California course

Math I

Build connected understanding across quantities, equations, functions, coordinate geometry, congruence, and data.

Problem types
659
Practice variants
2,636
Problem types

Page 4 of 19

Each problem type has four distinct practice variants. Open a preview to move among all four.

A-REI.3.1 M1-010-A08-V01

Choose an absolute-value model from tolerance language

Solve and graph one-variable absolute-value equations and inequalities; interpret solutions in context.

Tolerance language describes how far an actual measurement may lie from a target. We’ll subtract the target from the variable, take absolute value so deviations in either direction count positively, …

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A-REI.3.1 M1-010-A09-V01

Solve an absolute-value equation with a variable outside the absolute value

Solve and graph one-variable absolute-value equations and inequalities; interpret solutions in context.

When the other side contains the variable, the usual fixed-target two-case shortcut does not apply directly. We’ll first require that side to be nonnegative, then use the piecewise definition of …

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A-REI.3.1 M1-010-A10-V01

Solve an absolute-value inequality after isolating the absolute value

Solve and graph one-variable absolute-value equations and inequalities; interpret solutions in context.

The distance condition cannot be interpreted until the absolute-value expression is alone. We’ll undo the outside addition and multiplication in reverse order, watching the divisor’s sign, and then translate the …

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A-REI.3.1 M1-010-A11-V01

Determine endpoint inclusion in an absolute-value graph

Solve and graph one-variable absolute-value equations and inequalities; interpret solutions in context.

An absolute-value inequality describes distances from a center, so its graph begins with two boundary points equally far from that center. We’ll move the stated distance left and right to …

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A-REI.3.1 M1-010-A12-V01

Choose an absolute-value statement from a number-line graph

Solve and graph one-variable absolute-value equations and inequalities; interpret solutions in context.

A symmetric number-line interval can be rewritten as a distance condition. We’ll average the two endpoints to find the center, measure from that midpoint to either endpoint for the radius, …

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A-REI.5 M1-011-A01-V01

Verify that replacing an equation in a system preserves a solution

Justify elimination: replacing one equation in a two-variable system with a linear combination preserves solutions.

A solution of a system must satisfy every equation that remains after a transformation. We’ll preserve the ordered-pair roles, substitute into the retained equation and the replacement equation separately, and …

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A-REI.5 M1-011-A04-V01

Decide whether an elimination step is valid

Justify elimination: replacing one equation in a two-variable system with a linear combination preserves solutions.

A proposed elimination step must use one consistent equation operation and simplify every column correctly. We’ll align like terms, check whether one pair of coefficients are opposites, and add the …

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A-REI.5 M1-011-A05-V01

Find a multiplier to eliminate a variable

Justify elimination: replacing one equation in a two-variable system with a linear combination preserves solutions.

The useful multiplier is the one that makes a chosen pair of variable coefficients opposites. We’ll compare the coefficients, set their sum after scaling equal to zero, and apply the …

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A-REI.5 M1-011-A06-V01

Compare system transformations and decide which preserves solutions

Justify elimination: replacing one equation in a two-variable system with a linear combination preserves solutions.

A combined equation is a consequence of a system, but it may not contain all of the system’s information by itself. We’ll compare how many independent conditions each transformation retains, …

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A-REI.5 M1-011-A08-V01

Find the first invalid step in an elimination solution

Justify elimination: replacing one equation in a two-variable system with a linear combination preserves solutions.

Finding the first invalid step means auditing each transition in order, not merely noticing that the final value is wrong. We’ll verify that the equations were copied correctly, confirm the …

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A-REI.5 M1-011-A09-V01

Show that an original system and transformed system share the same ordered-pair solution

Justify elimination: replacing one equation in a two-variable system with a linear combination preserves solutions.

Sharing one derived equation is not enough to prove two systems have the same ordered-pair solution. We’ll solve the transformed system from its simplest equation, substitute to recover the second …

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A-REI.5 M1-011-A12-V01

Decide whether an elimination action loses system information

Justify elimination: replacing one equation in a two-variable system with a linear combination preserves solutions.

Information is preserved only when an equation transformation can be reversed and still imposes the original constraint. We’ll carry the proposed multiplier through every coefficient and constant, inspect the equation …

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A-REI.5 M1-011-R02-V01

Decide whether a system replacement is equivalent

Justify elimination: replacing one equation in a two-variable system with a linear combination preserves solutions.

A row combination preserves a system only when the replacement keeps enough information to reverse the operation. We’ll form the proposed sum term by term, retain the untouched equation, and …

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A-REI.5 M1-011-R03-V01

Form and verify an elimination row combination

Justify elimination: replacing one equation in a two-variable system with a linear combination preserves solutions.

The requested row combination should eliminate one variable while preserving equality. We’ll add corresponding terms and constants, confirm that the opposite coefficients cancel, and solve the resulting one-variable equation. Back-substitution …

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A-REI.6 M1-012-A01-V01

Solve a system by substitution when one variable is already isolated

Solve systems of two linear equations exactly and approximately using algebraic and graphical methods.

One equation already gives a complete expression for a variable, making substitution the shortest route. We’ll replace that variable in the other equation with the entire equivalent expression, simplify to …

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A-REI.6 M1-012-A02-V01

Solve a system by isolating a variable first, then substituting

Solve systems of two linear equations exactly and approximately using algebraic and graphical methods.

Substitution begins by isolating the variable whose coefficient makes the rearrangement simplest. We’ll create an equivalent expression from one equation, place that whole expression in parentheses when replacing the variable …

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A-REI.6 M1-012-A03-V01

Solve a system by elimination when coefficients already cancel

Solve systems of two linear equations exactly and approximately using algebraic and graphical methods.

Elimination is immediate when one variable already has opposite coefficients. We’ll align the equations, add each matching column and both constants, and solve the one-variable equation left after cancellation. Substituting …

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A-REI.6 M1-012-A04-V01

Solve a system by elimination after deciding whether a multiplier is needed

Solve systems of two linear equations exactly and approximately using algebraic and graphical methods.

Before multiplying either equation, inspect the coefficient pairs to see whether cancellation is already available. We’ll compare their magnitudes and signs, scale only if necessary, and combine complete equations with …

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A-REI.6 M1-012-A05-V01

Solve a system by elimination after multiplying both equations

Solve systems of two linear equations exactly and approximately using algebraic and graphical methods.

When neither coefficient pair cancels, use a common multiple to create opposites efficiently. We’ll determine one multiplier for each equation, apply it to every variable term and constant, and add …

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A-REI.6 M1-012-A06-V01

Build and solve a system for a break-even context

Solve systems of two linear equations exactly and approximately using algebraic and graphical methods.

Each plan needs a total-cost model built from its fixed fee and its rate per month. We’ll use the same input and output variables for both plans, place fixed amounts …

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A-REI.6 M1-012-A07-V01

Estimate the solution of a system from a graph

Solve systems of two linear equations exactly and approximately using algebraic and graphical methods.

A graphical system solution is the single point shared by both lines. We’ll locate the crossing, read the horizontal and vertical axis scales independently, and record the horizontal coordinate first. …

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A-REI.6 M1-012-A08-V01

Estimate the solution of a system from a table

Solve systems of two linear equations exactly and approximately using algebraic and graphical methods.

A table shows an exact system solution only when both output columns match in the same row. We’ll compare the outputs row by row and look for adjacent inputs where …

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A-REI.6 M1-012-A09-V01

Classify a linear system as one solution, no solution, or infinitely many solutions

Solve systems of two linear equations exactly and approximately using algebraic and graphical methods.

A system’s classification comes from how its two lines relate, not from their intercepts alone. We’ll read each slope and intercept from slope-intercept form, compare the slopes to distinguish crossing, …

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A-REI.6 M1-012-A10-V01

Interpret a system solution in context

Solve systems of two linear equations exactly and approximately using algebraic and graphical methods.

A contextual ordered pair has meaning only after each coordinate is matched to its variable and unit. We’ll read the input first and the output second, use the fact that …

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A-REI.6 M1-012-A12-V01

Solve a system with fractions or decimals

Solve systems of two linear equations exactly and approximately using algebraic and graphical methods.

Decimal coefficients do not change the logic of elimination. We’ll inspect the coefficient signs for immediate cancellation, add matching terms while combining the decimal coefficients exactly, and solve the remaining …

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A-REI.6 M1-012-A14-V01

Write and solve a system from two context constraints

Solve systems of two linear equations exactly and approximately using algebraic and graphical methods.

The context provides two independent constraints on two ticket counts. We’ll define each variable with its ticket type, translate the total count and total cost separately, and solve the resulting …

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A-REI.6 M1-012-A15-V01

Verify whether an ordered pair solves a system

Solve systems of two linear equations exactly and approximately using algebraic and graphical methods.

An ordered pair solves a system only if it makes every equation true. We’ll preserve the x-then-y coordinate roles, substitute into each equation exactly as written, and compare each resulting …

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A-SSE.1.a M1-013-A01-V01

Interpret a coefficient as a rate or per-unit change

Interpret terms, factors, and coefficients in linear and exponential expressions.

A coefficient in context describes the output change attached to one unit of its variable. We’ll identify the number multiplying the ticket count, attach dollars per ticket as its compound …

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A-SSE.1.a M1-013-A02-V01

Interpret a constant term as a starting value or fixed amount

Interpret terms, factors, and coefficients in linear and exponential expressions.

The constant term is the part of a linear expression that does not depend on the input. We’ll identify the standalone addend, set the ticket count to zero, and see …

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A-SSE.1.a M1-013-A03-V01

Interpret each term in a multi-term expression

Interpret terms, factors, and coefficients in linear and exponential expressions.

Each variable term is one contribution to the total, formed by a per-item rate times the matching item count. We’ll pair every coefficient with its variable meaning, track the units …

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A-SSE.1.a M1-013-A04-V01

Interpret a factor outside parentheses as repeated groups

Interpret terms, factors, and coefficients in linear and exponential expressions.

Parentheses mark one complete group, while the outside factor tells how many identical copies of that group occur. We’ll interpret the contents inside first, then read the factor as a …

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A-SSE.1.a M1-013-A05-V01

Interpret the base in an exponential expression

Interpret terms, factors, and coefficients in linear and exponential expressions.

In an exponential model, the coefficient in front is the initial amount, the base is the repeated multiplier, and the exponent counts equal periods. We’ll identify those roles, convert the …

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A-SSE.1.a M1-013-A06-V01

Interpret the initial value in an exponential expression

Interpret terms, factors, and coefficients in linear and exponential expressions.

For an exponential model written as an amount times a powered factor, the starting value is exposed by testing time zero. We’ll separate the leading coefficient from the growth factor …

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A-SSE.1.a M1-013-A07-V01

Interpret the exponent in an exponential expression

Interpret terms, factors, and coefficients in linear and exponential expressions.

An exponent counts repeated uses of its base, so its contextual meaning comes from the model’s period scale. We’ll isolate the powered factor, connect one multiplication to one equal time …

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A-SSE.1.a M1-013-A08-V01

Choose the correct contextual meaning of a coefficient

Interpret terms, factors, and coefficients in linear and exponential expressions.

A number’s contextual role depends on where it appears in the formula. We’ll identify the input and output quantities first, locate the number multiplying the input, and require its units …

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A-SSE.1.a M1-013-A10-V01

Identify units for each part of a linear expression

Interpret terms, factors, and coefficients in linear and exponential expressions.

Units must stay consistent across a contextual equation. We’ll begin with the units supplied for the input and output, require each added term to match the output unit, and infer …

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