Systems of Two Linear Equations

A system of two linear equations is a pair of equations that must be true at the same time. Its solution is the point $(x, y)$ where the two lines cross. The SAT asks you to solve systems, to build them from word problems (two unknowns, two facts), and to decide when a system has no solution or infinitely many solutions.

• Substitution works best when one equation already says $y = \ldots$ or $x = \ldots$.
• Elimination works best when the coefficients line up: add or subtract the equations so one variable cancels. Multiply an equation first if needed.
• A word problem with a count and a total usually gives one "how many" equation and one "how much" equation: $a + c = 150$ and $12a + 7c = 1400$.
• If a question asks for $x + y$ or $3x - y$, look for a way to get it directly by adding or subtracting the equations.
• Two lines meet once, never (parallel), or everywhere (the same line).

For $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:

In plain words: no solution means same slope, different intercept. Infinitely many means one equation is a multiple of the other.

Solve the system $4x + 3y = 26$ and $2x - 3y = 4$.

1. The $y$-terms are opposites, so add the equations: $6x = 30$, and $x = 5$.
2. Substitute into the first: $20 + 3y = 26$, so $3y = 6$ and $y = 2$.

Check in the second: $2(5) - 3(2) = 10 - 6 = 4$. Correct. The solution is $(5, 2)$.

In the system $6x - 9y = 15$ and $4x + ky = c$, $k$ and $c$ are constants. If the system has infinitely many solutions, what is the value of $k + c$?

1. Infinitely many solutions means the second equation is a multiple of the first.
2. Compare the $x$-coefficients: $4 = 6 \cdot \frac{2}{3}$, so the multiplier is $\frac{2}{3}$.
3. Apply it to the rest: $k = -9 \cdot \frac{2}{3} = -6$ and $c = 15 \cdot \frac{2}{3} = 10$.
4. So $k + c = -6 + 10 = 4$.

Check: $\frac{2}{3}(6x - 9y) = 4x - 6y$ and $\frac{2}{3}(15) = 10$, so the second equation is $4x - 6y = 10$. That's the same line, so it checks out.

• Stopping after one variable. If the question asks for $y$ or $x + y$, finish the job.
• Subtracting only some terms. When you subtract equations, subtract every term, including the constants: $(3x + y) - (3x - y) = 2y$.
• Mixing up the solution-count conditions. Matching slopes alone gives no solution; you need the constants to match too for infinitely many.
• Setting up a word problem with the totals swapped. The count equation uses $1$ as each coefficient; the money equation uses the prices.
• Multiplying only one side when scaling an equation for elimination.