Solve systems of two linear equations in two variables algebraically by substitution and by elimination, and identify systems with no solution or infinitely many solutions (Ohio A-REI.6, A-REI.5).

What this topic is asking

Ohio standard A-REI.6 asks you to solve a system of two linear equations in two variables exactly, and A-REI.5 explains why elimination is valid. A solution is an ordered pair $(x, y)$ that makes both equations true at once, the point where the two lines cross. The two exact methods are substitution and elimination. Systems sit in the Expressions and Equations reporting category and can appear on either part of the test.

The substitution method

Substitution works best when one equation already gives a variable alone, or one is easy to isolate.

The elimination method

Elimination adds the equations so one variable disappears. Scale first if needed so the coefficients of one variable are opposites.

When no variable cancels directly, multiply one or both equations first. To solve $2x + 3y = 7$ and $x - y = -1$, multiply the second by $2$ to get $2x - 2y = -2$, then subtract to cancel $x$.

Choosing a method

No solution and infinitely many

Just as with one-variable equations, the variables can cancel.

How Ohio examines this topic

• Equation response. Type the ordered pair $(x, y)$, or the value of a single variable the item asks for.
• Multiple choice. Pick the solution, or decide how many solutions the system has.
• Drag and drop. Order the solving steps, or match a system to "one / none / infinitely many."

Because many items are scored by exact match, write the solution as an ordered pair and check it in both equations, not just one.

Why a checked answer must satisfy both equations

A common error is to verify a solution in only one equation. The defining feature of a system is that the ordered pair lies on both lines, so it must satisfy both equations at once. A point can easily satisfy one and fail the other, which is exactly what happens if you make a sign slip during back-substitution. The two-equation check is therefore not optional polish, it is what proves the point is the intersection. On a no-calculator part this ten-second substitution into each equation is the surest way to catch an arithmetic mistake before it costs the point.

Why elimination is allowed

A-REI.5 justifies elimination: replacing one equation with the sum of that equation and a multiple of the other produces an equivalent system with the same solution set. Adding equal quantities to equal quantities keeps a true statement true, so any pair that satisfied the original pair of equations still satisfies the new pair, and none is gained or lost. That is why you may scale an equation and add, the solution set is preserved. Understanding this is what lets you trust elimination and explain it on a reasoning item rather than treating it as a trick.

Try this

Q1. Solve $y = 3x$ and $x + y = 8$. [2 points]

• Cue. $x + 3x = 8$, so $x = 2$ and $y = 6$; $(2, 6)$.

Q2. Solve $x + y = 10$ and $x - y = 2$ by elimination. [2 points]

• Cue. Add: $2x = 12$, $x = 6$; then $y = 4$; $(6, 4)$.