Reference sheet

SAT Math reference sheet

The definitions, formulas and conditions of the SAT Math course on one page, unit by unit, with links to the lessons that explain each topic in full.

For SAT Math

This sheet follows the six units of the course in order. Each unit starts with the big idea in a few sentences, then gives the definitions, formulas and rules you will look up most often, with the conditions they need. When you want the full explanation or more examples, follow the lesson links under each topic.

A reference sheet of common formulas is available during the test, but it cannot tell you which formula a question needs or what its letters mean there. This page collects the rules the lessons rely on, whether or not they appear on the test’s own sheet, so learn the ones you use often well enough to recognize them quickly.

Unless a question says otherwise, every variable and expression on the SAT is a real number, a function’s inputs must give real outputs, and figures lie in a plane and are drawn to scale. On this page angles are in degrees unless they are marked in radians, and a formula holds wherever both sides are defined unless a condition says otherwise.

Essential Skills

Nearly every SAT question uses fractions, signs, ratios, percents or units somewhere along the way. This unit makes those steps reliable, then reads expressions precisely and turns a situation described in words into an equation.

Fractions, signs and order of operations

Work in this order. Grouping symbols first, then powers and roots, then multiplication and division from left to right, and last addition and subtraction from left to right. A fraction bar groups its whole numerator and its whole denominator. An exponent applies only to the symbol or group just before it, so (−3)2=9(-3)^2 = 9 while −32=−9-3^2 = -9.

RuleCondition
ab+cd=ad+bcbd\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad + bc}{bd}b≠0b \ne 0 and d≠0d \ne 0
ab⋅cd=acbd\dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{ac}{bd}b≠0b \ne 0 and d≠0d \ne 0
ab÷cd=ab⋅dc\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \cdot \dfrac{d}{c}b≠0b \ne 0, c≠0c \ne 0 and d≠0d \ne 0
−ab=−ab=a−b-\dfrac{a}{b} = \dfrac{-a}{b} = \dfrac{a}{-b}b≠0b \ne 0

Division by zero is undefined, while 0b=0\frac{0}{b} = 0 for b≠0b \ne 0. A product or quotient of two numbers with the same sign is positive, and with opposite signs it is negative. Subtracting a negative number adds its size, as in 5−(−2)=75 - (-2) = 7. For example

23+34=8+912=1712.\frac{2}{3} + \frac{3}{4} = \frac{8 + 9}{12} = \frac{17}{12}.

Lessons Fractions, signs and order of operations

Ratios, proportions and rates

A ratio a:ba : b compares two quantities, and multiplying both parts by the same positive number gives an equivalent ratio. Read whether a ratio compares part to part or part to whole. If boys and girls in a class are in the ratio 3:53 : 5, then the boys are 38\frac{3}{8} of the class, not 35\frac{3}{5}. With 4040 students that means 1515 boys and 2525 girls.

A proportion states that two ratios are equal. For b≠0b \ne 0 and d≠0d \ne 0,

ab=cd  ⟺  ad=bc.\frac{a}{b} = \frac{c}{d} \iff ad = bc.

So 38=x56\frac{3}{8} = \frac{x}{56} gives 8x=1688x = 168 and x=21x = 21. Two quantities are directly proportional when y=kxy = kx for a constant k≠0k \ne 0, so the quotient yx\frac{y}{x} stays equal to kk for every x≠0x \ne 0 and the graph is a line through the origin. A unit rate is the amount per one unit of the second quantity, such as $12 for 44 pounds, which is $3 per pound.

Lessons Ratios and proportions

Percents and percent change

A percent is a number out of 100100, so p%p\% of a quantity QQ is p100⋅Q\frac{p}{100} \cdot Q. The percent change from an original value a≠0a \ne 0 to a new value bb is

b−aa⋅100%,\frac{b - a}{a} \cdot 100\%,

always divided by the original value. A change of rr percent is a multiplication. An increase by r%r\% multiplies by 1+r1001 + \frac{r}{100}, and a decrease by r%r\% multiplies by 1−r1001 - \frac{r}{100}.

SituationCalculation
From 8080 to 929292−8080=0.15\dfrac{92 - 80}{80} = 0.15, an increase of 15%15\%
Up 20%20\%, then down 20%20\%1.2⋅0.8=0.961.2 \cdot 0.8 = 0.96, a net decrease of 4%4\%
Price $150 after a 25%25\% increaseoriginal 1501.25=120\dfrac{150}{1.25} = 120

Successive percent changes multiply, so equal increases and decreases do not cancel. To undo a change, divide by its multiplier rather than subtracting the same percent. A percentage point is a difference of two percentages. A rise from 40%40\% to 46%46\% is 66 percentage points, but a relative increase of 15%15\%.

Lessons Percentages and percent change

Units and conversions

Multiply by conversion factors written as fractions equal to 11, arranged so the unwanted units cancel. For area square the length factor, and for volume cube it.

LengthAreaVolume
1 ft=12 in1 \text{ ft} = 12 \text{ in}1 ft2=144 in21 \text{ ft}^2 = 144 \text{ in}^21 ft3=1728 in31 \text{ ft}^3 = 1728 \text{ in}^3
1 yd=3 ft1 \text{ yd} = 3 \text{ ft}1 yd2=9 ft21 \text{ yd}^2 = 9 \text{ ft}^21 yd3=27 ft31 \text{ yd}^3 = 27 \text{ ft}^3
1 m=100 cm1 \text{ m} = 100 \text{ cm}1 m2=10,000 cm21 \text{ m}^2 = 10{,}000 \text{ cm}^21 m3=1,000,000 cm31 \text{ m}^3 = 1{,}000{,}000 \text{ cm}^3

The denominator of a rate names the reference unit. With 11 mile =5280= 5280 feet and 11 hour =3600= 3600 seconds,

60 mih=60⋅5280 ft3600 s=88 fts.\begin{aligned} 60 \ \frac{\text{mi}}{\text{h}} &= \frac{60 \cdot 5280 \text{ ft}}{3600 \text{ s}} \\[0.4em] &= 88 \ \frac{\text{ft}}{\text{s}}. \end{aligned}

Keep exact values until the last step and round only as the question asks.

Lessons Units, rates and conversions

Expressions and word problems

An expression such as 3x−73x - 7 has no equals sign and is simplified or evaluated, while an equation such as 3x−7=113x - 7 = 11 makes a claim that is solved. Terms are joined by addition or subtraction, factors by multiplication, and a coefficient is the number multiplying a variable. When you substitute, put the value in parentheses, so x=−2x = -2 in x2−3xx^2 - 3x gives (−2)2−3(−2)=10(-2)^2 - 3(-2) = 10.

To write an equation from words, name each unknown with its unit, then translate whole relationships rather than single words.

WordsMeaning
at least 1010≥10\ge 10
at most 1010≤10\le 10
more than 1010, fewer than 1010>10> 10, <10< 10
55 less than xxx−5x - 5, not 5−x5 - x
$4 per item for nn items4n4n dollars

For example, a gym charges a $25 fee plus $12 per month, so mm months cost 25+12m25 + 12m dollars, and a budget of at most $150 means 25+12m≤15025 + 12m \le 150, so m≤12512≈10.4m \le \frac{125}{12} \approx 10.4 and at most 1010 whole months. Check that an answer fits the context, since a count must be a whole number and a length cannot be negative.

Lessons Expressions and substitution, Word problems into equations

Algebra

Linear equations and linear functions are the largest single part of SAT Math. A linear relationship changes by the same amount for every unit step in the input, so its graph is a line, and almost every question here comes down to slope, an intercept, or a point that satisfies several conditions at once.

Linear equations in one variable

Adding or subtracting the same quantity on both sides, and multiplying or dividing both sides by the same nonzero number, give an equivalent equation with the same solutions. Distribute first, collect the variable terms on one side, and check the answer in the original equation. For example

3(x−2)+4=2x+73x−2=2x+7x=9.\begin{aligned} 3(x - 2) + 4 &= 2x + 7 \\[0.4em] 3x - 2 &= 2x + 7 \\[0.4em] x &= 9. \end{aligned}

Read what the question asks for. If 3x+2=113x + 2 = 11, the value of 6x+46x + 4 is 2⋅11=222 \cdot 11 = 22 without finding xx first.

The equation ax+b=cx+dax + b = cx + d has

CaseSolutions
a≠ca \ne cexactly one, x=d−ba−cx = \dfrac{d - b}{a - c}
a=ca = c and b≠db \ne dnone, the equation reduces to a false statement
a=ca = c and b=db = dinfinitely many, every real xx

So 2(x+3)=2x+62(x + 3) = 2x + 6 holds for every real xx, while 2x+5=2x+12x + 5 = 2x + 1 has no solution. When the variable cancels, the answer is never automatically x=0x = 0. A question asking for a constant that makes an equation have no solution is asking for equal coefficients of xx and unequal constants.

Lessons Solving linear equations, No or infinitely many solutions

Rearranging formulas

To make a variable the subject, undo the operations applied to it in reverse order. Dividing by an expression requires that expression to be nonzero, and the zero case is a separate question. For example, solving A=12bhA = \frac{1}{2}bh for hh gives

h=2Ab,b≠0,h = \frac{2A}{b}, \qquad b \ne 0,

and in geometry b>0b > 0 anyway, since a length is positive. Solving y=mx+by = mx + b for xx gives x=y−bmx = \frac{y - b}{m} only when m≠0m \ne 0.

Lessons Rearranging formulas

Functions and graphs

A function ff assigns exactly one output f(x)f(x) to each allowed input xx. The symbol f(x)f(x) is a value, not ff times xx, and evaluating at an expression substitutes the whole expression, so for f(x)=x2+1f(x) = x^2 + 1 the value f(x+1)f(x + 1) is (x+1)2+1(x + 1)^2 + 1. The point (a,b)(a, b) lies on the graph of ff exactly when f(a)=bf(a) = b. The yy-intercept is (0,f(0))(0, f(0)), and the xx-intercepts are the points (r,0)(r, 0) with f(r)=0f(r) = 0. A question asking for “the value of xx when f(x)=7f(x) = 7” reads the graph or table backward, from output to input.

Lessons Coordinates, functions and graphs

Slope and the forms of a line

The slope of the line through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), with x1≠x2x_1 \ne x_2, is

m=y2−y1x2−x1,m = \frac{y_2 - y_1}{x_2 - x_1},

the change in yy for each increase of 11 in xx, in units of yy per unit of xx. A positive slope rises from left to right, a horizontal line has slope 00, and a vertical line has undefined slope.

FormEquationShows
Slope-intercepty=mx+by = mx + bslope mm and yy-intercept (0,b)(0, b)
Point-slopey−y1=m(x−x1)y - y_1 = m(x - x_1)slope mm and the point (x1,y1)(x_1, y_1)
StandardAx+By=CAx + By = C, with AA and BB not both 00slope −AB-\dfrac{A}{B} when B≠0B \ne 0
Horizontaly=cy = cslope 00
Verticalx=cx = cundefined slope

In standard form with A≠0A \ne 0 and B≠0B \ne 0, the xx-intercept is (CA,0)\left(\frac{C}{A}, 0\right) and the yy-intercept is (0,CB)\left(0, \frac{C}{B}\right). For example the points (1,2)(1, 2) and (4,11)(4, 11) give m=93=3m = \frac{9}{3} = 3, and then 2=3+b2 = 3 + b gives y=3x−1y = 3x - 1.

In a linear model y=mx+by = mx + b the slope is the rate of change and bb is the starting value, the value of yy when x=0x = 0. If a plumber charges C=45h+60C = 45h + 60 dollars for hh hours, then $45 is the charge per hour and $60 is the fixed charge. A model is valid only on the inputs that make sense in the situation.

Lessons Slope and rate of change, Intercepts and linear models, Writing the equation of a line

Parallel and perpendicular lines

Two distinct nonvertical lines are parallel exactly when their slopes are equal. Two lines with slopes m1m_1 and m2m_2 are perpendicular exactly when

m1m2=−1,m2=−1m1,m_1 m_2 = -1, \qquad m_2 = -\frac{1}{m_1},

which needs both slopes defined and nonzero. A horizontal line and a vertical line are also perpendicular. Lines with equal slopes and equal intercepts are the same line, not two parallel lines. For example, the line through (4,1)(4, 1) perpendicular to y=2x+1y = 2x + 1 has slope −12-\frac{1}{2}, so it is y=−12x+3y = -\frac{1}{2}x + 3.

Lessons Parallel and perpendicular lines

Systems of linear equations

A solution of a system is an ordered pair that satisfies every equation, and on a graph it is a point where the lines meet. Substitution solves one equation for a variable and puts that expression into the other. Elimination adds a multiple of one whole equation, both sides, to the other so that a variable cancels. For example

x+y=10x−y=4\begin{aligned} x + y &= 10 \\[0.2em] x - y &= 4 \end{aligned}

adds to 2x=142x = 14, so x=7x = 7 and y=3y = 3, the solution (7,3)(7, 3). Sometimes the question asks for a combination such as x+yx + y, which one addition or subtraction of the equations may give directly.

For the system a1x+b1y=c1a_1 x + b_1 y = c_1 and a2x+b2y=c2a_2 x + b_2 y = c_2,

ConditionSolutionsLines
a1b2−a2b1≠0a_1 b_2 - a_2 b_1 \ne 0exactly onemeet at one point
a1b2−a2b1=0a_1 b_2 - a_2 b_1 = 0, one equation a multiple of the otherinfinitely manythe same line
a1b2−a2b1=0a_1 b_2 - a_2 b_1 = 0, otherwisenoneparallel and distinct

For example kx+3y=6kx + 3y = 6 and 2x+y=52x + y = 5 have no solution exactly when k=6k = 6, because then the first equation is 6x+3y=66x + 3y = 6 while three times the second is 6x+3y=156x + 3y = 15. Equal coefficient ratios alone do not separate the last two cases, so compare the constants too, and avoid ratio tests that divide by a zero coefficient.

Lessons Graphs and substitution, Elimination, One, no or infinite solutions

Linear inequalities

Adding the same number to both sides of an inequality, or multiplying or dividing both sides by a positive number, keeps its direction. Multiplying or dividing by a negative number reverses it. So −3x+5>11-3x + 5 > 11 gives −3x>6-3x > 6 and x<−2x < -2. A strict inequality (<< or >>) excludes its boundary, drawn as a hollow dot or a dashed line, and an inclusive one (≤\le or ≥\ge) includes it, drawn as a filled dot or a solid line.

The graph of y>mx+by > mx + b is the half-plane above the line y=mx+by = mx + b, and y<mx+by < mx + b is the half-plane below it. When the inequality is not solved for yy, test a point not on the boundary line, such as (0,0)(0, 0). The solution set of a system of inequalities is the region where every inequality holds at the same time, and a point on a dashed boundary is not in it.

Lessons Solving linear inequalities, Graphing inequalities

Advanced Math

Advanced Math is the nonlinear algebra on the test. The key skill is choosing the form of an expression that answers the question, such as factored form for zeros, vertex form for a maximum or minimum, and the exponential form a⋅bta \cdot b^t for a growth rate, while keeping track of every value a denominator or a root rules out.

Exponent rules

For a positive integer nn the power ana^n is the product of nn factors of aa. A negative exponent means a reciprocal, never a negative number.

RuleCondition
a0=1a^0 = 1a≠0a \ne 0
a−n=1ana^{-n} = \dfrac{1}{a^n}a≠0a \ne 0
aman=am+na^m a^n = a^{m+n}
aman=am−n\dfrac{a^m}{a^n} = a^{m-n}a≠0a \ne 0
(am)n=amn(a^m)^n = a^{mn}
(ab)n=anbn(ab)^n = a^n b^n and (ab)n=anbn\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}b≠0b \ne 0 for the quotient

These rules hold for all real exponents when the bases are positive, and for integer exponents with any nonzero bases. They apply to products, never to sums, since (a+b)2≠a2+b2(a + b)^2 \ne a^2 + b^2 in general. For example 23⋅2−5=2−2=142^3 \cdot 2^{-5} = 2^{-2} = \frac{1}{4}, and x7x3=x4\frac{x^7}{x^3} = x^4 for x≠0x \ne 0, where the restriction stays after canceling.

Lessons Exponent rules

Radicals and rational exponents

For a≥0a \ge 0 the principal square root a\sqrt{a} is the number b≥0b \ge 0 with b2=ab^2 = a. An even root needs a nonnegative radicand and is never negative, while an odd root accepts every real number, as in −83=−2\sqrt[3]{-8} = -2.

RuleCondition
ab=a b\sqrt{ab} = \sqrt{a}\,\sqrt{b}a≥0a \ge 0 and b≥0b \ge 0
ab=ab\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}a≥0a \ge 0 and b>0b > 0
x2=∣x∣\sqrt{x^2} = \lvert x \rvertevery real xx
a1/n=ana^{1/n} = \sqrt[n]{a}a>0a > 0 and nn a positive integer
am/n=(an)m=amna^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}a>0a > 0, mm an integer, nn a positive integer

For example 50=25 2=52\sqrt{50} = \sqrt{25}\,\sqrt{2} = 5\sqrt{2}, 82/3=48^{2/3} = 4 and 27−1/3=1327^{-1/3} = \frac{1}{3}. The equation x2=9x^2 = 9 has the two solutions ±3\pm 3, but 9=3\sqrt{9} = 3 only.

Lessons Radicals and rational exponents

Polynomials and factoring

To multiply polynomials, multiply every term of one by every term of the other, then combine like terms. For all real aa and bb,

(a+b)2=a2+2ab+b2,(a−b)2=a2−2ab+b2,a2−b2=(a−b)(a+b).\begin{aligned} (a + b)^2 &= a^2 + 2ab + b^2, \\[0.4em] (a - b)^2 &= a^2 - 2ab + b^2, \\[0.4em] a^2 - b^2 &= (a - b)(a + b). \end{aligned}

A sum of squares x2+c2x^2 + c^2 with c≠0c \ne 0 does not factor over the real numbers. To factor, take out the greatest common factor first, then look for these patterns. A trinomial x2+bx+cx^2 + bx + c factors as (x+p)(x+q)(x + p)(x + q) when p+q=bp + q = b and pq=cpq = c, so x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x - 2)(x - 3). Grouping handles four terms, as in x3+2x2−9x−18=(x+2)(x−3)(x+3)x^3 + 2x^2 - 9x - 18 = (x + 2)(x - 3)(x + 3). Look for structure too, since x4−16x^4 - 16 is a difference of squares in x2x^2 and factors as (x2+4)(x−2)(x+2)(x^2 + 4)(x - 2)(x + 2). Two polynomials are equal for every xx exactly when their coefficients of each power match, which is how questions with unknown constants such as (2x+k)2=4x2+12x+9(2x + k)^2 = 4x^2 + 12x + 9 are solved, here with k=3k = 3. Expand a factored answer to check it.

Lessons Expanding polynomials, Factoring

Solving quadratic equations

A quadratic equation can be written as ax2+bx+c=0ax^2 + bx + c = 0 with a≠0a \ne 0. The zero product property says that AB=0AB = 0 exactly when A=0A = 0 or B=0B = 0, so move everything to one side before factoring. For example x2=5xx^2 = 5x gives x(x−5)=0x(x - 5) = 0, so x=0x = 0 or x=5x = 5, and dividing both sides by xx would lose the solution 00. The square root method gives z2=cz^2 = c the two solutions z=±cz = \pm\sqrt{c} when c>0c > 0, one when c=0c = 0, and none when c<0c < 0.

The quadratic formula gives the real solutions of ax2+bx+c=0ax^2 + bx + c = 0, for a≠0a \ne 0 and b2−4ac≥0b^2 - 4ac \ge 0,

x=−b±b2−4ac2a,x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},

with the whole numerator divided by 2a2a. The discriminant D=b2−4acD = b^2 - 4ac tells how many real solutions there are.

DiscriminantReal solutionsGraph of y=ax2+bx+cy = ax^2 + bx + c
D>0D > 0two distinctcrosses the xx-axis twice
D=0D = 0one repeated, −b2a-\dfrac{b}{2a}touches the xx-axis at its vertex
D<0D < 0nonemisses the xx-axis

For example 2x2−4x−3=02x^2 - 4x - 3 = 0 has D=16+24=40D = 16 + 24 = 40, so

x=4±404=1±102.x = \frac{4 \pm \sqrt{40}}{4} = 1 \pm \frac{\sqrt{10}}{2}.

Completing the square rewrites x2+bxx^2 + bx as (x+b2)2−(b2)2\left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2. With a leading coefficient other than 11, factor it out of the xx terms first.

Lessons Solving by factoring, Completing the square, The quadratic formula

Quadratic graphs and their forms

The graph of y=ax2+bx+cy = ax^2 + bx + c with a≠0a \ne 0 is a parabola that opens up when a>0a > 0 and down when a<0a < 0. Its three common forms describe the same graph.

FormShows
y=ax2+bx+cy = ax^2 + bx + cthe yy-intercept (0,c)(0, c)
y=a(x−h)2+ky = a(x - h)^2 + kthe vertex (h,k)(h, k) and the axis of symmetry x=hx = h
y=a(x−r1)(x−r2)y = a(x - r_1)(x - r_2)the xx-intercepts (r1,0)(r_1, 0) and (r2,0)(r_2, 0)

The vertex has

h=−b2a,k=f(h).h = -\frac{b}{2a}, \qquad k = f(h).

Watch the sign inside the square, since (x+3)2(x + 3)^2 means h=−3h = -3. When the zeros are real, the axis of symmetry lies halfway between them, at x=r1+r22x = \frac{r_1 + r_2}{2}. If a>0a > 0 the minimum value is kk, and if a<0a < 0 the maximum value is kk, both attained at x=hx = h. On a restricted domain, such as a time interval, check the endpoints as well. For example

y=x2−6x+5=(x−3)2−4=(x−1)(x−5)\begin{aligned} y &= x^2 - 6x + 5 \\[0.4em] &= (x - 3)^2 - 4 \\[0.4em] &= (x - 1)(x - 5) \end{aligned}

has vertex (3,−4)(3, -4), minimum value −4-4, yy-intercept (0,5)(0, 5) and xx-intercepts (1,0)(1, 0) and (5,0)(5, 0).

Lessons Quadratic graphs

Sum and product of the roots

If a≠0a \ne 0 and r1r_1 and r2r_2 are the solutions of ax2+bx+c=0ax^2 + bx + c = 0, counted twice when repeated, then

r1+r2=−ba,r1r2=ca.r_1 + r_2 = -\frac{b}{a}, \qquad r_1 r_2 = \frac{c}{a}.

Read aa, bb and cc from the standard form with 00 on one side. For example 3x2−12x+7=03x^2 - 12x + 7 = 0 has two real roots, since D=60>0D = 60 > 0, with sum 44 and product 73\frac{7}{3}.

A parameter in a quadratic usually enters through the discriminant. The equation x2+kx+9=0x^2 + kx + 9 = 0 has exactly one real solution when k2−36=0k^2 - 36 = 0, that is when k=6k = 6 or k=−6k = -6. If the parameter multiplies x2x^2, check separately the value that makes the leading coefficient 00, because the equation is then linear and none of these formulas apply.

Lessons Parameters and root relationships

Polynomial zeros and graphs

For a polynomial pp, the number cc is a zero exactly when x−cx - c is a factor, and dividing p(x)p(x) by x−cx - c leaves the remainder p(c)p(c). So p(x)=x3−2x2+x−5p(x) = x^3 - 2x^2 + x - 5 leaves the remainder p(2)=−3p(2) = -3, and x−2x - 2 is not a factor. A real zero is an xx-intercept of the graph. A zero cc of multiplicity mm, meaning the factor x−cx - c appears exactly mm times, is crossed by the graph when mm is odd and touched when mm is even. For example y=(x+2)(x−1)2y = (x + 2)(x - 1)^2 crosses the xx-axis at −2-2 and touches it at 11. The zeros alone do not fix the whole graph, since multiplying by a nonzero constant keeps them.

Lessons Polynomial zeros and graph features

Rational and radical equations

A rational expression is undefined wherever its original denominator is 00, and those values stay excluded after any simplification. Cancel common factors, never terms. For example

x2−9x−3=x+3,x≠3,\frac{x^2 - 9}{x - 3} = x + 3, \qquad x \ne 3,

so the graph is the line y=x+3y = x + 3 with a hole at (3,6)(3, 6).

To solve a rational equation, list the excluded values, multiply both sides by the common denominator, solve, and reject any candidate that makes an original denominator 00. For example

xx−2=2x−2+3\frac{x}{x - 2} = \frac{2}{x - 2} + 3

leads to x=2+3(x−2)x = 2 + 3(x - 2) and the single candidate x=2x = 2, which is excluded, so the equation has no solution.

To solve a radical equation, isolate the radical, square both sides, and check every candidate in the original equation, because squaring can add extraneous solutions. For example 2x+3=x\sqrt{2x + 3} = x squares to x2−2x−3=0x^2 - 2x - 3 = 0, with candidates 33 and −1-1. Only x=3x = 3 works, since x=−1x = -1 gives 11 on the left and −1-1 on the right.

Lessons Rational expressions, Rational equations, Radical equations

Absolute value

The absolute value ∣x∣\lvert x \rvert is the distance from xx to 00, so it is never negative, and ∣x−a∣\lvert x - a \rvert is the distance between xx and aa. For a number dd,

StatementSolutions
∣x−a∣=d\lvert x - a \rvert = dx=a−dx = a - d or x=a+dx = a + d if d>0d > 0, only x=ax = a if d=0d = 0, none if d<0d < 0
∣x−a∣<d\lvert x - a \rvert < da−d<x<a+da - d < x < a + d, for d>0d > 0
∣x−a∣>d\lvert x - a \rvert > dx<a−dx < a - d or x>a+dx > a + d, for d≥0d \ge 0

For example ∣2x−3∣=7\lvert 2x - 3 \rvert = 7 gives 2x−3=72x - 3 = 7 or 2x−3=−72x - 3 = -7, so x=5x = 5 or x=−2x = -2, and ∣x−4∣<3\lvert x - 4 \rvert < 3 gives 1<x<71 < x < 7. When the other side contains xx, check each candidate, because that side must not be negative. The graph of y=a∣x−h∣+ky = a\lvert x - h \rvert + k with a≠0a \ne 0 is a V with vertex (h,k)(h, k), opening up when a>0a > 0.

Lessons Absolute value equations and graphs

Systems with lines and curves

Solve a system of a line and a curve by substituting the line into the curve, then find both coordinates of each solution. For example y=x2y = x^2 and y=2x+3y = 2x + 3 give x2−2x−3=0x^2 - 2x - 3 = 0, so x=3x = 3 or x=−1x = -1, and the solutions are (3,9)(3, 9) and (−1,1)(-1, 1). When the substitution leads to a quadratic, its discriminant counts the intersections, with D=0D = 0 meaning the line touches the curve at exactly one point.

Lessons Systems with lines and curves

Exponential growth and decay

An exponential model has the form

f(t)=a⋅bt,f(t) = a \cdot b^t,

with a≠0a \ne 0 and a base b>0b > 0, b≠1b \ne 1. In applications a>0a > 0 is the initial value f(0)f(0) and tt counts time intervals. Over each interval the quantity is multiplied by bb, so it grows when b>1b > 1 and decays when 0<b<10 < b < 1.

SituationModel
Grows by r%r\% per intervala(1+r100)ta\left(1 + \dfrac{r}{100}\right)^t
Decreases by r%r\% per interval, 0<r<1000 < r < 100a(1−r100)ta\left(1 - \dfrac{r}{100}\right)^t
Multiplied by bb every kk time unitsa⋅bt/ka \cdot b^{t/k}
Half-life kka(12)t/ka\left(\dfrac{1}{2}\right)^{t/k}

A percent rate always belongs to a stated time interval. For example $2000 growing by 5%5\% per year is worth 2000⋅1.053=2315.252000 \cdot 1.05^3 = 2315.25 dollars after 33 years, and 8080 milligrams with a half-life of 66 hours leaves 80⋅(12)18/6=1080 \cdot \left(\frac{1}{2}\right)^{18/6} = 10 milligrams after 1818 hours. Equivalent forms show other rates, since 1.44t=(1.22)t=1.22t1.44^t = (1.2^2)^t = 1.2^{2t}, so growth of 44%44\% every 22 years is growth of 20%20\% per year. An exponential equation with a common base is solved by comparing exponents, as 2x+1=32=252^{x + 1} = 32 = 2^5 gives x=4x = 4.

Linear models add the same amount over equal steps of input, and exponential models multiply by the same factor. In a table with equally spaced inputs, constant differences of outputs point to a linear model and constant ratios, with no output 00, point to an exponential one. A short table supports a model but cannot prove it.

Lessons Exponential growth and decay, Models from tables

Shifting, reflecting and stretching graphs

Here hh and kk are real numbers and c>0c > 0. The graph of each new function comes from the graph of y=f(x)y = f(x) as follows.

New functionEffect on the graph
f(x)+kf(x) + kup by kk, or down by ∣k∣\lvert k \rvert if k<0k < 0
f(x−h)f(x - h)right by hh, or left by ∣h∣\lvert h \rvert if h<0h < 0
−f(x)-f(x)reflection across the xx-axis
f(−x)f(-x)reflection across the yy-axis
cf(x)c f(x)vertical stretch by cc if c>1c > 1, compression if 0<c<10 < c < 1
f(cx)f(cx)horizontal compression by the factor 1c\dfrac{1}{c} if c>1c > 1, stretch if 0<c<10 < c < 1

A change inside the parentheses acts on the input, so it moves the graph horizontally and the opposite way to how it looks. The point (u,f(u))(u, f(u)) lies on y=f(x)y = f(x), and y=f(x−h)y = f(x - h) reaches the same height at x=u+hx = u + h. For example y=(x+2)2−5y = (x + 2)^2 - 5 is y=x2y = x^2 moved 22 left and 55 down, with vertex (−2,−5)(-2, -5). The new function is defined only where the whole new input lies in the domain of ff.

Lessons Shifting, reflecting and stretching graphs

Problem-Solving and Data Analysis

This part of the test asks you to read data from tables and graphs, summarize it, and judge what it supports. Most errors come from using the wrong group as the total, so always ask what the denominator of a rate, percent or probability is, and how the data were collected before deciding what they show.

Data displays

A dot plot shows every value, and a histogram groups values into intervals, so it shows how many fall in each interval but not the exact values. A box plot marks the minimum, the first quartile Q1Q_1, the median, the third quartile Q3Q_3 and the maximum. About half of the data lie between Q1Q_1 and Q3Q_3, and the interquartile range is

IQR=Q3−Q1.\text{IQR} = Q_3 - Q_1.

Read every axis label and scale before comparing heights or lengths. Quartile conventions differ, so use the quartiles a question gives.

Lessons Reading data displays

Mean, median and weighted averages

The mean of nn values is

xˉ=x1+x2+⋯+xnn,\bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n},

so the total of the values equals nxˉn \bar{x}, which helps when a value is added or removed. The median is the middle value of the ordered data, or the mean of the two middle values when nn is even. For the data 33, 77, 88, 1212, 2020, 4141 the median is 8+122=10\frac{8 + 12}{2} = 10 and the mean is 916≈15.2\frac{91}{6} \approx 15.2. An extreme value pulls the mean toward it but barely moves the median, so in data skewed to the right the mean is usually larger than the median.

To combine groups, weight each group mean by its size. A class of 2020 with mean 8080 and a class of 3030 with mean 7070 have the combined mean

20⋅80+30⋅7050=74,\frac{20 \cdot 80 + 30 \cdot 70}{50} = 74,

not 7575. A frequency table works the same way, with each value weighted by its frequency.

Lessons Mean, median and weighted averages

Spread and standard deviation

The range is the maximum minus the minimum. The standard deviation measures how far the values typically lie from the mean, and it is 00 only when all values are equal. For a population of nn values with mean μ\mu,

σ=∑i=1n(xi−μ)2n,\sigma = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \mu)^2}{n}},

and the sample standard deviation ss divides by n−1n - 1 instead, for n≥2n \ge 2. SAT questions usually ask you to compare spreads rather than compute them. Data bunched near the mean have a smaller standard deviation than data spread far from it, even with the same mean and range.

Change to every valueMeanMedianStandard deviation and range
add kkadds kkadds kkunchanged
multiply by cctimes cctimes cctimes ∣c∣\lvert c \rvert

For example 22, 44, 44, 44, 55, 55, 77, 99 has mean 55 and population standard deviation 22. Removing an outlier usually lowers the standard deviation and the range, and it usually changes the mean more than the median.

Lessons Spread, outliers and standard deviation

Scatterplots and models

A scatterplot shows pairs of values. A line of best fit summarizes the trend without passing through every point, and its slope is the predicted change in yy for each increase of 11 in xx. The residual of a data point is

residual=actual−predicted,\text{residual} = \text{actual} - \text{predicted},

positive for a point above the model and negative for one below. For example, with the model y=2.5x+10y = 2.5x + 10 the point (8,27)(8, 27) has predicted value 3030 and residual −3-3. A prediction inside the range of the data is an interpolation, and one outside it is an extrapolation, which is less reliable the farther it goes. An association between two variables does not show that one causes the other.

Lessons Scatterplots, models and predictions

Probability rules

When all outcomes are equally likely,

P(A)=outcomes in Aall outcomes,P(A) = \frac{\text{outcomes in } A}{\text{all outcomes}},

and every probability lies between 00 and 11.

RuleFormulaCondition
ComplementP(not A)=1−P(A)P(\text{not } A) = 1 - P(A)
OrP(A or B)=P(A)+P(B)−P(A and B)P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)
Or, no overlapP(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B)AA and BB cannot both happen
GivenP(A∣B)=P(A and B)P(B)P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}P(B)>0P(B) > 0
IndependentP(A and B)=P(A) P(B)P(A \text{ and } B) = P(A)\,P(B)AA and BB independent

A conditional probability restricts the total to the given group, and P(A∣B)P(A \mid B) usually differs from P(B∣A)P(B \mid A). In this two-way table of 150150 students,

YesNoTotal
Grade 10303020205050
Grade 1145455555100100
Total75757575150150

the probability that a grade 10 student said yes is 3050=0.6\frac{30}{50} = 0.6, while the probability that a student who said yes is in grade 10 is 3075=0.4\frac{30}{75} = 0.4. The probability that a student is in grade 10 or said yes is 50+75−30150=1930\frac{50 + 75 - 30}{150} = \frac{19}{30}. Treat events as independent only when the question says so or the numbers show it.

Lessons Probability and two-way tables, Conditional probability

Samples and margin of error

A sample statistic estimates a population value but is almost never exactly equal to it. Results generalize only to the population that was sampled, and only when the sample was chosen at random from it. A sample that favors some members, such as volunteers or people who are easy to reach, is biased, and a larger sample does not fix that bias.

A margin of error EE describes the uncertainty from random sampling, and an estimate with margin EE gives the interval

estimate−E  to  estimate+E.\text{estimate} - E \ \text{ to } \ \text{estimate} + E.

It is plausible that the population value lies in this interval, but it is not guaranteed, and values outside it are less plausible rather than impossible. For example a poll result of 54%54\% with a margin of 33 percentage points gives 51%51\% to 57%57\%. Under comparable conditions a larger random sample gives a smaller margin of error, and under the usual formula the margin shrinks in proportion to 1n\frac{1}{\sqrt{n}}, so four times the sample size roughly halves it. A higher confidence level gives a wider interval from the same data.

Lessons Samples, populations and bias, Margin of error

Studies and conclusions

DesignSupports
Random sample from a populationgeneralizing to that population
Random assignment of treatmentsa cause and effect conclusion for the subjects studied
Botha cause and effect conclusion for the population
Neitheronly a description of the subjects studied

An observational study records what happens without assigning treatments, so a confounding variable can explain an association, and it cannot show cause and effect. Random sampling and random assignment answer different questions.

Lessons Experiments, observations and causal claims

Geometry and Trigonometry

SAT geometry rests on a short list of facts about angles, triangles and circles. Most questions combine two of them, such as similar triangles with a proportion, or a right triangle inside a circle, so it pays to know the facts below well enough to spot them in a figure.

Angles, lines and triangles

FactStatement
Vertical anglesequal
Angles on a straight lineadd to 180∘180^\circ
Angles around a pointadd to 360∘360^\circ
Triangleinterior angles add to 180∘180^\circ
Exterior angle of a triangleequals the sum of the two remote interior angles
Polygon with nn sidesinterior angles add to (n−2)⋅180∘(n - 2) \cdot 180^\circ
Isosceles trianglethe angles opposite the equal sides are equal

When a line crosses two parallel lines, corresponding angles are equal, alternate interior angles are equal, and same-side interior angles add to 180∘180^\circ. These hold only when the lines are known to be parallel. In any triangle the longest side lies opposite the largest angle, and each side is shorter than the sum of the other two.

Lessons Angles, parallel lines and triangles

Similar and congruent triangles

Similar triangles have equal corresponding angles and proportional corresponding sides. Congruent triangles are similar with scale factor 11. List the vertices in matching order so that corresponding parts line up.

Shows similarityShows congruence
AA, two pairs of equal anglesSSS, three pairs of equal sides
SSS, three proportional pairs of sidesSAS, two equal sides and the angle between them
SAS, two proportional sides and the equal angle between themASA and AAS, two equal angles and a corresponding side

Equal angles alone never prove congruence, and SSA is not a general test. A line parallel to one side of a triangle cuts off a smaller triangle similar to the whole. For example, if △ABC∼△DEF\triangle ABC \sim \triangle DEF with AB=6AB = 6, DE=9DE = 9 and BC=8BC = 8, then

EFBC=DEAB=32,EF=12.\frac{EF}{BC} = \frac{DE}{AB} = \frac{3}{2}, \qquad EF = 12.

Lessons Similar and congruent triangles

Right triangles

In a right triangle with legs aa and bb and hypotenuse cc, the side opposite the right angle,

a2+b2=c2.a^2 + b^2 = c^2.

Conversely, a triangle whose sides satisfy this equation has a right angle opposite cc. Common whole-number triples are 3,4,53, 4, 5 and 5,12,135, 12, 13 and 8,15,178, 15, 17, and any positive multiple of a triple is again a right triangle. The two special right triangles have fixed side ratios.

AnglesSides opposite the angles, in order
45∘,45∘,90∘45^\circ, 45^\circ, 90^\circx, x, x2x, \ x, \ x\sqrt{2}
30∘,60∘,90∘30^\circ, 60^\circ, 90^\circx, x3, 2xx, \ x\sqrt{3}, \ 2x

The shortest side of a 30∘30^\circ-60∘60^\circ-90∘90^\circ triangle lies opposite the 30∘30^\circ angle and is half the hypotenuse. For example a square with side 55 has diagonal 525\sqrt{2}, and an equilateral triangle with side 1010 has height 535\sqrt{3}.

Lessons Pythagorean theorem and special right triangles

Sine, cosine and tangent

For an acute angle θ\theta in a right triangle, SOH CAH TOA gives

sin⁡θ=oppositehypotenuse,cos⁡θ=adjacenthypotenuse,tan⁡θ=oppositeadjacent=sin⁡θcos⁡θ.\begin{aligned} \sin\theta &= \frac{\text{opposite}}{\text{hypotenuse}}, \\[0.6em] \cos\theta &= \frac{\text{adjacent}}{\text{hypotenuse}}, \\[0.6em] \tan\theta &= \frac{\text{opposite}}{\text{adjacent}} = \frac{\sin\theta}{\cos\theta}. \end{aligned}

The ratios depend only on the angle, because all right triangles with that angle are similar. The two acute angles of a right triangle are complementary, so

sin⁡θ=cos⁡(90∘−θ),cos⁡θ=sin⁡(90∘−θ).\begin{aligned} \sin\theta &= \cos(90^\circ - \theta), \\[0.4em] \cos\theta &= \sin(90^\circ - \theta). \end{aligned}

For every angle, sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1. In the 3,4,53, 4, 5 triangle the angle AA opposite the side 33 has sin⁡A=35\sin A = \frac{3}{5}, cos⁡A=45\cos A = \frac{4}{5} and tan⁡A=34\tan A = \frac{3}{4}. A ladder of length 1010 at 30∘30^\circ to the ground reaches a height of 10sin⁡30∘=510 \sin 30^\circ = 5. Set a calculator to degrees or radians to match the question.

Lessons Sine, cosine and tangent

Radians and the unit circle

A full turn is 360∘360^\circ or 2π2\pi radians, so

180∘=π radians.180^\circ = \pi \text{ radians}.

To convert degrees to radians multiply by π180\frac{\pi}{180}, and to convert radians to degrees multiply by 180π\frac{180}{\pi}. For example 150∘=5π6150^\circ = \frac{5\pi}{6} radians. An angle of 11 radian at the center of a circle cuts off an arc as long as the radius. The point at angle θ\theta on the unit circle, measured counterclockwise from the positive xx-axis, is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta), with the cosine first.

θ\theta00π6\dfrac{\pi}{6}π4\dfrac{\pi}{4}π3\dfrac{\pi}{3}π2\dfrac{\pi}{2}
Degrees0∘0^\circ30∘30^\circ45∘45^\circ60∘60^\circ90∘90^\circ
sin⁡θ\sin\theta0012\dfrac{1}{2}22\dfrac{\sqrt{2}}{2}32\dfrac{\sqrt{3}}{2}11
cos⁡θ\cos\theta1132\dfrac{\sqrt{3}}{2}22\dfrac{\sqrt{2}}{2}12\dfrac{1}{2}00
tan⁡θ\tan\theta0033\dfrac{\sqrt{3}}{3}113\sqrt{3}undefined

Lessons Degrees, radians and the unit circle

Circle facts

A circle of radius rr has diameter 2r2r, circumference C=2πrC = 2\pi r and area A=πr2A = \pi r^2.

FactStatement
Inscribed anglehalf the central angle that cuts off the same arc
Angle in a semicircle90∘90^\circ
Tangent lineperpendicular to the radius at the point of contact
Two tangents from one outside pointequal lengths
Radius perpendicular to a chordbisects the chord
Equal chords of one circleequal central angles and equal arcs

The measure of an arc in degrees equals its central angle. Name the arc you mean, since a chord that is not a diameter cuts off a minor arc and a major arc that add to 360∘360^\circ. For example, an inscribed angle that cuts off an arc of 110∘110^\circ measures 55∘55^\circ.

Lessons Circle angles, chords and tangents

Arc length and sector area

An arc or sector with central angle θ\theta is the fraction θ2π\frac{\theta}{2\pi} of the whole circle, or θ360∘\frac{\theta}{360^\circ} when θ\theta is in degrees. With θ\theta in radians,

s=rθ,Asector=12r2θ,s = r\theta, \qquad A_{\text{sector}} = \frac{1}{2} r^2 \theta,

and with θ\theta in degrees,

s=θ360∘⋅2πr,Asector=θ360∘⋅πr2.\begin{aligned} s &= \frac{\theta}{360^\circ} \cdot 2\pi r, \\[0.6em] A_{\text{sector}} &= \frac{\theta}{360^\circ} \cdot \pi r^2. \end{aligned}

An arc length is a length and a sector area is an area, so they have different units. For example r=6r = 6 and θ=π3\theta = \frac{\pi}{3}, which is 60∘60^\circ, give the arc length 2π2\pi and the sector area 6π6\pi. The perimeter of that sector also includes the two radii, 2π+122\pi + 12.

Lessons Arc length and sector area

Distance, midpoint and circle equations

For points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) the distance and the midpoint are

d=(x2−x1)2+(y2−y1)2,M=(x1+x22,y1+y22).\begin{gathered} d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, \\[0.6em] M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right). \end{gathered}

For example (1,2)(1, 2) and (7,10)(7, 10) are 1010 apart with midpoint (4,6)(4, 6). The circle with center (h,k)(h, k) and radius r>0r > 0 is

(x−h)2+(y−k)2=r2.(x - h)^2 + (y - k)^2 = r^2.

The right side is r2r^2, not rr, and the signs inside the squares are the opposite of the center’s coordinates. To find the center and radius from an expanded equation, complete the square in xx and in yy.

x2+y2−6x+4y−12=0(x−3)2+(y+2)2=25\begin{gathered} x^2 + y^2 - 6x + 4y - 12 = 0 \\[0.4em] (x - 3)^2 + (y + 2)^2 = 25 \end{gathered}

This circle has center (3,−2)(3, -2) and radius 55. If completing the square leaves 00 on the right the equation describes a single point, and a negative number means no points at all. A point lies inside, on or outside the circle as its distance from the center is less than, equal to or greater than rr.

Lessons Circle equations, centers and radii

Area and perimeter

FigureArea
Rectangleℓw\ell w
Parallelogrambhbh
Triangle12bh\dfrac{1}{2}bh
Trapezoid with parallel sides b1b_1 and b2b_212(b1+b2)h\dfrac{1}{2}(b_1 + b_2)h
Circleπr2\pi r^2

The height hh is always perpendicular to the chosen base, so a slanted side is not a height. The perimeter is the total length around a figure. For a composite figure, add the areas of pieces that do not overlap, or subtract a cut-out piece from the whole. For example a 1010 by 66 rectangle with a semicircle of diameter 66 removed from one end has area 60−9π260 - \frac{9\pi}{2}.

Lessons Area, perimeter and composite figures

Volume and surface area

SolidVolumeSurface area
Rectangular boxℓwh\ell w h2(ℓw+ℓh+wh)2(\ell w + \ell h + w h)
Prism with base area BBBhBhthe sum of the areas of its faces
Cylinderπr2h\pi r^2 h2πr2+2πrh2\pi r^2 + 2\pi r h when closed
Pyramid with base area BB13Bh\dfrac{1}{3}Bhthe base plus the triangular faces
Cone13πr2h\dfrac{1}{3}\pi r^2 hπr2+πrℓ\pi r^2 + \pi r \ell when closed, with slant height ℓ\ell
Sphere43πr3\dfrac{4}{3}\pi r^34πr24\pi r^2

The height hh of a pyramid or cone is perpendicular to the base, and the slant height of a cone is ℓ=r2+h2\ell = \sqrt{r^2 + h^2}. Check whether a surface is open, such as a can without a lid. For example a closed cylinder with r=3r = 3 and h=5h = 5 has volume 45π45\pi and surface area 18π+30π=48π18\pi + 30\pi = 48\pi.

Lessons Volume and surface area

Scaling

If every length of a figure is multiplied by k>0k > 0, then

lengths×k,areas×k2,volumes×k3.\begin{gathered} \text{lengths} \times k, \qquad \text{areas} \times k^2, \\[0.4em] \text{volumes} \times k^3. \end{gathered}

For example a model built at 1.51.5 times the size has 2.252.25 times the surface area and 3.3753.375 times the volume. Going backward, an area ratio of 94\frac{9}{4} means a length ratio of 32\frac{3}{2}. These laws need the whole figure scaled by the same factor, and they are different from converting units, which changes the numbers but not the figure.

Lessons Scaling lengths, areas and volumes

Test Strategy

Knowing the math is most of the work, and the rest is using the time and tools well and entering answers in a form the test accepts. The facts below describe the current format of the digital SAT.

The format of the Math section

The Math section has two modules of 2222 questions each, with 3535 minutes for each module, and the second module adapts to performance on the first. Review is possible only within the current module. About three quarters of the questions are multiple choice with four options, and the rest are student-produced responses. Wrong answers are not penalized, so answer every question. The Desmos graphing calculator is built into the testing app, you may also bring an approved calculator, and a reference sheet of common formulas is available.

Entering a student-produced response

RuleDetail
Lengthup to 55 characters for a positive answer, up to 66 for a negative answer including the minus sign
Fractions and decimalsboth accepted
Mixed numbersnot accepted, enter 7/27/2 or 3.53.5 for 3123\frac{1}{2}
A decimal that does not fitcut off or rounded at the last digit that fits
Symbolsdo not enter percent signs, dollar signs or commas
More than one correct answerenter only one

A fraction avoids rounding, so 23\frac{2}{3} is safest entered as 2/32/3. An answer of 1,2501{,}250 dollars is entered as 12501250, and 35%35\% as 3535. Before entering, compare your number with what the question asks for, such as 2x2x instead of xx or a value in other units.

The graphing calculator and choosing a method

In the Desmos graphing calculator, type an equation such as y=x2−2x−3y = x^2 - 2x - 3 to graph it, and click a marked intersection or intercept to read its coordinates. To solve an equation, graph each side as its own expression and find where they meet. A decimal read from a graph may only approximate an exact answer such as 2\sqrt{2}, and a viewing window can hide a solution, so zoom out and check the restrictions of the original equation.

Algebra by hand is often faster for linear equations, for a requested expression such as 6x+46x + 4, and for questions about structure. Graphing helps with nonlinear systems, with checking answer choices, and with models. Whatever the method, check that the answer satisfies every condition in the question, its units and its context.

Lessons Desmos for equations, Desmos tables and parameters