Linear Equations | SATMath800 Algebra 01

SAT Math · Algebra · Area 01

Linear Equations

Build, rearrange, solve, and interpret linear equations in one variable. The goal is not just to get x—it is to understand what the equation says and why each step works.

Observe → Identify → Decompose → Reason → Solve → Verify

The big idea

A linear equation states that two expressions have the same value. Solving means finding the value of the variable that makes that statement true.

\[3x+5=26\]

There is one unknown quantity, x. Our job is to isolate it without changing the truth of the equation.

Start$3x+5=26$

Undo +5$3x=21$

Undo ×3$x=7$

Verify$3*(7)+5=26$

Key idea: Every operation you perform to one side of an equation must also be performed to the other side.

Read an equation before solving it

On the SAT, the equation itself may contain information the question wants you to interpret.

\[7x-12=30\]

left side$7x-12$an expression

right side$30$a constant

Variable

$x$ represents an unknown quantity.

Coefficient

In $7x$, the coefficient of $x$ is $7$.

Constant

$-12$ and $30$ are constants because they do not contain the variable.

Solution

The value of $x$ that makes the entire equation true.

An equation is a balance

Think of the equals sign as a balance point. Legal algebraic moves preserve that balance.

Whatever keeps the two sides equal is a valid equation-preserving move.

Undo operations in reverse

If the equation contains addition, subtraction, multiplication, or division, use the inverse operation to isolate the variable.

\[2x+3=11\]

Blue marks the operation applied to both sides.

\[2x+3\color{#1769aa}{-3}=11\color{#1769aa}{-3}\]

\[2x=8\]

\[\frac{2x}{\color{#1769aa}{2}}=\frac{8}{\color{#1769aa}{2}}\]

\[x=4\]

Subtract 3 from both sides first. Then divide both sides by 2. The colored operation shows exactly what is being done to each side.

Do not “move” numbers across the equals sign. Instead, perform the same operation on both sides. That habit prevents many sign errors.

Choose the cleanest solving strategy

The SAT rewards fluency. You do not need to follow one rigid sequence if the structure gives you a faster path.

One-step equations

\[x-9=14\]

Undo subtraction by adding 9 to both sides.

\[x-9\color{#1769aa}{+9}=14\color{#1769aa}{+9}\]

\[x=23\]

Two-step equations

\[4x+7=31\]

Undo addition first, then undo multiplication.

\[4x+7\color{#1769aa}{-7}=31\color{#1769aa}{-7}\]

\[4x=24\]

\[\frac{4x}{\color{#1769aa}{4}}=\frac{24}{\color{#1769aa}{4}}\]

\[x=6\]

Distribute when needed

\[3*(x-4)=18\quad\Rightarrow\quad 3x-12=18\]

Then continue solving, or recognize an easier structure if one is available.

Variables on both sides

\[5x+2=3x+18\]

Subtract the same variable term from both sides, then isolate the variable.

\[5x\color{#1769aa}{-3x}+2=3x\color{#1769aa}{-3x}+18\]

\[2x+2=18\]

\[2x+2\color{#1769aa}{-2}=18\color{#1769aa}{-2}\]

\[2x=16\]

\[\frac{2x}{\color{#1769aa}{2}}=\frac{16}{\color{#1769aa}{2}}\]

\[x=8\]

Structure shortcut: Before expanding everything, look for cancellation, common factors, or matching expressions. Sometimes the fastest solution is the one that changes the equation the least.

Turn a situation into an equation

A large part of SAT Algebra is deciding what the symbols mean before doing the algebra. The variable does not have to be x.

Example

Temperature conversion

The Celsius-to-Fahrenheit relationship is

\[F=\frac{9}{5}C+32\]

Suppose the temperature is 95°F. What is the corresponding temperature in Celsius?

Here, C is the unknown. The equation is already a model, so our job is to solve it.

\[95=\frac{9}{5}C+32\]

\[95\color{#1769aa}{-32}=\frac{9}{5}C+32\color{#1769aa}{-32}\]

\[63=\frac{9}{5}C\]

\[\frac{\color{#1769aa}{5}}{\color{#1769aa}{9}}*63=\frac{\color{#1769aa}{5}}{\color{#1769aa}{9}}*\frac{9}{5}C\]

\[C=35\]

The important idea is not the letter used for the variable. The same solving principles work for C, F, p, t, or any other symbol.

A variable can represent temperature, length, weight, time, or any other quantity.

Fractions and decimals are still linear

Do not let awkward-looking numbers hide a simple linear equation.

Fractions

\[\frac{3}{4}x+5=17\]

Subtract 5 from both sides.

\[\frac{3}{4}x+5\color{#1769aa}{-5}=17\color{#1769aa}{-5}\]

\[\frac{3}{4}x=12\]

\[\frac{\color{#1769aa}{4}}{\color{#1769aa}{3}}*\frac{3}{4}x=\frac{\color{#1769aa}{4}}{\color{#1769aa}{3}}*12\]

\[x=16\]

Decimals

\[0.6x+4.8=12\]

Subtract 4.8, then divide both sides by 0.6.

\[0.6x+4.8\color{#1769aa}{-4.8}=12\color{#1769aa}{-4.8}\]

\[0.6x=7.2\]

\[\frac{0.6x}{\color{#1769aa}{0.6}}=\frac{7.2}{\color{#1769aa}{0.6}}\]

\[x=12\]

SAT habit: If decimals are making the arithmetic uncomfortable, consider multiplying the entire equation by a power of 10 before solving.

One solution, no solution, or infinitely many?

The SAT can ask you to recognize what happens when the variable terms disappear.

One solution

\[2x+5=17\]

The variable remains after simplification, so one value of $x$ works.

No solution

\[4x+3=4x-5\]

Subtract $4x$: $3=-5$. This is impossible.

Infinitely many

\[3*(x+4)=3x+12\]

Expand: $3x+12=3x+12$. Every value of $x$ works.

Common SAT traps

Most errors are not caused by difficult algebra. They come from changing the meaning of the equation.

Sign errors

When subtracting a term, apply the subtraction to both sides. Do not simply “move” the term and hope the sign changes correctly.

Distributing incorrectly

For $a(b+c)$, multiply both terms: $ab+ac$.

Ignoring the context

If $x$ represents hours, people, dollars, or miles, interpret the final value using those units.

Doing too much work

Look for structure before expanding. A matching expression may disappear immediately.

Not checking

Substitute the answer into the original equation, especially when the problem involves negatives or fractions.

Confusing expression and equation

An expression such as $3x+5$ has no equals sign. An equation makes a claim of equality.

Original SAT-style practice

These questions are newly written for SATMath800. None of the equations or contexts below repeats an example from the lesson.

What is the value of $x$ in the equation $7x-11=45$?

A) 6
B) 8
C) 7
D) 9

Solution: Add 11 to both sides, then divide both sides by 7.

\[7x-11\color{#1769aa}{+11}=45\color{#1769aa}{+11}\]

\[7x=56\]

\[\frac{7x}{\color{#1769aa}{7}}=\frac{56}{\color{#1769aa}{7}}\]

\[x=8\]

Answer: B

If $4*(x+6)=44$, what is the value of $x$?

A) 5
B) 16
C) 11
D) 17

Solution: Divide both sides by 4, then subtract 6 from both sides.

\[\frac{4*(x+6)}{\color{#1769aa}{4}}=\frac{44}{\color{#1769aa}{4}}\]

\[x+6=11\]

\[x+6\color{#1769aa}{-6}=11\color{#1769aa}{-6}\]

\[x=5\]

Answer: A

What is the value of $x$ in the equation $9-3x=-12$?

A) −3
B) 3
C) 7
D) −7

Solution: Subtract 9 from both sides, then divide by $-3$.

\[9-3x\color{#1769aa}{-9}=-12\color{#1769aa}{-9}\]

\[-3x=-21\]

\[\frac{-3x}{\color{#1769aa}{-3}}=\frac{-21}{\color{#1769aa}{-3}}\]

\[x=7\]

Answer: C

What is the value of $x$ in $\frac{2}{5}x+6=18$?

A) 25
B) 30
C) 24
D) 35

Solution: Subtract 6, then multiply both sides by $\frac{5}{2}$.

\[\frac{2}{5}x+6\color{#1769aa}{-6}=18\color{#1769aa}{-6}\]

\[\frac{2}{5}x=12\]

\[\frac{\color{#1769aa}{5}}{\color{#1769aa}{2}}*\frac{2}{5}x=\frac{\color{#1769aa}{5}}{\color{#1769aa}{2}}*12\]

\[x=30\]

Answer: B

A student solves $0.75x-3=9$. What value of $x$ should the student obtain?

A) 12
B) 14
C) 16
D) 18

Solution: Add 3 to both sides, then divide both sides by 0.75.

\[0.75x-3\color{#1769aa}{+3}=9\color{#1769aa}{+3}\]

\[0.75x=12\]

\[\frac{0.75x}{\color{#1769aa}{0.75}}=\frac{12}{\color{#1769aa}{0.75}}\]

\[x=16\]

Answer: C

What is the solution to $7x+5=3x+33$?

A) 6
B) 7
C) 8
D) 9

Solution: Subtract $3x$ from both sides, subtract 5 from both sides, then divide by 4.

\[7x\color{#1769aa}{-3x}+5=3x\color{#1769aa}{-3x}+33\]

\[4x+5=33\]

\[4x+5\color{#1769aa}{-5}=33\color{#1769aa}{-5}\]

\[4x=28\]

\[\frac{4x}{\color{#1769aa}{4}}=\frac{28}{\color{#1769aa}{4}}\]

\[x=7\]

Answer: B

A bike rental service charges a $15 equipment fee plus $6 for each hour of rental. If the total charge is $57, how many hours was the bike rented?

A) 6
B) 8
C) 9
D) 7

Solution: Let $h$ be the number of hours. The equation is $15+6h=57$.

\[15+6h\color{#1769aa}{-15}=57\color{#1769aa}{-15}\]

\[6h=42\]

\[\frac{6h}{\color{#1769aa}{6}}=\frac{42}{\color{#1769aa}{6}}\]

\[h=7\]

Answer: D

Which value of $k$ makes $5x+k=2x+21$ have the solution $x=4$?

A) 6
B) 9
C) 12
D) 7

Solution: Substitute the given solution $x=4$, then solve for $k$.

\[5*(4)+k=2*(4)+21\]

\[20+k=29\]

\[20+k\color{#1769aa}{-20}=29\color{#1769aa}{-20}\]

\[k=9\]

Answer: B

How many solutions does the equation $6x+14=6x+14$ have?

A) 0
B) 1
C) 6
D) Infinitely many

Solution: Subtract $6x$ from both sides. The variable disappears and the resulting statement is always true.

\[6x+14\color{#1769aa}{-6x}=6x+14\color{#1769aa}{-6x}\]

\[14=14\]

Answer: D

How many solutions does the equation $8x-3=8x+5$ have?

A) 1
B) 0
C) 8
D) Infinitely many

Solution: Subtract $8x$ from both sides. The resulting statement is impossible.

\[8x-3\color{#1769aa}{-8x}=8x+5\color{#1769aa}{-8x}\]

\[-3=5\]

Answer: B

Check yourself: Before moving on, make sure you can solve a linear equation, interpret a variable in context, work with fractions or decimals, and recognize one-solution, no-solution, and infinitely-many-solution cases.

Mastery check

Before moving to Linear Functions, make sure these skills feel automatic.

Identify the variable, coefficient, constant, terms, and solution.
Solve one-step and multi-step linear equations accurately.
Use the distributive property without losing signs.
Solve equations with variables on both sides.
Work comfortably with fractions and decimals.
Translate a real-world situation into a linear equation.
Interpret the solution in the original context.
Recognize one solution, no solution, and infinitely many solutions.
Use algebraic structure to avoid unnecessary work.
Verify a solution in the original equation.

Next: Linear Functions. Once you can solve a linear equation, the next step is to understand how a linear rule describes a relationship between an input and an output.

Latest Updates

Categories

The big idea

Read an equation before solving it

Variable

Coefficient

Constant

Solution

An equation is a balance

Undo operations in reverse

Choose the cleanest solving strategy

One-step equations

Two-step equations

Distribute when needed

Variables on both sides

Turn a situation into an equation

Temperature conversion

Fractions and decimals are still linear

Fractions

Decimals

One solution, no solution, or infinitely many?

One solution

No solution

Infinitely many

Common SAT traps

Sign errors

Distributing incorrectly

Ignoring the context

Doing too much work

Not checking

Confusing expression and equation

Original SAT-style practice

What is the value of \(x\) in the equation \(7x-11=45\)?

If \(4*(x+6)=44\), what is the value of \(x\)?

What is the value of \(x\) in the equation \(9-3x=-12\)?

What is the value of \(x\) in \(\frac{2}{5}x+6=18\)?

A student solves \(0.75x-3=9\). What value of \(x\) should the student obtain?

What is the solution to \(7x+5=3x+33\)?

A bike rental service charges a $15 equipment fee plus $6 for each hour of rental. If the total charge is $57, how many hours was the bike rented?

Which value of \(k\) makes \(5x+k=2x+21\) have the solution \(x=4\)?

How many solutions does the equation \(6x+14=6x+14\) have?

How many solutions does the equation \(8x-3=8x+5\) have?

Mastery check

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