Solution
- $x$: —
- $y$: —
Step-by-step
Explanation:
This tool solves a system of two linear equations with two variables and shows the work step by step using Cramer’s rule. Enter the coefficients for $a_1x+b_1y=c_1,\quad a_2x+b_2y=c_2$ and the solver will compute the determinants $D$, $D_x$, and $D_y$ to determine whether the system has a unique solution, no solution, or infinitely many solutions. For more algebra tools, explore Algebra.
Solve a 2x2 system in the form:
A 2x2 system is two linear equations in the same two variables, usually $x$ and $y$. Geometrically, each equation is a line. The solution is the intersection point of the two lines. Depending on the lines, a system can have one intersection (unique solution), no intersection (no solution), or infinitely many intersections (the same line).
This solver uses Cramer’s rule. It computes three determinants: the main determinant $D$, plus $D_x$ and $D_y$ where the constants replace one column at a time. These values tell you both the solution type and, when it exists, the exact values of $x$ and $y$.
Cramer’s rule gives a clean classification:
If you only need to solve a single equation in one variable, use the Linear Equation Solver. If your equation includes an $x^2$ term, use the Quadratic Solver.
Coefficients can be integers or decimals. The tool formats results using the configured decimal precision. If the determinants are extremely close to zero (because of rounding), the solver treats them as zero to keep the classification stable.
Solve a system with one intersection point.
Solve: $$\begin{cases} 2x + 3y = 13\\ x - y = 1 \end{cases}$$
Compute determinants: $$D=(2)(-1)-(1)(3)=-2-3=-5$$ $$D_x=(13)(-1)-(1)(3)=-13-3=-16$$ $$D_y=(2)(1)-(1)(13)=2-13=-11$$
Since $D\ne 0$: $$x=\frac{D_x}{D}=\frac{-16}{-5}=3.2,\quad y=\frac{D_y}{D}=\frac{-11}{-5}=2.2$$
The equations are multiples of each other, so they represent the same line.
Solve: $$\begin{cases} x + 2y = 4\\ 2x + 4y = 8 \end{cases}$$
Compute determinants: $$D=(1)(4)-(2)(2)=4-4=0$$ $$D_x=(4)(4)-(8)(2)=16-16=0$$ $$D_y=(1)(8)-(2)(4)=8-8=0$$
Since $D=0$ and $D_x=0$ and $D_y=0$, the system has infinitely many solutions.
The lines are parallel, so they never intersect.
Solve: $$\begin{cases} x + y = 2\\ 2x + 2y = 5 \end{cases}$$
Compute determinants: $$D=(1)(2)-(2)(1)=2-2=0$$ $$D_x=(2)(2)-(5)(1)=4-5=-1$$ $$D_y=(1)(5)-(2)(2)=5-4=1$$
Since $D=0$ but at least one of $D_x$ or $D_y$ is nonzero, the system has no solution.
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Quick answers about determinants, Cramer's rule, and solution types.
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