Digital SAT Prep · Digital SAT Reading, Writing & Math
Algebra (Math)
Linear equations, systems, and linear inequalities in one and two variables.
Seven concepts. SAT Algebra is linear equations, systems, and inequalities in one and two variables — the skills that show up on roughly a third of Math questions, usually as a quick solve or a slope-and-intercept read from a word problem.
- Digital SAT Prep
- Medium level
- 7 concepts
- 5 practice questions
1Solving a system by substitution
A linear system asks for the ordered pair that satisfies every equation at once. On the SAT the fastest route is usually substitution: solve one equation for a variable — pick the one whose coefficient is 1 if you can — and plug that expression into the other. Elimination works too, but substitution is fastest when one equation is already nearly solved for a variable.
Figure. Each line is one equation; the marked point is the only pair on both. Parallel lines never meet — that is the geometric face of no solution.
How it works
- IsolateFrom the easier equation, write one variable in terms of the other — x = y + 1 from x - y = 1 is already done.
- SubstituteReplace that variable in the second equation with the expression, so one equation in one unknown remains.
- Back-substituteSolve for the first unknown, then plug back to recover the second.
A two-equation system
Solve for x: \begin{cases} 2x + 3y = 12 \\ x - y = 1 \end{cases}
- x = y + 1from second equation
- 2(y + 1) + 3y = 125y = 10
- y = 2, then x = y + 1x = 3
Pro tip. Isolate the variable with coefficient 1 first. If both equations need rearranging, compare slopes — parallel lines (same slope, different intercept) mean no solution, and matching multiples mean infinitely many.
For \begin{cases} x + y = 5 \\ x - y = 1 \end{cases}, what is x?
- 2
- 3
- 4
From the second equation, x = y + 1. Substituting into the first gives (y + 1) + y = 5, so y = 2 and x = 3. Adding the two original equations also gives 2x = 6 immediately.
2Slope-intercept form and rate of change
A line in slope-intercept form is y = mx + b, where m is the slope — rise over run, \dfrac{y_2 - y_1}{x_2 - x_1} — and b is the y-intercept. On the SAT, the slope usually names a per-unit rate and the intercept a starting or fixed charge (dollars per mile, texts per day). Read them directly from the equation before doing any algebra.
Figure. Slope 2 means each step right of 1 raises y by 2; the line crosses the y-axis at (0, 2).
Reading a line
- Identify mThe coefficient of x is the slope. Between two points, compute \dfrac{\Delta y}{\Delta x}.
- Identify bThe constant term is the y-intercept — the value when x = 0.
- Write y = mx + bSubstitute the slope and intercept, or use point-slope form if only one point is given.
From two points to an equation
Find the equation of the line through (1, 4) and (3, 8).
- m = \dfrac{8 - 4}{3 - 1}m = 2
- y - 4 = 2(x - 1)point-slope form
- y = 2x + 2slope-intercept form
Pro tip. When answer choices are equations, plug in a given point before expanding — one substitution often eliminates three options.
What is the slope of the line 3y = 6x - 9?
- 2
- 3
- -3
Divide by 3: y = 2x - 3, so m = 2. Reading the coefficient of x before the equation is in slope-intercept form is a common trap.
3Parallel and perpendicular lines
Parallel lines share the same slope: if one line is y = 3x + 1, any parallel line has m = 3. Perpendicular lines have slopes whose product is negative one: a line perpendicular to slope 3 has slope -\dfrac{1}{3}. In standard form Ax + By = C, the slope is -\dfrac{A}{B} — useful when the SAT hides m inside coefficients.
Figure. Both lines rise 3 for every 1 right — same slope, different intercepts, so they never meet.
Matching slopes
- Read mRewrite in y = mx + b or read -\dfrac{A}{B} from Ax + By = C.
- ParallelKeep the same m and solve for a new b through the given point.
- PerpendicularTake the negative reciprocal of m, then fit the point.
A line parallel to a given one
Write the equation of the line through (0, 2) that is parallel to y = 3x + 1.
- Parallel \Rightarrow m = 3same slope as y = 3x + 1
- y = 3x + b through (0, 2)b = 2
- y = 3x + 2answer
Pro tip. For perpendicular, flip the slope and change the sign: slope -\dfrac{2}{5} pairs with \dfrac{5}{2}. A horizontal line (m = 0) is perpendicular to a vertical one (undefined slope).
Which slope is perpendicular to a line with slope \dfrac{4}{3}?
- -\dfrac{4}{3}
- -\dfrac{3}{4}
- \dfrac{3}{4}
Perpendicular slopes multiply to -1: \dfrac{4}{3} \times \left(-\dfrac{3}{4}\right) = -1. The negative reciprocal, not the negative, is the trap.
4Solving a system by elimination
When the coefficients of one variable are opposites or easy multiples, add or subtract the equations to eliminate that variable in one step. Elimination is often faster than substitution when both equations are already in standard form and the coefficients line up — including after you multiply one equation through by a constant.
Figure. Adding the equations removes y because its coefficients are +1 and -1; the crossing point is the shared solution.
Eliminating a variable
- Line upWrite both equations so like terms stack. Multiply through if one variable needs matching coefficients.
- Add or subtractCombine the equations so one variable cancels.
- Back-substituteSolve the remaining equation, then plug into either original.
Add to eliminate y
Solve \begin{cases} 2x + y = 7 \\ x - y = 2 \end{cases}.
- Add the equations3x = 9
- x = 3first unknown
- 3 - y = 2y = 1
Pro tip. If coefficients are already equal, subtract instead of add. When one equation is a multiple of the other, elimination exposes infinitely many solutions immediately.
Adding which pair eliminates x instantly? \begin{cases} 5x + 2y = 11 \\ -5x + y = 4 \end{cases}
- Add the equations
- Subtract the second from the first
- Multiply the first by 2
The x coefficients are opposites (5 and -5), so adding gives 3y = 15 directly. Multiplying first is unnecessary work.
5No solution and infinitely many solutions
A system has no solution when the lines are parallel — same slope, different intercepts — or infinitely many when one equation is a scalar multiple of the other. For a parameter question, match the ratios \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} for parallel slopes, then check whether \dfrac{c_1}{c_2} matches; equal constant ratios mean the same line, not a contradiction.
Figure. Same slope (-\dfrac{3}{2}) but different intercepts — the lines stay the same distance apart and never intersect.
Coefficient ratios
- Match slopesSet \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} to force parallel (or identical) lines.
- Check constantsIf slope ratios match but \dfrac{c_1}{c_2} differs, the lines are parallel — no solution.
- ConfirmIf all three ratios match, every point on one line satisfies the other — infinitely many solutions.
Identifying no solution
For what value of k does \begin{cases} 3x + ky = 7 \\ 6x + 4y = 5 \end{cases} have no solution?
- Parallel lines: \dfrac{3}{6} = \dfrac{k}{4}k = 2
- Check constants: \dfrac{7}{5} \neq \dfrac{1}{2}different intercepts
- k = 2no solution confirmed
Pro tip. Match x- and y-coefficient ratios for parallelism, then confirm the constant ratio differs to rule out infinitely many solutions.
The system \begin{cases} 2x + 4y = 6 \\ x + 2y = 3 \end{cases} has
- No solution
- Exactly one solution
- Infinitely many solutions
The second equation is half of the first — same line twice — so every point on the line satisfies both.
6Linear inequalities
Solve a linear inequality like an equation — same moves on both sides — except multiply or divide by a negative flips the direction: -2x < 6 \Rightarrow x > -3. For two variables, graph the boundary line (dashed if strict < or >) and shade the side that satisfies the inequality; on the SAT the boundary alone is often enough to pick among choices.
Figure. Boundary for y > x - 1: points above this dashed line satisfy the inequality; the region above cannot be shaded in-app.
Solving and graphing
- IsolateCollect x (or y) on one side the same way you would in an equation.
- Flip when neededReverse < and > only when multiplying or dividing both sides by a negative number.
- GraphDraw y = mx + b as the boundary; test a point like (0, 0) to see which side to shade.
Flip the sign
Solve -2x + 4 \ge 10.
- Subtract 4: -2x + 4 - 4 \ge 10 - 4-2x \ge 6
- Divide by -2; flip \ge to \lex \le -3
- Check x = -3-2(-3) + 4 = 10 — boundary included
Pro tip. Plug a test value from your answer back into the original inequality — sign flips are the step most students miss under time pressure.
Which is equivalent to -3x > 9?
- x > -3
- x < -3
- x < 3
Dividing by -3 flips > to <, giving x < -3. Forgetting the flip turns a valid solution set into its opposite.
7Interpreting linear models
Word problems on the SAT often hide a line in plain language: a fixed fee plus a per-unit charge becomes C = mx + b, where m is the rate per unit and b is the starting value. Read the question carefully — some ask for the slope (rate of change), some for the intercept (initial amount), and some for a value after a given number of units.
Figure. In a fee-plus-rate story, b is the starting cost on the axis and m is the per-unit rise. Read the question for which piece it wants.
Building the model
- Name variablesLet x be the independent quantity (time, miles, items) and y the dependent total.
- Find bThe flat fee or starting amount is the intercept — what you pay or have when x = 0.
- Find mEach additional unit adds m dollars (or whatever unit the problem uses).
Membership plus monthly charge
A gym charges a $50 signup fee and $25 per month. Which equation models total cost C after m months?
- Signup fee is the intercept bb = 50
- Monthly charge is the slope (not the month-count variable m)slope = 25
- C = (\text{slope})\,m + bC = 25m + 50
Pro tip. In y = mx + b, subtract 1 from a growth factor only when the base is a multiplier (1.08 means 8\% growth). A plain '+25 per month' is already the slope — do not rewrite it. Do not set the month-count variable equal to the slope.
In P = 120 - 4t, where t is hours, what does -4 represent?
- The starting value
- The amount lost each hour
- The time when P reaches zero
The coefficient of t is the slope: P drops 4 units for each additional hour. The starting value is 120 (when t = 0).
Notes
- Slope-Intercept Form: A line is y = mx + b, where m is the slope (rate of change) and b is the y-intercept; parallel lines share m and perpendicular lines have slopes with product -1.
- Systems of Equations: Solve by substitution or elimination; a system has no solution when the lines are parallel (same slope, different intercept) and infinitely many when the equations are multiples of each other.
- Linear Inequalities: Solve like equations, but flip the inequality sign when multiplying or dividing by a negative, e.g., -2x < 6 \Rightarrow x > -3.
- Interpreting Linear Models: In a word problem y=mx+b, the slope m is the per-unit rate and b is the starting or fixed value.
Formulas
- Slope between two points: m = \dfrac{y_2 - y_1}{x_2 - x_1}
- Slope-intercept form: y = mx + b
- Point-slope form: y - y_1 = m(x - x_1)
- No solution condition for a system: equal slopes, unequal intercepts
- Standard form of a line: Ax + By = C with slope -\dfrac{A}{B}
Exam traps & shortcuts
- For 'no solution' or 'infinitely many solutions' questions, compare the coefficients: set the ratios \frac{a_1}{a_2}=\frac{b_1}{b_2} and check the constant to decide instantly.
- When answer choices are numbers, plug them into the equation and test rather than solving algebraically to save time.
- For a system, add or subtract the equations to eliminate a variable in one step when coefficients line up.
Reference tables
| Form | What it tells you | SAT move |
|---|---|---|
| y = mx + b | Slope m, intercept b | Read rate and starting value in word problems |
| Ax + By = C | Slope -\dfrac{A}{B} | Compare coefficients for parallel / identical systems |
| y - y_1 = m(x - x_1) | Line through (x_1, y_1) with slope m | Build an equation from a point and slope |
| \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} | Same slope (parallel or same line) | Parameter questions for no / infinite solutions |
| Flip on negative divide | -2x < 6 \Rightarrow x > -3 | Only step that differs from solving an equation |
Recap
Night-before pegs for SAT Algebra.
- Slope
- m = \dfrac{\Delta y}{\Delta x} in y = mx + b. Parallel lines share m; perpendicular slopes multiply to -1.
- Systems
- Substitution when one variable is already isolated; elimination when coefficients line up. Parallel slopes + different intercepts = no solution.
- Parameters
- Match \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}, then check constants — equal all three ratios means infinitely many.
- Inequalities
- Same algebra as equations; flip the sign only when multiplying or dividing by a negative.
- Word models
- m is the per-unit rate, b is the flat or starting amount — read the question before computing.
Practise Algebra (Math)
Reading is free and needs no account. Practice, mocks and progress live in the app.
- 5 exam-style questions on this topic, with explanations
- A 4-question practice set that ends the chapter
- Timed mocks scored with the real marking scheme
- Readiness tracked per topic, kept on your device