Systems of Equations Solver
Use this Systems of Equations Calculator to solve supported linear or nonlinear systems and view clear solution steps.
A system can have two equations, three equations, four equations, or more. It can also have a different number of equations and variables, such as three equations in two variables or two equations in three variables.
How to Solve Systems of Equations
Solving a system means finding values that make every equation true at the same time. The best method depends on the equations: some are linear, some are nonlinear, and some are easier to solve with substitution or elimination than with graphing.
Substitution Method
Substitution works well when one equation already has a variable by itself, or when isolating that variable is simple. Consider:
\[ \begin{aligned} y&=7-x\\ 2x+y&=11 \end{aligned} \]
Replace \(y\) in the second equation with \(7-x\):
\[ 2x+(7-x)=11 \]
\[ x=4 \]
Now substitute \(x=4\) into \(y=7-x\):
\[ y=7-4=3 \]
The solution is:
\[ (x,y)=(4,3) \]
This method is often a good choice for two equations in two variables. It can also work in larger systems when one equation gives a clear expression for a variable.
If you only need to isolate a variable in one equation, use the Solve for x Calculator.
Elimination Method
Elimination removes one variable by adding or subtracting equations. You may first multiply one or both equations so that the coefficients of one variable become opposites.
For example:
\[ \begin{aligned} x+y&=8\\ x-y&=2 \end{aligned} \]
Add the equations:
\[ (x+y)+(x-y)=8+2 \]
\[ 2x=10 \]
\[ x=5 \]
Substitute \(x=5\) into \(x+y=8\):
\[ 5+y=8 \]
\[ y=3 \]
Elimination is useful when coefficients line up cleanly. The same idea can be repeated to reduce a larger linear system one variable at a time.
Graphing Method
To solve a system by graphing, graph each equation and find where the graphs intersect. Every intersection represents values that satisfy both equations.
For two linear equations:
One intersection means one solution.
Parallel distinct lines mean no solution.
The same line means infinitely many solutions.
Graphing is most helpful for equations in two variables. It becomes less practical for systems with three or more variables because those systems require three-dimensional or higher-dimensional representations.
Matrix and Gaussian Elimination
A system of linear equations can be organized in an augmented matrix, which places coefficients and constants into rows. Row operations then simplify the system without changing its solution set.
Gaussian elimination is especially useful for three-variable systems, four-equation systems, and larger linear systems. It can show whether the equations lead to one solution, a contradiction, or one or more variables that are free to vary.
For linear systems, the important question is whether the equations are compatible and whether each equation adds new information. Equations that repeat the same relationship do not narrow the answer further, while contradictory equations make the system inconsistent.
Larger and Nonlinear Systems
Systems can contain \(2\) equations with \(2\) variables, \(3\) equations with \(3\) variables, \(4\) equations with \(4\) variables, or many other combinations. More equations than variables can still be consistent if the equations agree. Fewer equations than variables do not automatically mean infinitely many solutions.
A nonlinear system contains at least one equation that is not linear. It may include \(x^2\), \(xy\), powers such as \(y^3\), or other nonlinear expressions.
When a nonlinear system includes a quadratic equation, the Quadratic Equation Calculator can help with the quadratic part of the work.
How to solve a nonlinear system depends on its structure. Substitution, factoring, graphing, elimination, or another algebraic method may be useful. The number of solutions can also depend on the domain being considered.
One Solution, No Solution, or Infinitely Many Solutions
A system has one solution when exactly one value combination satisfies every equation. It has no solution when no value combination can satisfy all equations; this is called an inconsistent system.
A system has infinitely many solutions when infinitely many value combinations satisfy every equation. In a linear system, this often happens when one equation is dependent on the others, meaning it adds no new independent condition.
Nonlinear systems can behave differently. Depending on the equations and the domain, they may have no real solutions, one solution, several distinct solutions, or infinitely many solutions.
Checking a Solution
Always substitute the final values into every original equation. A value that works in only one equation is not a solution to the system.
For example, check \((x,y)=(5,3)\) in:
\[ \begin{aligned} x+y&=8\\ x-y&=2 \end{aligned} \]
\[ 5+3=8 \]
\[ 5-3=2 \]
Both equations are true, so \((5,3)\) is the solution.
Concept Explainer and Worked Examples
A system of equations is a group of equations involving shared variables. One equation describes one relationship, while a system requires several relationships to hold at the same time.
For example:
\[ \begin{aligned} x+y&=9\\ x-y&=3 \end{aligned} \]
The values that satisfy both equations are:
\[ x=6,\qquad y=3 \]
Checking confirms the result:
\[ 6+3=9 \]
\[ 6-3=3 \]
A two-variable solution is often written as an ordered pair, such as \((6,3)\). A three-variable solution is written as an ordered triple, such as \((2,-1,4)\). The solution set contains every value combination that satisfies every equation.
Three-Variable Elimination Example
Solve:
\[ \begin{aligned} 2x+y+z&=11\\ x-2y+3z&=9\\ 3x+y-2z&=4 \end{aligned} \]
Eliminate \(y\) by subtracting the first equation from the third:
\[ (3x+y-2z)-(2x+y+z)=4-11 \]
\[ x-3z=-7 \]
This gives one equation involving only \(x\) and \(z\).
Eliminate \(y\) again by adding twice the first equation to the second:
\[ 2(2x+y+z)+(x-2y+3z)=22+9 \]
\[ 5x+5z=31 \]
\[ x+z=\frac{31}{5} \]
Now solve the resulting two-variable system.
Solve \(x-3z=-7\) for \(x\):
\[ x=3z-7 \]
Substitute this into \(x+z=\frac{31}{5}\).
Find \(z\):
\[ (3z-7)+z=\frac{31}{5} \]
\[ 4z=\frac{66}{5} \]
\[ z=\frac{33}{10} \]
Find \(x\):
\[ x=3\left(\frac{33}{10}\right)-7 \]
\[ x=\frac{29}{10} \]
Substitute into the first original equation to find \(y\):
\[ 2\left(\frac{29}{10}\right)+y+\frac{33}{10}=11 \]
\[ \frac{58}{10}+y+\frac{33}{10}=\frac{110}{10} \]
\[ y=\frac{19}{10} \]
The solution is:
\[ \left(x,y,z\right)=\left(\frac{29}{10},\frac{19}{10},\frac{33}{10}\right) \]
Check every original equation:
\[ 2\left(\frac{29}{10}\right)+\frac{19}{10}+\frac{33}{10}=11 \]
\[ \frac{29}{10}-2\left(\frac{19}{10}\right)+3\left(\frac{33}{10}\right)=9 \]
\[ 3\left(\frac{29}{10}\right)+\frac{19}{10}-2\left(\frac{33}{10}\right)=4 \]
All three equations are true, so the solution checks.
Nonlinear System Example
Consider:
\[ \begin{aligned} y&=x^2\\ y&=2x+3 \end{aligned} \]
Because both expressions equal \(y\), set them equal:
\[ x^2=2x+3 \]
\[ x^2-2x-3=0 \]
\[ (x-3)(x+1)=0 \]
\[ x=3\quad\text{or}\quad x=-1 \]
Substitute each value into \(y=2x+3\):
\[ (x,y)=(3,9) \]
\[ (x,y)=(-1,1) \]
The two solutions are the two points where the parabola \(y=x^2\) intersects the line \(y=2x+3\).