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How do you create and solve linear equations and inequalities, including literal equations and contextual models?

Create equations and inequalities in one variable and use them to solve problems; solve linear equations and inequalities including those with variables on both sides; rearrange literal equations (formulas) to isolate a chosen variable; and graph the solution set of an inequality on a number line.

A NY Regents Algebra I answer on creating and solving linear equations and inequalities: variables on both sides, literal equations, contextual modeling, the sign-flip rule for inequalities, and graphing solutions on a number line.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. Solving linear equations
  3. Inequalities and the sign-flip rule
  4. Literal equations (rearranging formulas)
  5. Modeling a context
  6. Try this

What this topic is asking

The Regents Algebra I exam (the Creating Equations, A-CED, and Reasoning with Equations and Inequalities, A-REI, clusters) wants you to build a linear equation or inequality from a situation, solve it (even with variables on both sides), rearrange a formula to isolate any variable, and graph an inequality's solution on a number line. Linear reasoning is the single most tested idea on the exam, appearing in Part I and across the constructed-response parts.

Solving linear equations

The strategy is to isolate the variable by undoing operations in reverse. With variables on both sides, first gather them together.

7xβˆ’4=2x+11β€…β€Šβ‡’β€…β€Š5xβˆ’4=11β€…β€Šβ‡’β€…β€Š5x=15β€…β€Šβ‡’β€…β€Šx=3.7x - 4 = 2x + 11 \;\Rightarrow\; 5x - 4 = 11 \;\Rightarrow\; 5x = 15 \;\Rightarrow\; x = 3.

Always distribute first if there are parentheses, then combine like terms, then move variables to one side. A linear equation has exactly one solution unless the variable cancels: if you reach a true statement like 5=55 = 5 the equation has infinitely many solutions, and if you reach a false statement like 5=85 = 8 there is no solution.

Inequalities and the sign-flip rule

Inequalities are solved with the same moves as equations, with one crucial difference.

The solution set is graphed on a number line with an open circle for << or >> (endpoint not included) and a closed circle for ≀\leq or β‰₯\geq (endpoint included), shading toward all values that satisfy it.

Literal equations (rearranging formulas)

A literal equation has several letters, and you solve for one in terms of the others using the same inverse operations. To solve P=2l+2wP = 2l + 2w for ww: subtract 2l2l to get Pβˆ’2l=2wP - 2l = 2w, then divide by 2 to get w=Pβˆ’2l2w = \frac{P - 2l}{2}. Treat every other letter as a constant. This appears regularly as a 2-credit constructed-response item, and full credit requires the target variable alone on one side.

Modeling a context

The hardest part of a word problem is the translation, not the algebra. Name the unknown, turn each phrase into an expression, set up the equation or inequality, then interpret the answer in context. A subtle but heavily tested point is that the domain matters: a number of months, people, or items must be a whole number and cannot be negative, so a raw solution like m>5m > 5 may need rounding to "6 whole months". Regents graders award a specific credit for that contextual interpretation, so always finish a modeling problem with a sentence that answers the actual question.

Try this

Q1. Solve 2(xβˆ’3)=102(x - 3) = 10. [1 credit]

  • Cue. Distribute or divide first: xβˆ’3=5x - 3 = 5, so x=8x = 8.

Q2. Solve V=Ο€r2hV = \pi r^2 h for hh. [2 credits]

  • Cue. Divide both sides by Ο€r2\pi r^2: h=VΟ€r2h = \frac{V}{\pi r^2}.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (style)2 marksPart I (multiple choice). What is the solution to 5βˆ’2(x+3)=3xβˆ’165 - 2(x + 3) = 3x - 16? (1) x=3x = 3 (2) x=5x = 5 (3) x=βˆ’3x = -3 (4) x=βˆ’1x = -1
Show worked answer β†’

The correct answer is (1).

Distribute: 5βˆ’2xβˆ’6=3xβˆ’165 - 2x - 6 = 3x - 16, so βˆ’1βˆ’2x=3xβˆ’16-1 - 2x = 3x - 16. Add 2x2x to both sides: βˆ’1=5xβˆ’16-1 = 5x - 16. Add 1616: 15=5x15 = 5x, so x=3x = 3. The common error is distributing βˆ’2-2 only to xx and not to 33, which mishandles the constant. Substituting x=3x = 3 into both sides gives 5βˆ’2(6)=βˆ’75 - 2(6) = -7 and 3(3)βˆ’16=βˆ’73(3) - 16 = -7, confirming.

Regents (style)4 marksPart III (constructed response). A gym charges a 50signβˆ’upfeeplus50 sign-up fee plus 25 per month. A rival gym charges no sign-up fee but 35permonth.(a)Writeaninequalitytofindthenumberofmonths35 per month. (a) Write an inequality to find the number of months m$ for which the first gym costs less than the second. (b) Solve it and state, in context, after how many whole months the first gym becomes the cheaper option.
Show worked answer β†’

A 4-credit question: roughly 2 credits for the correct inequality and 2 for solving and interpreting.

(a) First gym cost is 50+25m50 + 25m; rival is 35m35m. The first costs less when 50+25m<35m50 + 25m < 35m.
(b) Subtract 25m25m: 50<10m50 < 10m, so m>5m > 5. Since mm must be a whole number of months, the first gym is cheaper once mβ‰₯6m \geq 6, that is, from the 6th month onward. Stating m>5m > 5 without translating to "6 whole months" typically costs the final interpretation credit.

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