Systems of Linear Equations

What a System of Linear Equations Actually Asks

A system of linear equations is two (or more) linear equations that share the same variables. On the SAT Math section, a system almost always looks like two equations in two unknowns, and a "solution" is an ordered pair $(x, y)$ that makes both equations true at the same time. Graphically, each linear equation is a straight line, and the solution is the point where the two lines cross.

That single idea — the solution is where the lines intersect — is the key to every systems question, whether the test asks you to solve the system, count how many solutions exist, or find a value of an unknown coefficient.

The Two Reliable Methods

You do not need to graph on the SAT. Two algebraic methods handle everything, and choosing the faster one for a given problem is a real time-saver.

Method 1 — Substitution

Substitution works best when one variable is already isolated (or is easy to isolate).

Solve the system:

$y = 2x - 1$ $3x + y = 14$

Step 1. The first equation already gives $y$ in terms of $x$, so substitute $2x - 1$ in place of $y$ in the second equation:

$3x + (2x - 1) = 14$

Step 2. Combine like terms and solve the resulting one-variable equation:

$5x - 1 = 14 \;\Rightarrow\; 5x = 15 \;\Rightarrow\; x = 3$

Step 3. Back-substitute $x = 3$ into the isolated equation to get the other coordinate:

$y = 2(3) - 1 = 5$

The solution is $(3, 5)$. Always check it in the other equation: $3(3) + 5 = 14$. ✓

Method 2 — Elimination

Elimination shines when both equations are in the standard $ax + by = c$ form and no variable is isolated.

Solve the system:

$4x + 3y = 10$ $2x - 3y = 8$

Step 1. The $y$-terms are $+3y$ and $-3y$ — already opposites. Add the two equations so the $y$-terms cancel:

$(4x + 3y) + (2x - 3y) = 10 + 8 \;\Rightarrow\; 6x = 18 \;\Rightarrow\; x = 3$

Step 2. Substitute $x = 3$ into either original equation:

$2(3) - 3y = 8 \;\Rightarrow\; 6 - 3y = 8 \;\Rightarrow\; -3y = 2 \;\Rightarrow\; y = -\tfrac{2}{3}$

The solution is $\left(3, -\tfrac{2}{3}\right)$. When the coefficients are not already opposites, multiply one (or both) equations by a constant first so that one variable's coefficients become opposites — then add.

How Many Solutions? The Question the SAT Loves

A frequent SAT twist gives you a system with an unknown constant and asks how many solutions it has. Compare the equations after writing both in slope-intercept form $y = mx + b$:

Relationship between the lines	Slopes and intercepts	Number of solutions
Lines cross once	Different slopes	Exactly one solution
Same line	Same slope and same intercept	Infinitely many solutions
Parallel, never touching	Same slope, different intercept	No solution

Worked example. For what value of $c$ does the system below have no solution?

$y = 3x + 4$ $6x - 2y = c$

Rewrite the second equation in slope-intercept form: $-2y = -6x + c$, so $y = 3x - \tfrac{c}{2}$. Both lines now have slope $3$, so they are parallel. They will never intersect — giving no solution — as long as the intercepts differ: $-\tfrac{c}{2} \neq 4$, i.e. $c \neq -8$. If $c = -8$ the two equations describe the same line and there would be infinitely many solutions instead. So the system has no solution for every value except $c = -8$.

Choosing a Method Under Time Pressure

• A variable is already isolated (like $y = \dots$) → use substitution.
• Both equations are in $ax + by = c$ form → use elimination.
• The question asks "how many solutions" → don't solve at all; compare slopes and intercepts.

Common Mistakes to Avoid

• Solving for only one variable and stopping — the answer is an ordered pair.
• Distributing a negative sign incorrectly during substitution.
• Forgetting to multiply an entire equation (both sides) when scaling for elimination.
• Confusing "infinitely many" (same line) with "no solution" (parallel lines).