Linear Equations in One Variable

A linear equation in one variable has a single unknown, usually $x$, that is never squared, square-rooted, or in a denominator. You solve it by undoing operations until $x$ is alone. The SAT also asks you to build the equation from a short story, and to decide whether an equation has one solution, no solution, or infinitely many solutions, often with a missing constant you must find.

• Whatever you do to one side, do to the other. The goal is always $x = \text{number}$.
• Distribute first, then combine like terms on each side, then move all $x$-terms to one side.
• Clear fractions early by multiplying every term on both sides by the least common denominator. Decimals work the same way: multiply by $10$ or $100$.
• In word problems, a one-time amount is a constant and a per-unit amount multiplies the variable: "a $20 fee plus $8 per hour" becomes $20 + 8h$.
• After simplifying, every linear equation looks like $ax + b = cx + d$. Comparing the coefficients tells you how many solutions there are.

• Distributing a negative flips every sign inside: $-2(x - 3) = -2x + 6$.
• To clear $\frac{x}{4}$ and $\frac{x}{6}$, multiply by $12$, not by $24$ (either works, but the smaller number keeps the arithmetic easy).

Solve $7x + 4 = 3x + 24$.

1. Subtract $3x$ from both sides: $4x + 4 = 24$.
2. Subtract $4$: $4x = 20$.
3. Divide by $4$: $x = 5$.

Check: $7(5) + 4 = 39$ and $3(5) + 24 = 39$. Correct.

In the equation $\frac{1}{2}(8x - 6) + kx = 3 - 2x$, $k$ is a constant. If the equation has no solution, what is the value of $k$?

1. Distribute: $4x - 3 + kx = 3 - 2x$.
2. Group the left side: $(4 + k)x - 3 = -2x + 3$.
3. No solution means the $x$-coefficients match but the constants do not. Set $4 + k = -2$, so $k = -6$.
4. Confirm the constants differ: $-3 \ne 3$. Good.

Check: with $k = -6$ the equation becomes $-2x - 3 = -2x + 3$, which simplifies to $-3 = 3$. That is never true, so there is no solution, as required.

If the question had said infinitely many solutions, you would need the constants to match too. That is impossible here, so no value of $k$ would work.

• Distributing to only the first term. $4(x + 6)$ is $4x + 24$, not $4x + 6$.
• Dropping the sign with a subtraction. In $10 - 2(x - 4)$, the result is $10 - 2x + 8$, not $10 - 2x - 8$.
• Clearing fractions on one term only. Multiply every term, including whole numbers: $\frac{x}{3} + 2 = 5$ times $3$ is $x + 6 = 15$.
• Mixing up "no solution" and "infinitely many." Same $x$-coefficient plus different constants means none; everything the same means infinitely many.
• Answering the wrong quantity. If the question asks for $2x - 1$ or for the number of hours, compute that after you find $x$.