📚 New SAT Math Key Concepts and Practice | 新SAT数学知识点精析与配套练习
The redesigned SAT Math test focuses on a core set of math skills that are essential for college and career readiness. It covers topics from algebra, problem solving and data analysis, advanced math, and additional topics such as geometry and trigonometry. This article provides a comprehensive review of key concepts, illustrated with examples, and concludes with a set of practice questions to help you prepare effectively.
新版SAT数学测试围绕大学和职业准备所必需的核心数学技能展开，涵盖代数、问题解决与数据分析、高等数学以及几何、三角等附加主题。本文对这些关键知识点进行系统梳理，并辅以例题精讲，最后提供配套练习与解析，助你高效备考。
1. Overview of New SAT Math | 新SAT数学概览
The SAT Math section consists of two portions: a 25-minute No Calculator section with 20 questions, and a 55-minute Calculator section with 38 questions. The total raw score is converted to a scaled score ranging from 200 to 800. Question types include multiple-choice and student-produced responses (grid-ins). The test emphasizes fluency, conceptual understanding, and real-world applications.
SAT数学部分分为两部分：25分钟不可用计算器的部分包含20道题，55分钟可用计算器的部分包含38道题。原始分数会转换为200-800的标准分。题型包括选择题和填空题（grid-in）。考试注重运算流畅性、概念理解以及在实际情境中的应用能力。
2. Heart of Algebra – Linear Equations and Inequalities | 代数核心——线性方程与不等式
Linear equations in one variable can be solved by isolating the variable using inverse operations. For instance, to solve 2x + 5 = 13, subtract 5 from both sides to get 2x = 8, then divide by 2 to find x = 4. When facing inequalities, remember that multiplying or dividing by a negative number reverses the inequality sign. For example, -3x ≤ 9 becomes x ≥ -3 after dividing by -3.
一元线性方程可通过逆运算分离变量求解。例如，解 2x + 5 = 13，两边减去5得 2x = 8，再除以2得到 x = 4。处理不等式时，切记乘以或除以负数要反转不等号方向。例如 -3x ≤ 9，除以 -3 后变为 x ≥ -3。
Interpreting linear functions is crucial. A linear function y = mx + b has slope m and y-intercept b. The slope represents the rate of change, while the y-intercept is the starting value. In word problems, identify the independent and dependent variables and translate the situation into a linear expression or equation.
解读线性函数十分关键。线性函数 y = mx + b 中，斜率 m 表示变化率，y截距 b 是初始值。在文字应用题中，要识别自变量和因变量，并将情境转化为线性表达式或方程。
| Example: A taxi charges a base fare of $3.00 plus $0.50 per mile. The linear model is C = 0.5m + 3, where C is cost and m is miles. |
| 例题：出租车起步价3美元，每英里0.5美元。线性模型为 C = 0.5m + 3，C为费用，m为英里数。 |
3. Heart of Algebra – Systems of Equations | 代数核心——方程组
A system of two linear equations can be solved by substitution, elimination, or graphing. The solution is the ordered pair (x, y) that satisfies both equations. When the system has no solution, the lines are parallel; when it has infinitely many solutions, the equations represent the same line. The SAT often asks you to determine the number of solutions from the coefficients.
二元一次方程组可通过代入法、消元法或图像法求解。解是满足两个方程的有序对 (x, y)。当方程组无解时，两条直线平行；当有无穷多解时，两方程代表同一直线。SAT常要求根据系数判断解的个数。
For example, consider: 2x + y = 7 and x – y = 2. Adding the equations eliminates y: 3x = 9, so x = 3. Substituting back gives y = 1. The solution is (3, 1).
例如，方程组：2x + y = 7，x – y = 2。两式相加消去 y 得 3x = 9，故 x = 3。回代得 y = 1。解为 (3, 1)。
4. Heart of Algebra – Functions and Graphs | 代数核心——函数与图像
A function f(x) assigns exactly one output for each input. The graph of a linear function is a straight line. Key features include intercepts, slope, domain, and range. The SAT may test transformations such as translations: f(x + h) shifts the graph left by h units, while f(x) + k shifts it up by k units.
函数 f(x) 为每个输入值指定唯一输出值。线性函数的图像是直线。关键特征包括截距、斜率、定义域和值域。SAT可能考察图像变换，如平移：f(x + h) 表示图像向左平移 h 个单位，f(x) + k 表示向上平移 k 个单位。
Understanding function notation is also key. If f(x) = 3x – 5, then f(2) = 3(2) – 5 = 1. The graph of y = f(x) represents all points (x, f(x)).
理解函数符号同样重要。若 f(x) = 3x – 5，则 f(2) = 3(2) – 5 = 1。y = f(x) 的图像表示所有点 (x, f(x)) 的集合。
5. Problem Solving and Data Analysis – Ratios, Proportions, and Percentages | 问题解决与数据分析——比、比例与百分比
Ratios compare two quantities, while proportions state that two ratios are equal. Direct variation y = kx indicates that as x increases, y increases proportionally. Percent change is calculated as (new – original) / original × 100%. The SAT frequently embeds these concepts in multi-step problems with tables or graphs.
比用于比较两个量，比例则表明两个比相等。正比例关系 y = kx 表示 y 随 x 等比例增加。百分比变化计算公式为（新值 – 原值）/ 原值 × 100%。SAT常在多步骤问题中结合表格或图表考查这些概念。
Example: A shirt originally costs $40, now on sale for $30. The percent decrease is (40 – 30)/40 × 100% = 25%.
例题：一件衬衫原价40美元，现售价30美元。降价百分比为 (40 – 30)/40 × 100% = 25%。
6. Problem Solving and Data Analysis – Data Interpretation and Statistics | 问题解决与数据分析——数据解读与统计
Data interpretation involves reading bar graphs, line graphs, pie charts, and two-way tables. Measures of center (mean, median, mode) and spread (range, standard deviation) help summarize data. The mean is the average, the median is the middle value when ordered, and the mode is the most frequent value. The SAT also tests basic probability: probability = favorable outcomes / total outcomes.
数据解读要求读懂条形图、折线图、饼图和双向表。集中量数（平均数、中位数、众数）和离散量数（极差、标准差）用于概括数据。平均数是算术平均值，中位数是排序后的中间值，众数是出现频率最高的值。SAT还考查基本概率：概率 = 有利结果数 / 总结果数。
For example, given the data set {2, 3, 3, 5, 7}, the mean is (2+3+3+5+7)/5 = 4, the median is 3, and the mode is 3.
例如，数据集 {2, 3, 3, 5, 7}，平均数 (2+3+3+5+7)/5 = 4，中位数是3，众数是3。
7. Passport to Advanced Math – Quadratic Equations and Functions | 高等数学——二次方程与函数
Quadratic equations have the form ax² + bx + c = 0, with a ≠ 0. They can be solved by factoring, completing the square, or using the quadratic formula: x = [-b ± √(b² – 4ac)] / (2a). The discriminant b² – 4ac determines the nature of the roots: positive discriminant → two real roots; zero → one real root; negative → two complex roots.
二次方程的形式为 ax² + bx + c = 0，且 a ≠ 0。可以通过因式分解、配方法或求根公式求解：x = [-b ± √(b² – 4ac)] / (2a)。判别式 b² – 4ac 决定根的性质：判别式为正 → 两个实根；为零 → 一个实根；为负 → 两个复根。
Quadratic functions f(x) = a(x – h)² + k graph as parabolas with vertex (h, k). If a > 0, the parabola opens upward; if a < 0, it opens downward. The axis of symmetry is x = h.
二次函数 f(x) = a(x – h)² + k 的图像是抛物线，顶点为 (h, k)。若 a > 0，抛物线开口向上；若 a < 0，开口向下。对称轴为 x = h。
8. Passport to Advanced Math – Polynomials and Rational Expressions | 高等数学——多项式与有理式
Polynomials are expressions made up of variables and coefficients using addition, subtraction, and multiplication. Operations include addition, subtraction, multiplication, and factoring. Rational expressions are ratios of polynomials. To simplify a rational expression, factor numerator and denominator and cancel common factors, noting restrictions on the variable.
多项式是由变量、系数通过加减乘运算构成的表达式。运算包括加、减、乘和因式分解。有理式是多项式的比。化简有理式时，需对分子分母因式分解并约去公因式，同时注明变量的限制条件。
For example, simplify (x² – 4) / (x – 2). Factor: (x – 2)(x + 2) / (x – 2) = x + 2, provided x ≠ 2.
例如，化简 (x² – 4) / (x – 2)。因式分解：(x – 2)(x + 2) / (x – 2) = x + 2，其中 x ≠ 2。
9. Passport to Advanced Math – Radicals and Exponents | 高等数学——根式与指数
Exponent rules are fundamental: xᵃ · xᵇ = xᵃ⁺ᵇ, (xᵃ)ᵇ = xᵃᵇ, and x⁻ⁿ = 1/xⁿ. Radical expressions involve roots; the principal square root is √, and higher roots use the radical sign with an index. Rational exponents connect radicals and powers: x^(m/n) = ⁿ√(xᵐ). Simplifying radicals often involves extracting perfect squares.
指数法则是基础：xᵃ · xᵇ = xᵃ⁺ᵇ，(xᵃ)ᵇ = xᵃᵇ，以及 x⁻ⁿ = 1/xⁿ。根式涉及方根；算术平方根用 √ 表示，高次方根用带指数的根号。有理指数将根式与幂联系起来：x^(m/n) = ⁿ√(xᵐ)。化简根式常需提取完全平方因子。
For instance, √(48) = √(16·3) = 4√3. And 8^(2/3) = ³√(8²) = ³√64 = 4.
例如，√(48) = √(16·3) = 4√3。8^(2/3) = ³√(8²) = ³√64 = 4。
10. Additional Topics in Math – Geometry and Trigonometry | 附加数学——几何与三角
Geometry on the SAT includes lines, angles, triangles, circles, and volume of solids. Triangle properties such as the Pythagorean theorem (a² + b² = c² for right triangles) and special right triangles (30°-60°-90° and 45°-45°-90°) are frequently tested. Circle concepts include circumference = 2πr, area = πr², and arc length.
SAT几何涵盖直线、角度、三角形、圆以及立体体积。三角形性质如勾股定理（直角三角形中 a² + b² = c²）和特殊直角三角形（30°-60°-90° 与 45°-45°-90°）常考。圆的概念包括周长 = 2πr，面积 = πr²，以及弧长。
Basic trigonometry focuses on sine, cosine, and tangent in right triangles: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent. The SAT also tests the complementary angle relationship: sin(θ) = cos(90° – θ).
基础三角学关注直角三角形中的正弦、余弦、正切：sin θ = 对边/斜边，cos θ = 邻边/斜边，tan θ = 对边/邻边。SAT还考查余角关系：sin(θ) = cos(90° – θ)。
11. Additional Topics – Complex Numbers | 附加主题——复数
Complex numbers are of the form a + bi, where i² = -1. Operations with complex numbers follow the same rules as binomials. To add, combine like terms: (3 + 2i) + (1 – 4i) = 4 – 2i. To multiply, use FOIL and replace i² with -1: (2 + i)(3 – i) = 6 – 2i + 3i – i² = 6 + i + 1 = 7 + i.
复数的形式为 a + bi，其中 i² = -1。复数的运算遵循与二项式相同的规则。加法合并同类项：(3 + 2i) + (1 – 4i) = 4 – 2i。乘法使用 FOIL 并将 i² 替换为 -1：(2 + i)(3 – i) = 6 – 2i + 3i – i² = 6 + i + 1 = 7 + i。
The complex conjugate of a + bi is a – bi, and multiplying a complex number by its conjugate yields a real number: (a + bi)(a – bi) = a² + b². This is useful for simplifying division of complex numbers.
复数 a + bi 的共轭为 a – bi，复数与其共轭相乘得到一个实数：(a + bi)(a – bi) = a² + b²。这对于简化复数除法非常有用。
12. Practice Questions with Explanations | 配套练习与解析
Question 1: If 4x – 7 = 2x + 9, what is the value of x?
问题1：若 4x – 7 = 2x + 9，求 x 的值。
Solution: Subtract 2x from both sides: 2x – 7 = 9. Add 7: 2x = 16. Divide by 2: x = 8.
解析：两边减 2x 得 2x – 7 = 9，加7得 2x = 16，除以2得 x = 8。
Question 2: A line passes through points (1,3) and (4,9). Find its equation in slope-intercept form.
问题2：一条直线经过点 (1,3) 和 (4,9)，求其斜截式方程。
Solution: Slope m = (9 – 3)/(4 – 1) = 6/3 = 2. Using point (1,3): y – 3 = 2(x – 1) → y = 2x + 1.
解析：斜率 m = (9 – 3)/(4 – 1) = 6/3 = 2。用点 (1,3)：y – 3 = 2(x – 1) → y = 2x + 1。
Question 3: Simplify (x² – 9)/(x + 3), where x ≠ -3.
问题3：化简 (x² – 9)/(x + 3)，其中 x ≠ -3。
Solution: Factor numerator: (x – 3)(x + 3)/(x + 3) = x – 3, for x ≠ -3.
解析：分子因式分解为 (x – 3)(x + 3)，约去 (x + 3) 得 x – 3，x ≠ -3。
Question 4: In triangle ABC, angle C is 90°, AB = 13, BC = 5. Find sin A.
问题4：在三角形ABC中，角C为90°，AB = 13，BC = 5。求 sin A。
Solution: AB is hypotenuse = 13, BC is opposite to angle A = 5. sin A = opposite/hypotenuse = 5/13.
解析：AB为斜边 = 13，BC为角A的对边 = 5。sin A = 对边/斜边 = 5/13。
Question 5: If f(x) = 2x² – 3x + 1, find the value(s) of x for which f(x) = 0.
问题5：若 f(x) = 2x² – 3x + 1，求使得 f(x) = 0 的 x 值。
Solution: Set 2x² – 3x + 1 = 0. Factor: (2x – 1)(x – 1) = 0 → x = 1/2 or x = 1.
解析：解 2x² – 3x + 1 = 0，因式分解 (2x – 1)(x – 1) = 0，得 x = 1/2 或 x = 1。
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