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Foundations Reference

Systems of Equations and Inequalities

A system asks for values that satisfy multiple rules at once. Equations usually meet at points; inequalities usually overlap in regions.

Fact Table

System TypeUseResult
Two linear equationsGraph, substitution, or elimination.One point, no solution, or infinitely many solutions.
One equation already solved for a variableSubstitution.Replace that variable in the other equation.
Opposite or matching coefficientsElimination.Add or subtract equations to remove a variable.
Two-variable inequalitiesBoundary lines and shading.The overlap is the solution region.
Linear system in matrix form$AX=B$.If $A^{-1}$ exists, then $X=A^{-1}B$.

Content Formulas

Substitution
$$y=mx+b\quad \text{goes into the other equation}$$
Elimination
$$\text{add equations to cancel one variable}$$
Matrix Form
$$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}e\\f\end{bmatrix}$$
Inequality Boundary
$$<,\ >:\ \text{dashed}\qquad \le,\ \ge:\ \text{solid}$$
For inequality systems, test a point after drawing each boundary. Shade the side where the test point makes the inequality true.

Classic Examples

Substitution

Solve $y=2x+1$ and $x+y=10$.

System MoveSubstitute the expression for $y$ into the other equation.
$$x+(2x+1)=10$$$$3x=9$$$$x=3$$$$y=2(3)+1=7$$$$(3,7)$$

Elimination

Solve $3x+2y=16$ and $3x-2y=8$.

System MoveAdd the equations because the $y$ terms are opposites.
$$(3x+2y)+(3x-2y)=16+8$$$$6x=24$$$$x=4$$$$3(4)+2y=16$$$$y=2$$

No Solution

Classify $y=2x+3$ and $y=2x-5$.

System MoveSame slope with different intercepts means parallel lines.
$$m_1=2,\quad m_2=2$$$$b_1=3,\quad b_2=-5$$$$\text{No solution.}$$

Infinitely Many Solutions

Classify $2x+4y=8$ and $x+2y=4$.

System MoveIf one equation is a multiple of the other, they are the same line.
$$2(x+2y=4)\quad \Longrightarrow \quad 2x+4y=8$$$$\text{Infinitely many solutions.}$$

Inequality Region

Describe the solution to $y\ge x-1$ and $y< -2x+5$.

System MoveUse a solid boundary for $\ge$, a dashed boundary for $<$, and keep the overlap.
$$y=x-1:\ \text{solid line, shade above}$$$$y=-2x+5:\ \text{dashed line, shade below}$$$$\text{Solution: the overlapping region.}$$