Solving Linear Equations on the SAT
Where I would start
Undo what was done to $x$, in reverse order, and do the same thing to both sides every time. Clear parentheses and fractions first, gather the $x$ terms on one side, and then divide.
The SAT adds two twists. Sometimes there is a constant like $k$, which you find by putting the given solution into the equation. And sometimes you are asked whether an equation has one solution, no solution or infinitely many.
Linear equations are the most common question type in SAT Math. Most of them are quick, so when students lose points here it is usually a small slip: a sign dropped while distributing, a fraction that was not cleared from every term, or an answer for $x$ when the question asked for something else.
The question types
| You see | Do this |
|---|---|
| Parentheses | Distribute first, including any minus sign: $-2(x - 4) = -2x + 8$. |
| Fractions | Multiply every term on both sides by the least common denominator. |
| $x$ on both sides | Move all $x$ terms to one side and all numbers to the other. |
| A constant $k$ and a given solution | Substitute the solution for $x$, then solve for $k$. |
| "How many solutions?" | Simplify to $ax + b = cx + d$. If $a \neq c$: one solution. If $a = c$ and $b \neq d$: no solution. If $a = c$ and $b = d$: infinitely many. |
| "What is the value of $2x - 3$?" | Look for that expression inside the equation before solving for $x$. It is often a single step away. |
How I work through them
Example 1: finding a constant
The equation $5(x + k) - 5 = 20$ has the solution $x = 3$. What is $k$?
- Substitute $x = 3$: $5(3 + k) - 5 = 20$.
- Add $5$: $5(3 + k) = 25$. Divide by $5$: $3 + k = 5$.
- $k = 2$.
Example 2: clearing fractions
Solve $\dfrac{x}{3} + \dfrac{x}{4} = 7$.
- The least common denominator is $12$. Multiply every term by $12$: $4x + 3x = 84$.
- $7x = 84$, so $x = 12$.
- Check: $\dfrac{12}{3} + \dfrac{12}{4} = 4 + 3 = 7$.
Example 3: no solution or infinitely many
For what value of $k$ does $3(2x - 1) = 6x + k$ have infinitely many solutions?
- Distribute: $6x - 3 = 6x + k$.
- The $x$ terms already match on both sides. If the constants match too, every $x$ works.
- So $k = -3$. Any other value of $k$ gives no solution at all.
Example 4: solving for an expression
If $4x - 6 = 10$, what is the value of $2x - 3$?
- Notice that $4x - 6$ is exactly $2(2x - 3)$.
- So $2(2x - 3) = 10$, and $2x - 3 = 5$.
- Solving for $x$ first ($x = 4$) also works, but it takes longer and gives you one more chance to slip.
Small slips that cost points
- Distributing a minus sign to only the first term. $-3(x - 2)$ is $-3x + 6$, not $-3x - 6$.
- Clearing fractions from only some terms. When you multiply by the denominator, every term on both sides gets multiplied, including whole numbers.
- Answering with $x$ when the question asks for something else. Read the last line of the question again before you choose an answer.
- Mixing up "no solution" and "infinitely many". Same $x$ terms with different constants means no solution. Same $x$ terms and same constants means infinitely many.
- Sign slips when moving terms. Moving $+7$ across the equals sign makes it $-7$. Doing the same operation to both sides avoids this.
Practice questions
Do these the way you would on test day, then open each solution to check your method, not only your answer.
- A$3$
- B$5$
- C$\dfrac{29}{3}$
- D$15$
Show solution
Show solution
- A$1$
- B$5$
- C$3$
- D$2$
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- A$4$
- B$8$
- C$5$
- D$13$
Show solution
- A$\dfrac{18}{3 - k}$
- B$\dfrac{26}{k - 3}$
- C$-\dfrac{18}{k + 3}$
- D$\dfrac{18}{k - 3}$
Show solution
- AExactly one
- BInfinitely many
- CZero
- DExactly two
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- A$-3$
- B$0$
- C$3$
- D$5$
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Show solution
Find the slips you keep making
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