Simultaneous equations calculator with steps

1. Multiply one or both equations so the $x$ (or $y$) coefficients match.
2. Add or subtract the equations to eliminate that letter.
3. Solve the equation that's left.
4. Substitute back to find the other letter.
5. Check both values in the other original equation.

Example: $3x + 2y = 16$ and $x - y = 2$. Multiply the second by 2: $2x - 2y = 4$. Add to the first: $5x = 20$, so $x = 4$. Substitute: $4 - y = 2$, so $y = 2$. Check: $3(4) + 2(2) = 16$. Correct.

Rearrange one equation for a letter and substitute it into the other. From $x - y = 2$, $x = y + 2$. Then $3(y + 2) + 2y = 16$, so $5y = 10$ and $y = 2$. Substitution is easiest when one equation already has a letter on its own, and it's the method you need when one equation is quadratic.

Each linear equation is a straight line. The solution is the point where they cross: $(4, 2)$ in the example.

On the Digital SAT, you can also graph both lines in Desmos and read off the intersection; see Desmos on the SAT.

Put each equation in the form $ax + by = c$ first. For $y = 2x - 3$, rearrange to $-2x + y = -3$ and enter $-2$, $1$ and $-3$. For $x - y = 2$, the $y$ coefficient is $-1$. Fractions like $1/2$ are fine.

Always substitute your solution back into both original equations. For $2x + y = 7$ and $x - y = 2$, the solution $x = 3$, $y = 1$ gives $2(3) + 1 = 7$ and $3 - 1 = 2$, so both work. If only one equation checks out, there's an arithmetic slip in the elimination or substitution step. In exams, this check takes seconds and catches most errors. See systems of equations for more practice.