Meet every condition

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Difficulty: GOAT Mode
Estimated time: 50 minutes
Domains: Algebra Advanced Math
Techniques: ConstraintsModelingCondition StackWitness ValuesInvarianceVerification

What you’ll learn

Turn a dense prompt into a condition stack of fixed values, free constants, dependent values, and the target.
Choose a legal witness that satisfies every stated condition.
Build function definitions and outputs in dependency order for a readable model.
Use a second witness as an error check, then prove that the target is invariant.
Stop using witness values when the target changes or the prompt requires candidate filtering.

Prerequisites

Evaluate, combine, and compose functions0%
Lesson 5: define functions and evaluate named outputs in Desmos.

Why this matters on the SAT

Organize the conditions before calculating

Some hard SAT function questions give several constants, inequalities, named outputs, and one complicated target. The constants may not have unique values, but the requested expression can still have one fixed answer.

The key is to make every condition true in one ordered Desmos model.

SAT example

The function $q$ is defined by

q(x)=ax+b,

where $a>b$ and $b>87$ . Let

H=q(639)\qquad\text{and}\qquad K=q(621).

What is the value of

H-K-18a?

Fast Desmos solution

The problem does not determine unique values of $a$ and $b$ . On separate lines, enter the legal pair b=88 and a=89.

Then enter q(x)=a*x+b, H=q(639), K=q(621), and H-K-18a, with each definition on a new line.

Desmos returns $0$ . Now change the legal values to $b=90$ and $a=92$ . The target remains $0$ .

Try it yourself:

In the opening calculator, test $(a,b)=(92,90)$ . Check both inequalities before reading the target, then reset the example.

Two examples support the pattern, but they do not prove it. Algebra explains why the target is fixed:

\begin{aligned} H-K-18a &=(639a+b)-(621a+b)-18a\\ &=18a-18a\\ &=0. \end{aligned}

The calculator helps you organize and test the model. The cancellation proves the result for every allowed pair.

Calculator loads as you approach

Dependent definitions appear after their inputs, and the target remains 0 for another legal pair.

When should you use this workflow?

Use a condition stack with witness values when:

the prompt defines one or more functions;
some constants are constrained only by inequalities or simple relationships;
there is not enough information to determine every constant uniquely;
the question asks for a combination of outputs and constants; and
the combination may simplify to one value regardless of the legal constants.

Do not assign witness values when:

the question asks for the constant itself;
another exact condition determines the constant;
the target changes after a legal value changes;
the prompt asks for all solutions, a least or greatest possible value, or an integer count; or
the conditions uniquely determine several parameters, in which case custom regression with multiple conditions may be appropriate.

Check your understanding:

A problem gives $a>b>10$ , defines $f(x)=ax+b$ , and asks for $f(8)-f(3)-5a$ . The values of $a$ and $b$ are not otherwise determined. Is a legal-witness test appropriate, and what result should remain fixed?

1. Build a condition stack

Give every statement one job

A condition stack is an ordered list of everything that must be true before you evaluate the target.

For the opening problem, the stack is:

Layer	Condition	Meaning
Fixed values	$639$ , $621$ , and $18$	Copy these values exactly from the prompt.
Free constants	$a>b$ and $b>87$	$a$ and $b$ are not unique, but their values must be legal.
Function definition	$q(x)=ax+b$	Every later function output uses this rule.
Dependent values	$H=q(639)$ and $K=q(621)$	$H$ and $K$ depend on the function and constants.
Target	$H-K-18a$	Evaluate only after every earlier layer is defined.

The order is not decorative. A lower layer depends on the layers above it.

Before typing, ask:

Which quantities are fixed by the prompt?
Which constants are free but constrained?
Which values depend on a function or earlier definition?
What exact expression must be submitted?

Check your understanding:

In the opening problem, is $H$ a free constant or a dependent value? What must be defined before $H$ ?

Common mistake:

Skipping a condition that looks unimportant. A sign, order, integer, or domain condition can decide which candidate is valid. Record every condition before calculating, even if it does not appear in the final expression.

2. Choose one legal witness

Free does not mean arbitrary

A witness is one assignment that makes all the stated conditions true. A witness is legal only when it satisfies every restriction.

For

a>b\qquad\text{and}\qquad b>87,

the pair

(a,b)=(89,88)

is legal.

These pairs are not legal:

$(88,89)$ fails $a>b$ .
$(88,87)$ fails $b>87$ .
$(88,88)$ fails the strict inequality $a>b$ .

Choose simple values near the boundary because they reduce arithmetic. Avoid $0$ or $1$ when they could hide a term, sign, exponent, or coefficient.

Some restrictions protect the calculation itself. For example, if a target contains $\frac{1}{a-b}$ and the prompt states $a>b$ , choosing $a=b$ makes the expression undefined. An invariant may sometimes extend beyond the legal set, but you cannot assume that before checking every condition.

Check your understanding:

For $c>d>20$ , which pair is legal: $(c,d)=(21,22)$ , $(22,21)$ , or $(21,21)$ ?

3. Build in dependency order

Put the name being defined on the left

In Desmos, define an output with the new name on the left. Enter H=q(639) and K=q(621) on separate lines.

This creates reusable values. Entering q(639)=H treats the line as an equation instead of a direct definition and makes the dependency harder to read.

A clear build order is b=88, a=89, q(x)=a*x+b, H=q(639), K=q(621), and finally H-K-18a, with each entry on its own line.

Desmos resolves definitions and updates dependent expressions automatically, even when the expression list is not top to bottom. Use dependency order because it makes the model easier to audit and reduces transcription mistakes. If a name remains undefined, check that its definition exists and is valid.

Try it yourself:

Delete the last three lines, then rebuild H, K, and the target in dependency order. Reset the calculator when you finish.

Check your understanding:

Put these lines in dependency order: T=U+V-32c, U=p(d+4), p(x)=c(x-d)^2+7, c=2, d=51, and V=p(d-4).

Calculator loads as you approach

Each named value appears after the constants or function it depends on.

4. Example: Prove a symmetric quadratic stays fixed

One successful witness is not proof

An expression is invariant when its value does not change across the legal assignments allowed by the prompt.

Use this three-part check:

Evaluate the target with one legal witness.
Change to a clearly different legal witness.
Explain the unchanged result with cancellation, symmetry, or another exact relationship.

If the target changes, stop. Either:

the result is not uniquely determined;
a condition is missing from the model;
the changed witness is not legal; or
something was entered incorrectly.

Worked example

The function $p$ is defined by

p(x)=c(x-d)^2+7,

where

$c>0$ and $d>50$ .

Let

U=p(d+4)\qquad\text{and}\qquad V=p(d-4).

What is the value of

U+V-32c?

A
$0$
B
$7$
C
$14$
D
$32$

Step 1

Build the stack

The constants are free but constrained. The function depends on $c$ and $d$ , and the outputs depend on the function.

Choose a legal witness such as

c=2\qquad\text{and}\qquad d=51.

Step 2

Define the outputs and target

Enter c=2, d=51, p(x)=c*(x-d)^2+7, U=p(d+4), V=p(d-4), and U+V-32c on separate lines in that order.

Desmos returns $14$ .

Step 3

Change the witness, then explain

In the same calculator, change to

c=3\qquad\text{and}\qquad d=60.

The target remains $14$ .

Now verify the structure:

\begin{aligned} U&=c(4)^2+7=16c+7,\\ V&=c(-4)^2+7=16c+7. \end{aligned}

Therefore,

U+V-32c=(16c+7)+(16c+7)-32c=14.

The answer is C.

Calculator loads as you approach

Test c = 2, d = 51, then change to c = 3, d = 60 and compare the target.

Common mistake:

Treating two matching outputs as proof. A second witness is an error check, not a universal proof. Use algebraic cancellation or another exact relationship to explain why every legal witness gives the same result.

5. Stop when witness values are unsafe

A changing target is a warning

Suppose

q(x)=ax+b,\qquad a>b>2,

and

H=q(5),\qquad K=q(1).

Then

H-K=4a.

The value changes when $a$ changes. Choosing one legal witness and reporting its output would invent a unique answer that the prompt does not support.

Try it yourself:

Change a=4 to a=6. Then change only $b$ . Identify which constant controls $H-K$ , and reset the calculator.

Check your understanding:

Using $(a,b)=(4,3)$ gives $H-K=16$ . Using $(a,b)=(6,3)$ gives $H-K=24$ . What should you conclude?

Calculator loads as you approach

Change a from 4 to 6. The target changes from 16 to 24, so it is not invariant.

Some prompts require candidate filtering instead

Witness assignment is also unsafe when the question asks for a least possible integer, every solution, or a uniquely determined constant. In those cases, generate or test candidates and apply every condition.

For example, if $k$ must be the least positive integer satisfying several relationships, one legal value proves only that the value works. It does not prove that no smaller value works.

Use this decision:

Free constants and a fixed target: choose two legal witnesses, then verify the invariant.
A requested constant or extremum: solve or test all relevant candidates.
Several exact conditions that determine parameters: use a system or custom regression and verify every condition.

Common mistake:

Using one witness for a least-possible question. One legal value establishes possibility, not minimality. Check every smaller candidate allowed by the prompt before claiming a least value.

Finish the solution

The first legal witness and function are ready. Define the outputs, evaluate the target, then verify that another legal witness gives the same result.

Complete the linear condition stack

Finish the solution

The function $f$ is defined by

f(x)=m(x-n)+4,

where $m>n>10$ . Let

A=f(n+6)\qquad\text{and}\qquad B=f(n-2).

What is the value of

A-B-8m?

First steps

The pair $(m,n)=(12,11)$ is legal.
The function f(x)=m*(x-n)+4 is defined.
Define $A$ and $B$ from the function.
Evaluate the target only after both outputs exist.

Finish it

Your answer

Calculator loads as you approach

Add A, B, and the requested expression, then change to another legal pair.

Practice problems

SAT practice problems

For each problem, build the condition stack before calculating. Use two legal witnesses when the target should be fixed, then explain the invariant. If the question asks for a least value, test candidates instead.

Cancel a shifted quadratic

Practice problem

The function $g$ is defined by

g(x)=r(x-s)^2+5,

where $r>s>0$ . Let

U=g(s+5)\qquad\text{and}\qquad V=g(s-1).

What is the value of

U-V-24r?

Your answer

Calculator loads as you approach

Choose a legal pair, define the function and outputs, then test another pair.

Use opposite inputs

Practice problem

The function $h$ is defined by

h(x)=ax^2+bx+6,

where $a>b>10$ . Let

P=h(4)\qquad\text{and}\qquad Q=h(-4).

What is the value of

P+Q-32a?

Calculator loads as you approach

Use a legal pair, compare h(4) with h(-4), and evaluate the target.

Match exponential dependencies

Practice problem

The function $P$ is defined by

P(t)=ca^t+5,

where $c>a>1$ . Let

H=P(5)\qquad\text{and}\qquad K=P(3).

What is the value of

H-a^2K+5a^2?

Your answer

Calculator loads as you approach

Choose c greater than a greater than 1, then define H, K, and the target.

Recognize a changing target

Practice problem

The function $f$ is defined by

f(x)=a(x-b)^2+c,

where $a>0$ , $b>10$ , and $c>0$ . Which expression is equivalent to

f(b+3)+f(b-3)?

Calculator loads as you approach

Test two legal triples and notice that the target changes before simplifying it.

Finish the lesson

5 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Record every condition, then separate fixed values, free constants, dependent values, and the target.
Choose a legal witness and build definitions in dependency order so the model is easy to audit.
Test a clearly different legal witness as an error check, not as proof.
Prove invariance with cancellation, symmetry, or another exact relationship.
If the target changes, or the prompt asks for a least, greatest, integer, or complete solution set, switch to candidate filtering or exact solving.

Next lesson

Hidden functions inside quotients

Recover useful function information when the prompt hides it inside a rational expression.

Start next lesson

Free Practice

What you’ll learn

Prerequisites

Organize the conditions before calculating

Fast Desmos solution

When should you use this workflow?

1. Build a condition stack

Give every statement one job

2. Choose one legal witness

Free does not mean arbitrary

3. Build in dependency order

Put the name being defined on the left

4. Example: Prove a symmetric quadratic stays fixed

One successful witness is not proof

Build the stack

Define the outputs and target

Change the witness, then explain

5. Stop when witness values are unsafe

A changing target is a warning

Some prompts require candidate filtering instead

Finish the solution

Complete the linear condition stack

First steps

Finish it

Practice problems

SAT practice problems

Cancel a shifted quadratic

Use opposite inputs

Match exponential dependencies

Recognize a changing target

5 practice examples left

Quick recap

Hidden functions inside quotients

Free question bank practice problems