Linear Equations and Inequalities

Solve linear equations and inequalities - core SAT skill

🎯⭐ INTERACTIVE LESSON

Try the Interactive Version!

Learn step-by-step with practice exercises built right in.

Start Interactive Lesson →

Linear Equations and Inequalities on the SAT

Why This Topic Matters

Linear equations and inequalities appear in roughly 30-35% of SAT Math questions. Mastering this topic is the single highest-impact thing you can do to improve your math score.


What Is a Linear Equation?

A linear equation is any equation where the variable has an exponent of 1 (no x2x^2, x\sqrt{x}, etc.).

Standard form: Ax+By=CAx + By = C

Slope-intercept form: y=mx+by = mx + b

Point-slope form: yy1=m(xx1)y - y_1 = m(x - x_1)


Solving One-Step Equations

Use inverse operations to isolate the variable.

| Operation in Equation | Inverse Operation | |---|---| | Addition (++) | Subtraction (-) | | Subtraction (-) | Addition (++) | | Multiplication (×\times) | Division (÷\div) | | Division (÷\div) | Multiplication (×\times) |

Example: x+7=12x + 7 = 12 x=127=5x = 12 - 7 = 5


Solving Two-Step Equations

Strategy: Undo addition/subtraction first, then undo multiplication/division.

Example: 3x5=163x - 5 = 16

Step 1: Add 5 to both sides: 3x=213x = 21

Step 2: Divide both sides by 3: x=7x = 7


Multi-Step Equations

When equations have variables on both sides, parentheses, or fractions:

Step-by-Step Strategy

  1. Distribute any parentheses
  2. Combine like terms on each side
  3. Move variables to one side
  4. Move constants to the other side
  5. Divide to solve

Example: 4(2x3)=5x+64(2x - 3) = 5x + 6 8x12=5x+68x - 12 = 5x + 6 3x=183x = 18 x=6x = 6


Clearing Fractions

Multiply every term by the LCD (least common denominator).

Example: x3+x4=7\frac{x}{3} + \frac{x}{4} = 7

LCD = 12, so multiply everything by 12: 4x+3x=844x + 3x = 84 7x=847x = 84 x=12x = 12


Special Cases

No Solution

When simplifying leads to a false statement like 5=35 = 3, there is no solution.

Example: 2(x+1)=2x+52(x + 1) = 2x + 5 2x+2=2x+52x + 2 = 2x + 5 2=5(FALSE — no solution)2 = 5 \quad \text{(FALSE — no solution)}

Infinitely Many Solutions

When simplifying leads to a true statement like 3=33 = 3, there are infinitely many solutions.

Example: 3(x2)=3x63(x - 2) = 3x - 6 3x6=3x63x - 6 = 3x - 6 0=0(TRUE — infinite solutions)0 = 0 \quad \text{(TRUE — infinite solutions)}


Linear Inequalities

Inequalities use <<, >>, \leq, \geq instead of ==.

The One Critical Rule

When you multiply or divide by a negative number, FLIP the inequality sign.

Example: 3x>12-3x > 12 x<4(sign flipped!)x < -4 \quad \text{(sign flipped!)}

Compound Inequalities

2<3x+110-2 < 3x + 1 \leq 10 Subtract 1 from all parts: 3<3x9-3 < 3x \leq 9 Divide all parts by 3: 1<x3-1 < x \leq 3


SAT-Specific Question Types

Type 1: "What is the value of x?"

Straightforward solve — isolate the variable.

Type 2: "What is the value of an expression?"

Don't solve for x! Manipulate the equation to find the expression directly.

Example: If 3x+5=173x + 5 = 17, what is 6x+106x + 10? 6x+10=2(3x+5)=2(17)=346x + 10 = 2(3x + 5) = 2(17) = 34

Type 3: "Which value is NOT a solution?"

Test each answer choice — the one that makes the inequality false is your answer.

Type 4: "For what value of aa does the equation have no solution?"

Set up the equation so the variable terms cancel and constants don't match.

Type 5: "For what value of aa does the equation have infinitely many solutions?"

Set up the equation so both sides are completely identical.


Word Problem Translation

| English Phrase | Math Symbol | |---|---| | "is," "equals," "the result is" | == | | "more than," "increased by," "sum" | ++ | | "less than," "decreased by," "difference" | - | | "times," "product," "of" | ×\times | | "per," "quotient," "divided by" | ÷\div | | "at least" | \geq | | "at most" | \leq | | "more than" (comparison) | >> | | "fewer than" (comparison) | << |


Common SAT Mistakes

  1. Forgetting to distribute the negative sign: (x3)=x+3-(x - 3) = -x + 3, NOT x3-x - 3
  2. Not flipping the inequality sign when dividing by a negative
  3. Solving for xx when the question asks for an expression like 2x+12x + 1
  4. Arithmetic errors with fractions — always find the LCD first
  5. Misreading "less than" — "55 less than xx" means x5x - 5, NOT 5x5 - x

Quick Reference: 5-Step SAT Strategy

  1. Read the full question — what exactly are they asking for?
  2. Simplify — distribute, combine like terms
  3. Isolate — get the variable (or expression) alone
  4. Check — plug your answer back in
  5. Match — make sure your answer matches what was asked

Practice Tips

  • On the SAT, you see roughly 6-8 linear equation questions
  • Most can be solved in under 60 seconds with practice
  • Watch for "shortcut" questions where you find an expression, not xx
  • If stuck, try plugging in answer choices (backsolving)

📚 Practice Problems

1Problem 1easy

Question:

If 4x7=214x - 7 = 21, what is the value of xx?

💡 Show Solution

Solution:

4x7=214x - 7 = 21

Add 7 to both sides: 4x=284x = 28

Divide by 4: x=7x = 7

Answer: x=7x = 7

SAT Tip: Always check: 4(7)7=287=214(7) - 7 = 28 - 7 = 21

2Problem 2medium

Question:

Solve the system: {y=3x22x+y=13\begin{cases} y = 3x - 2 \\ 2x + y = 13 \end{cases}

💡 Show Solution

Solution:

Use substitution - plug first equation into second: 2x+(3x2)=132x + (3x - 2) = 13 5x2=135x - 2 = 13 5x=155x = 15 x=3x = 3

Find yy: y=3(3)2=7y = 3(3) - 2 = 7

Answer: (3,7)(3, 7) or x=3,y=7x = 3, y = 7

Check: 2(3)+7=132(3) + 7 = 13

3Problem 3hard

Question:

A phone plan costs \25permonthplusper month plus$0.10pertextmessage.Ifthetotalbillwasper text message. If the total bill was$37$, how many text messages were sent?

💡 Show Solution

Solution:

Let tt = number of text messages

Equation: 25+0.10t=3725 + 0.10t = 37

Subtract 25: 0.10t=120.10t = 12

Divide by 0.10: t=120t = 120

Answer: 120 text messages

SAT Tip: Set up word problems carefully - identify what the variable represents!

4Problem 4easy

Question:

If 5x3=225x - 3 = 22, what is the value of xx?

💡 Show Solution

Step 1: Add 3 to both sides 5x3+3=22+35x - 3 + 3 = 22 + 3 5x=255x = 25

Step 2: Divide both sides by 5 x=255=5x = \frac{25}{5} = 5

Check: 5(5)3=253=225(5) - 3 = 25 - 3 = 22

Answer: x=5x = 5

5Problem 5easy

Question:

If 2x14-2x \geq 14, what is the greatest possible integer value of xx?

💡 Show Solution

Step 1: Divide both sides by 2-2 and flip the inequality x142x \leq \frac{14}{-2} x7x \leq -7

Key: We flip the inequality because we divided by a negative number.

The greatest integer less than or equal to 7-7 is 7-7.

Answer: x=7x = -7

6Problem 6medium

Question:

If 3(2x4)+7=4x+53(2x - 4) + 7 = 4x + 5, what is the value of xx?

💡 Show Solution

Step 1: Distribute the 3 6x12+7=4x+56x - 12 + 7 = 4x + 5

Step 2: Combine like terms on the left 6x5=4x+56x - 5 = 4x + 5

Step 3: Subtract 4x4x from both sides 2x5=52x - 5 = 5

Step 4: Add 5 to both sides 2x=102x = 10

Step 5: Divide by 2 x=5x = 5

Check: 3(2(5)4)+7=3(6)+7=253(2(5) - 4) + 7 = 3(6) + 7 = 25 and 4(5)+5=254(5) + 5 = 25

Answer: x=5x = 5

7Problem 7hard

Question:

If 2x+34x13=2\frac{2x + 3}{4} - \frac{x - 1}{3} = 2, what is the value of 10x510x - 5?

💡 Show Solution

Step 1: Find the LCD of 4 and 3, which is 12. Multiply every term by 12. 122x+3412x13=12212 \cdot \frac{2x + 3}{4} - 12 \cdot \frac{x - 1}{3} = 12 \cdot 2 3(2x+3)4(x1)=243(2x + 3) - 4(x - 1) = 24

Step 2: Distribute 6x+94x+4=246x + 9 - 4x + 4 = 24

Step 3: Combine like terms 2x+13=242x + 13 = 24

Step 4: Solve for xx 2x=112x = 11 x=112x = \frac{11}{2}

Step 5: Find what the question asks: 10x510x - 5 10(112)5=555=5010\left(\frac{11}{2}\right) - 5 = 55 - 5 = 50

Answer: 10x5=5010x - 5 = 50

SAT Tip: The question asks for 10x510x - 5, not xx. Always read carefully!

8Problem 8expert

Question:

For what value of aa does the equation a(x+2)3=4x+2a7a(x + 2) - 3 = 4x + 2a - 7 have no solution?

💡 Show Solution

Step 1: Distribute aa on the left ax+2a3=4x+2a7ax + 2a - 3 = 4x + 2a - 7

Step 2: Subtract 2a2a from both sides ax3=4x7ax - 3 = 4x - 7

Step 3: For NO solution, the xx coefficients must be equal but the constants must differ. Set the xx coefficients equal: a=4a = 4

Step 4: Verify with a=4a = 4: 4x3=4x74x - 3 = 4x - 7 3=7(FALSE — no solution ✓)-3 = -7 \quad \text{(FALSE — no solution ✓)}

Answer: a=4a = 4

Why this works: When a=4a = 4, both sides have 4x4x, but the constants (3-3 vs 7-7) don't match, making the equation impossible.