8 and Algebra 1.
linear equation in two variables	An equation with two variables in which every term is first degree (no squares and no variables multiplied together). $x + y = 5$ is a linear equation in two variables. Its graph is a straight line.
solution	The values of the variables that make the equations true. For a system, the solution is a pair of values of $x$ and $y$, which on a graph is the point of intersection of the two lines.
elimination method	A way of solving by matching the coefficients of the two equations and then adding or subtracting them (left side to left side, right side to right side) to eliminate one variable. It is also called the addition method.
adding the equations	Adding the left sides together and the right sides together, and joining the two sums with "=" to make a new equation. Subtracting the equations works the same way.
substitution method	A way of solving by rewriting one equation in the form "$y = $ (an expression in $x$)" and substituting it for $y$ in the other equation to eliminate one variable.
substitute	To put a value or an expression in place of a letter. Substituting $y = 5 - x$ into $2x - y = 1$ gives $2x - (5 - x) = 1$.
eliminate	To remove one of the two variables from the equations, leaving an equation with only one variable. The elimination and substitution methods are both ways to do this.
moving a term	Moving a term of an equation to the other side and changing its sign. Moving the $3$ in $x + 3 = 7$ gives $x = 7 - 3$. It is a shortcut for "subtract the same number from both sides".
coefficient	The number in front of a letter. In $2x$, the coefficient is 2. When no number is written, as in $x$, the coefficient is 1, and the coefficient of $-y$ is $-1$.
properties of equality	An equation stays true when you add, subtract or multiply both sides by the same number, or divide both sides by the same nonzero number. It is like a balance scale that stays level when the same weight is put on both pans. Every step of elimination and substitution is based on these properties.
least common multiple (LCM)	The smallest number that is a multiple of both numbers. In the elimination method it is the target when matching coefficients (for coefficients 2 and 3, match them at 6).
parallel	Two lines that never meet, however far they go. In a system of equations, if the two graphs are parallel there is no intersection, so there is "no solution".
infinitely many solutions	When there is not one single solution but infinitely many. This happens when the two equations describe the same line, so every point on the line is a solution. Such a system is called dependent.
point of intersection	The point where two graphs cross. The solution of a system of equations is the coordinates of the point where the graphs of the two equations (two lines) cross.
check	Substituting the solution into both original equations to make sure the left and right sides really are equal. A system can be checked just by substituting, so making it a habit helps you catch your own mistakes.
Cramer's rule	A formula that calculates the solution directly from the coefficients: $x = \dfrac{ce - bf}{ae - bd},\ y = \dfrac{af - cd}{ae - bd}$. It is named after the Swiss mathematician Gabriel Cramer. It is usually taught with matrices in high school or linear algebra in college, but for two variables you can check it with basic algebra.

Good to know before you start

Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.
If you get stuck, going back to review these topics is the fastest way forward.

Expressions with variables (Grades 6–7)	Knowing that $x$ and $y$ stand in for numbers we do not know yet Reading $2x$ as short for "$2 \times x$", with 2 as its coefficient
Arithmetic with negative numbers (Grade 7)	Being able to multiply with the sign rules, as in $(-1) \times 3 = -3$ Being able to subtract a negative number, as in $5 - (-2) = 7$ (this comes up often when subtracting equations in the elimination method)
Properties of equality and linear equations (Grades 7–8)	Knowing that, like a balance scale, an equation stays true when you add, subtract, multiply or divide (by a nonzero number) both sides by the same number Being able to solve a linear equation such as $2x + 3 = 7$ by moving terms to the other side (and changing their signs)
Coordinates and graphs (Grades 6–8)	Being able to describe the position of a point with an ordered pair such as $(2,\ 3)$ Knowing that the points that satisfy an equation such as $y = 2x + 1$ form a straight line (used to understand no solution and infinitely many solutions on a graph)
Working with fractions (Grades 5–7)	Being able to find common denominators, simplify fractions, and add, subtract, multiply and divide fractions Being able to multiply both sides of an equation such as $\dfrac{1}{2}x + y = 4$ to turn the coefficients into whole numbers

How to calculate it in Excel

Copy the whole table below and paste it into cell A1 in Excel. It works as is.

Table to solve the system

(1) x coefficient a	1
(1) y coefficient b	1
(1) right side c	5
(2) x coefficient d	2
(2) y coefficient e	-1
(2) right side f	1
Denominator ae−bd	=B1B5-B2B4
Solution x ((ce−bf)÷(ae−bd))	=(B3B5-B2B6)/B7
Solution y ((af−cd)÷(ae−bd))	=(B1B6-B3B4)/B7

Table to check for no solution or infinitely many solutions

(1) x coefficient a	1
(1) y coefficient b	1
(1) right side c	3
(2) x coefficient d	2
(2) y coefficient e	2
(2) right side f	6
Denominator ae−bd	=B1B5-B2B4
Numerator ce−bf	=B3B5-B2B6
Numerator af−cd	=B1B6-B3B4
Result	=IF(B7<>0,"One solution",IF(AND(B8=0,B9=0),"Infinitely many solutions","No solution (parallel)"))

Table to check the solution (substitute into both equations)

(1) x coefficient a	1
(1) y coefficient b	1
(1) right side c	5
(2) x coefficient d	2
(2) y coefficient e	-1
(2) right side f	1
Solution x	2
Solution y	3
Left side of (1) ax+by (correct if equal to c)	=B1B7+B2B8
Left side of (2) dx+ey (correct if equal to f)	=B4B7+B5B8

After pasting, the upper rows (the coefficients and right sides) are your inputs and the lower rows are calculated automatically.
The first table solves x + y = 5, 2x − y = 1. The denominator ae−bd is −3, and the solution is x = 2, y = 3. If the denominator is 0, the division gives an error (#DIV/0!), which is the sign that there is no single solution.
The second table is for x + y = 3, 2x + 2y = 6, and the result is "Infinitely many solutions". Change the right side f of (2) to 10 and it becomes "No solution (parallel)".
The third table checks the solution by substituting it into both equations. The left side of (1) is 5 and the left side of (2) is 1, matching their right sides.

How to calculate it in Google Sheets

Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.

Table to solve the system

(1) x coefficient a	1
(1) y coefficient b	1
(1) right side c	5
(2) x coefficient d	2
(2) y coefficient e	-1
(2) right side f	1
Denominator ae−bd	=B1B5-B2B4
Solution x ((ce−bf)÷(ae−bd))	=(B3B5-B2B6)/B7
Solution y ((af−cd)÷(ae−bd))	=(B1B6-B3B4)/B7

Table to check for no solution or infinitely many solutions

(1) x coefficient a	1
(1) y coefficient b	1
(1) right side c	3
(2) x coefficient d	2
(2) y coefficient e	2
(2) right side f	6
Denominator ae−bd	=B1B5-B2B4
Numerator ce−bf	=B3B5-B2B6
Numerator af−cd	=B1B6-B3B4
Result	=IF(B7<>0,"One solution",IF(AND(B8=0,B9=0),"Infinitely many solutions","No solution (parallel)"))

Table to check the solution (substitute into both equations)

(1) x coefficient a	1
(1) y coefficient b	1
(1) right side c	5
(2) x coefficient d	2
(2) y coefficient e	-1
(2) right side f	1
Solution x	2
Solution y	3
Left side of (1) ax+by (correct if equal to c)	=B1B7+B2B8
Left side of (2) dx+ey (correct if equal to f)	=B4B7+B5B8

The same formulas as in Excel work as is. Copy the whole table, paste it into cell A1, and replace the coefficients and right sides with your own numbers.

How to calculate it in Python

from fractions import Fraction

# (1): a1*x + b1*y = c1
# (2): a2*x + b2*y = c2  (a fraction such as 3/4 can be written Fraction(3, 4))
a1, b1, c1 = Fraction(1), Fraction(1), Fraction(5)
a2, b2, c2 = Fraction(2), Fraction(-1), Fraction(1)

denominator = a1 * b2 - a2 * b1   # denominator of the formula, ae − bd
if denominator != 0:
    x = (c1 * b2 - c2 * b1) / denominator
    y = (a1 * c2 - a2 * c1) / denominator
    print(f"Solution: x = {x}, y = {y}")
    print(f"As decimals: x = {float(x)}, y = {float(y)}")
elif c1 * b2 - c2 * b1 == 0 and a1 * c2 - a2 * c1 == 0:
    print("Infinitely many solutions (both equations describe the same line)")
else:
    print("No solution (the two lines are parallel and never meet)")

With the fractions module from the standard library, you can calculate with exact fractions and no decimal rounding errors. This example solves x + y = 5, 2x − y = 1, and running it prints "Solution: x = 2, y = 3". Change the coefficients and right sides and run it.

How to write it in LaTeX and other math languages (copy and paste)

A system of two linear equations and its solution

ax + by = c, dx + ey = f

\begin{cases} ax + by = c \\ dx + ey = f \end{cases}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mo>{</mo>
    <mtable columnalign="left">
      <mtr><mtd>
        <mi>a</mi><mi>x</mi><mo>+</mo><mi>b</mi><mi>y</mi><mo>=</mo><mi>c</mi>
      </mtd></mtr>
      <mtr><mtd>
        <mi>d</mi><mi>x</mi><mo>+</mo><mi>e</mi><mi>y</mi><mo>=</mo><mi>f</mi>
      </mtd></mtr>
    </mtable>
  </mrow>
</math>

{(ax + by = c),(dx + ey = f):}

Solve[{a*x + b*y == c, d*x + e*y == f}, {x, y}]

solve({a*x + b*y = c, d*x + e*y = f}, {x, y});

S = solve(a*x + b*y == c, d*x + e*y == f, [x, y])

ax + by = c, dx + ey = f

Elimination method (match the coefficients and subtract the equations)

(ax + by) − (ax + ey) = c − f

(ax + by) - (ax + ey) = c - f

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mo>(</mo><mi>a</mi><mi>x</mi><mo>+</mo><mi>b</mi><mi>y</mi><mo>)</mo>
    <mo>&#x2212;</mo>
    <mo>(</mo><mi>a</mi><mi>x</mi><mo>+</mo><mi>e</mi><mi>y</mi><mo>)</mo>
    <mo>=</mo>
    <mi>c</mi><mo>&#x2212;</mo><mi>f</mi>
  </mrow>
</math>

(ax + by) - (ax + ey) = c - f

Simplify[(a*x + b*y) - (a*x + e*y) == c - f]

(a*x + b*y) - (a*x + e*y) = c - f;

simplify((a*x + b*y) - (a*x + e*y) == c - f)

(ax + by) − (ax + ey) = c − f

Substitution method (rewrite as $y = $ and substitute)

y = px + q, dx + e(px + q) = f

y = px + q,\quad dx + e(px + q) = f

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>y</mi><mo>=</mo><mi>p</mi><mi>x</mi><mo>+</mo><mi>q</mi>
    <mo>,</mo>
    <mi>d</mi><mi>x</mi><mo>+</mo><mi>e</mi>
    <mo>(</mo><mi>p</mi><mi>x</mi><mo>+</mo><mi>q</mi><mo>)</mo>
    <mo>=</mo><mi>f</mi>
  </mrow>
</math>

y = px + q, dx + e(px + q) = f

Solve[{y == p*x + q, d*x + e*y == f}, {x, y}]

subs(y = p*x + q, d*x + e*y = f);

subs(d*x + e*y == f, y, p*x + q)

y = px + q, dx + e(px + q) = f

A formula for the solution in one step, and how to spot no solution or infinitely many

x = (ce − bf)/(ae − bd), y = (af − cd)/(ae − bd)

x = \frac{ce - bf}{ae - bd},\quad y = \frac{af - cd}{ae - bd}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>x</mi><mo>=</mo>
    <mfrac>
      <mrow><mi>c</mi><mi>e</mi><mo>&#x2212;</mo><mi>b</mi><mi>f</mi></mrow>
      <mrow><mi>a</mi><mi>e</mi><mo>&#x2212;</mo><mi>b</mi><mi>d</mi></mrow>
    </mfrac>
    <mo>,</mo>
    <mi>y</mi><mo>=</mo>
    <mfrac>
      <mrow><mi>a</mi><mi>f</mi><mo>&#x2212;</mo><mi>c</mi><mi>d</mi></mrow>
      <mrow><mi>a</mi><mi>e</mi><mo>&#x2212;</mo><mi>b</mi><mi>d</mi></mrow>
    </mfrac>
  </mrow>
</math>

x = (ce - bf)/(ae - bd), y = (af - cd)/(ae - bd)

{x, y} = {(c*e - b*f)/(a*e - b*d), (a*f - c*d)/(a*e - b*d)}

x := (c*e - b*f)/(a*e - b*d); y := (a*f - c*d)/(a*e - b*d);

x = (c*e - b*f)/(a*e - b*d); y = (a*f - c*d)/(a*e - b*d);

x = (ce − bf)/(ae − bd), y = (af − cd)/(ae − bd)

How to have ChatGPT do the calculation

You are a calculation assistant for math (systems of equations). Do the following calculation by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing).

Solve the system of equations x + y = 5, 2x − y = 1.
Show each of the following:
1. The solution x, y (if they are fractions, give both fractions in lowest terms and decimals)
2. The steps with the elimination method (match the coefficients of one variable, then add or subtract the equations)
3. A check that substitutes the solution into both equations

In Python, calculate exactly with the fractions module from the standard library, and show the formulas you used and the numbers from the execution result. If there is no solution or there are infinitely many, also explain why.

How to Use

1

Enter your numbers

Type the numbers you want to calculate with into the input fields
2

Calculate

Press the "Calculate" button
3

Check the result

The result appears on the spot. The same page also explains the idea behind the calculation and the formula

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\(x,\ y\)	ex, why	The letters for the two numbers you want to find (the unknowns). By custom, letters near the end of the alphabet, \(x,\ y,\ z\), are used for numbers we do not know yet.
\(a,\ b,\ c,\ d,\ e,\ f\)	a, b, c, d, e, f	Letters that stand for the coefficients and the right-hand constants (the known numbers). By custom, letters near the start of the alphabet are used for fixed numbers. On this page, (1) is \(ax + by = c\) and (2) is \(dx + ey = f\).
(1), (2)	equation one, equation two	Numbers given to the equations. They are used to write briefly which operation was done to which equation, as in "(1) × 2" or "(1) + (2)". Textbooks and worked solutions use them too.
\(\{\)	brace	The large brace on the left of a system of equations. It means "these two equations must be true at the same time", showing that they belong together as a set.
\(ae - bd\)	a e minus b d	The value in the denominator of the one-step formula, found by cross-multiplying the coefficients and subtracting. Whether it is 0 decides whether there is exactly one solution. In the language of matrices, it is the determinant of the matrix of coefficients.
\((x,\ y)\)	the pair x, y	A way of writing two values as a set. The solution of a system is a pair of \(x\) and \(y\), so it is written as an ordered pair such as \((2,\ 3)\), the same way as the coordinates of a point on a graph.

Expressions with variables (Grades 6–7)	Knowing that \(x\) and \(y\) stand in for numbers we do not know yet Reading \(2x\) as short for "\(2 \times x\)", with 2 as its coefficient
Arithmetic with negative numbers (Grade 7)	Being able to multiply with the sign rules, as in \((-1) \times 3 = -3\) Being able to subtract a negative number, as in \(5 - (-2) = 7\) (this comes up often when subtracting equations in the elimination method)
Properties of equality and linear equations (Grades 7–8)	Knowing that, like a balance scale, an equation stays true when you add, subtract, multiply or divide (by a nonzero number) both sides by the same number Being able to solve a linear equation such as \(2x + 3 = 7\) by moving terms to the other side (and changing their signs)
Coordinates and graphs (Grades 6–8)	Being able to describe the position of a point with an ordered pair such as \((2,\ 3)\) Knowing that the points that satisfy an equation such as \(y = 2x + 1\) form a straight line (used to understand no solution and infinitely many solutions on a graph)
Working with fractions (Grades 5–7)	Being able to find common denominators, simplify fractions, and add, subtract, multiply and divide fractions Being able to multiply both sides of an equation such as \(\dfrac{1}{2}x + y = 4\) to turn the coefficients into whole numbers

System of Equations Calculator (Elimination and Substitution with Steps)

Table of Contents

What you can do on this page

What is this calculation used for?

Formulas and graphs

Symbols and terms

Symbols

Terms

Good to know before you start

How to calculate it in Excel

How to calculate it in Google Sheets

How to calculate it in Python

How to write it in LaTeX and other math languages (copy and paste)

How to have ChatGPT do the calculation

How to Use

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Related Features

Linear Diophantine Equation Solver (ax + by = c) with Euclidean Algorithm Steps

The same form ax+by=c, but only whole-number solutions are wanted. Find them with the Euclidean algorithm here

Quadratic Formula Calculator (Discriminant and Complex Roots)

Solve quadratic equations with the quadratic formula, with the discriminant and steps

Matrix Calculator (Add, Subtract, Multiply, Transpose, Power)

A system of equations can also be written with matrices. Matrix arithmetic and the basics of rows, columns and entries are here

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