Systems of Equations - ACT Math

A system with no solutions. Parallel lines that never meet have no common solution.

Infinite solutions. Same line represented twice has all points in common.

$x = 7, y = 3$. Add equations to get $2x = 8$, then substitute back.

$x = 2, y = 1$. Multiply first by 3, add to second to eliminate $y$.

Two or more linear equations involving the same variables. Multiple equations with shared variables that must be solved together.

Substitution method. First equation is already solved for $x$.

$x = 4, y = 2$. Add equations to get $2x = 8$, so $x = 4$; then $y = 2$.

$(1, 3)$. Set equations equal: $2x + 1 = -x + 4$, solve for $x = 1$.

Exactly one solution. Independent lines intersect at a single point.

$x = 2, y = 1$. Add equations to eliminate $y$ and solve for $x$.

$x = 3, y = 2$. Add equations to eliminate $y$, getting $5x = 15$.

Lines coincide on the graph. Same line appears twice, creating overlapping graphs.

Equations are multiples of each other. One equation is a scalar multiple of the other.

Equations simplify to the same equation. Both equations reduce to identical forms after simplification.

The lines are parallel and never intersect. Different slopes create parallel lines with no intersection.

Dependent system. Second equation is twice the first, creating infinite solutions.

The point where the graphs intersect. Lines cross at exactly one coordinate pair.

Elimination method. Coefficients don't align easily for substitution.

Solve one equation for a variable; substitute into the other. Isolate one variable and replace it in the second equation.

Combine equations to eliminate one variable, then solve. Add or subtract equations to cancel out one variable.

$x = 2, y = 1.5$. Multiply second by 2, add to first to eliminate $y$.

$(2, -1)$. Set equal: $3x - 7 = -2x + 3$, solve to get $x = 2$.

$x = 6, y = 4$. Add equations to eliminate $y$, giving $2x = 12$.

Dependent system. Second equation is double the first equation.

$(x, y) = (6, 0)$. From second: $x = 6 - 2y$, substitute into first equation.

$(x, y) = (1, 1)$. Substitute $x = y + 1$ from second equation into first, solve for $y$.

$(x, y) = (3, 2)$. Add equations to eliminate $y$, giving $2x = 6$, so $x = 3$.

Dependent system. Equations are equivalent and represent the same line.

$(x, y) = (2, 1)$. Subtract second from first to eliminate $y$, solve for $x$.