Solving Systems by Elimination

To solve a system by elimination, you must align the equations and manipulate the coefficients so that one variable "disappears" when the equations are combined.

1. The Standard Form

Most elimination problems on the SAT will be presented in Standard Form: $Ax + By = C$ $Dx + Ey = F$ Where $A, B, C, D, E,$ and $F$ are constants. Unlike the Slope-Intercept form ($y = mx + b$), the Standard Form keeps the variables on one side, making it easy to see which coefficients can be matched.

2. Creating Additive Inverses

The goal of elimination is to create additive inverses for one of the variables. Additive inverses are numbers that sum to zero, such as $5$ and $-5$.

If the equations are:

1. $3x + 2y = 10$
2. $5x - 2y = 6$

The $y$-coefficients ($2$ and $-2$) are already additive inverses. Adding the equations immediately eliminates $y$: $(3x + 5x) + (2y - 2y) = 10 + 6$ $8x = 16 \implies x = 2$

3. The Multiplication Step

On medium and hard SAT questions, the coefficients will rarely match perfectly. You must multiply one or both equations by a constant to create matching coefficients.

Scenario A: Multiplying one equation If you have: $x + 3y = 7$ $2x - 5y = 3$ Multiply the first equation by $-2$ to eliminate $x$: $-2(x + 3y = 7) \implies -2x - 6y = -14$ Now, add this to the second equation to eliminate $x$.

Scenario B: Multiplying both equations If you have: $3x + 4y = 10$ $2x + 3y = 7$ To eliminate $x$, find the Least Common Multiple (LCM) of $3$ and $2$, which is $6$. Multiply the first by $2$ and the second by $-3$: $2(3x + 4y = 10) \implies 6x + 8y = 20$ $-3(2x + 3y = 7) \implies -6x - 9y = -21$

4. Solving for the Second Variable

Once you find the value of the first variable (e.g., $x = 2$), you must back-substitute that value into one of the original equations to find the second variable ($y$).

5. Special Cases: No Solution vs. Infinite Solutions

The SAT frequently tests your understanding of systems that don't have a single $(x, y)$ solution.

• No Solution: When you attempt elimination and both variables cancel out, leaving a false statement (e.g., $0 = 5$), the lines are parallel and never intersect.
• Infinitely Many Solutions: When both variables and the constant cancel out, leaving a true statement (e.g., $0 = 0$), the two equations represent the same line.

Reference Sheet Note

Important: None of the formulas for solving systems of equations are provided on the SAT Reference Sheet. You must memorize the process of elimination and the conditions for "no solution" and "infinite solutions."

1. Look for the "Shortcut" Expression

The SAT loves to ask for $x + y$, $x - y$, or $10x + 10y$. Before you solve for $x$ and $y$ individually, look at the system. Can you add or subtract the equations to get the requested expression immediately? Example: If the question asks for $x + y$ and the system is $3x + 2y = 10$ and $x + 2y = 6$, subtracting the second from the first gives $2x = 4$, but adding them gives $4x + 4y = 16$. Dividing by $4$ gives $x + y = 4$ instantly.

2. Desmos vs. Hand Calculation

On the Digital SAT, you have access to the Desmos graphing calculator.

• Use Desmos when: The coefficients are large decimals or fractions, or when the question is a simple "find the solution" problem.
• Solve by hand when: The question contains constants like $k$ or $a$ (e.g., "For what value of $k$ does the system have no solution?"). Desmos cannot always solve for unknown constants as easily as you can algebraically.

3. Align Your Variables

The SAT will sometimes try to trick you by swapping the order of variables: $3x + 2y = 10$ $4y + 2x = 12$ Before performing elimination, rewrite the second equation as $2x + 4y = 12$ so the $x$ and $y$ terms line up vertically.

4. The "Subtraction Trap"

Instead of subtracting one equation from another, I recommend multiplying one equation by $-1$ and then adding them. Students frequently make sign errors when subtracting a whole expression (e.g., forgetting to distribute the negative to the constant on the right side).