Equation Solver

(Handles Linear and Quadratic Equations)

Equation Graph (Quadratic Only)

Linear Equations: Word Problem Tutorial

Linear equations (where the highest power of the variable is 1, e.g., ax + b = c) are fundamental in solving many everyday problems.

General Step-by-Step Guide (Linear)

  • Read and Understand: What is the problem asking for? Note all given numbers and relationships.
  • Define a Variable: Let 'x' (or another letter) represent the unknown quantity.
  • Formulate the Equation: Translate the problem's words into a mathematical equation involving 'x'.
  • Solve the Linear Equation: Isolate 'x' by performing the same operations on both sides of the equation (add, subtract, multiply, divide).
    Example: For 2x + 5 = 11
    1. Subtract 5 from both sides: 2x = 11 - 5 => 2x = 6
    2. Divide by 2: x = 6 / 2 => x = 3
  • Check Your Answer: Substitute the value of 'x' back into the original problem statement or equation to see if it makes sense.
  • Answer the Question: Clearly state the answer with appropriate units.

Number Problems (Linear)

These involve relationships between numbers, often consecutive integers or sums/differences.

Example: Sum of Consecutive Integers

Problem: The sum of three consecutive even integers is 78. Find the integers.

Solution Steps:

  1. Variables: First even integer = x. Next = x + 2. Third = x + 4.
  2. Equation: x + (x + 2) + (x + 4) = 78.
  3. Solve: 3x + 6 = 78 => 3x = 72 => x = 24.
  4. Answer: The integers are 24, 26, and 28.

Age Problems (Linear)

These compare ages of people at different points in time (past, present, future).

Example: Current and Future Ages

Problem: Sarah is twice as old as her younger brother Michael. In 7 years, the sum of their ages will be 38. How old are they now?

Solution Steps:

  1. Variables: Michael's current age = x. Sarah's current age = 2x.
    In 7 years: Michael = x + 7, Sarah = 2x + 7.
  2. Equation: (x + 7) + (2x + 7) = 38.
  3. Solve: 3x + 14 = 38 => 3x = 24 => x = 8.
  4. Answer: Michael is 8, Sarah is 16.

Mixture Problems (Linear)

These problems involve combining two or more substances with different values or concentrations to create a mixture with a desired overall value or concentration.

Example: Mixing Coffee Blends

Problem: A merchant wants to mix coffee beans worth $5 per pound with 30 pounds of coffee beans worth $8 per pound to get a mixture that can be sold for $6 per pound. How many pounds of the $5 coffee beans should be used?

Solution Steps:

  1. Variables: Let x = pounds of $5 coffee beans.
    We have 30 pounds of $8 coffee beans.
    Total pounds of mixture = x + 30.
  2. Value Equation: The total value of the individual components must equal the total value of the mixture.
    Value of $5 beans: 5x
    Value of $8 beans: 8 * 30 = 240
    Value of mixture: 6(x + 30)
  3. Equation: 5x + 240 = 6(x + 30).
  4. Solve:
    5x + 240 = 6x + 180
    240 - 180 = 6x - 5x
    60 = x.
  5. Answer: 60 pounds of the $5 coffee beans should be used.

Percent & Interest Problems (Linear)

These involve calculating percentages of numbers, discounts, markups, or simple interest (Interest = Principal × Rate × Time, or I = PRT).

Example: Simple Interest Investment

Problem: You invest a total of $12,000 in two accounts. Part of the money is invested at 5% simple interest per year, and the rest is invested at 7% simple interest per year. After one year, the total interest earned from both investments is $740. How much was invested at each rate?

Solution Steps:

  1. Variables: Let x = amount invested at 5% (0.05 rate).
    Then, amount invested at 7% (0.07 rate) = 12000 - x.
    Time (T) = 1 year.
  2. Interest Equation: Interest from 5% account + Interest from 7% account = Total Interest.
    Interest from 5% account: x * 0.05 * 1 = 0.05x
    Interest from 7% account: (12000 - x) * 0.07 * 1 = 0.07(12000 - x)
  3. Equation: 0.05x + 0.07(12000 - x) = 740.
  4. Solve:
    0.05x + 840 - 0.07x = 740
    -0.02x + 840 = 740
    -0.02x = 740 - 840
    -0.02x = -100
    x = -100 / -0.02
    x = 5000.
  5. Answer: $5,000 was invested at 5%.
    $12,000 - $5,000 = $7,000 was invested at 7%.

Geometry Problems (Perimeter - Linear)

These involve the sum of the lengths of the sides of a polygon. For a rectangle, Perimeter = 2(Length + Width).

Example: Perimeter of a Rectangle

Problem: The length of a rectangle is 3 cm more than twice its width. If the perimeter of the rectangle is 60 cm, find its dimensions.

Solution Steps:

  1. Variables: Let width = w cm.
    Then length = 2w + 3 cm.
  2. Perimeter Equation: 2(Length + Width) = Perimeter
    2((2w + 3) + w) = 60.
  3. Solve:
    2(3w + 3) = 60
    6w + 6 = 60
    6w = 54
    w = 9.
  4. Answer: Width = 9 cm.
    Length = 2(9) + 3 = 18 + 3 = 21 cm. (Check: Perimeter = 2(21+9) = 2(30) = 60 cm).

Tips for Linear Problems

  • Identify the unknown you're trying to find and assign it a variable.
  • Break down complex sentences into smaller mathematical pieces.
  • Remember the order of operations (PEMDAS/BODMAS) when simplifying.
  • Be consistent with units throughout the problem.

Quadratic Equations: Word Problem Tutorial

Many real-world problems can be modeled and solved using quadratic equations (where the highest power of the variable is 2, e.g., Ax² + Bx + C = 0). This tutorial will guide you through the general process and provide examples for common problem types.

General Step-by-Step Guide (Quadratic)

  • Read and Understand: Identify what's asked. Note given info. Diagrams help!
  • Define Variables: Let 'x' be the primary unknown. Express other unknowns in terms of 'x'.
  • Formulate Equation: Translate words into math. Look for an Ax² + Bx + C = 0 structure.
  • Standard Form: Rearrange your equation to Ax² + Bx + C = 0.
  • Solve: Use the solver on this page by typing your equation.
  • Interpret Solutions: Quadratic equations often yield two solutions. Check if they make sense in the context of the problem. Discard extraneous ones (e.g., negative length, negative time).
  • Answer the Question: Ensure your final answer directly addresses the problem. Include units.

Area & Geometry Problems (Quadratic)

These often involve formulas for areas of shapes where dimensions are related, leading to a product of terms involving the variable.

Example: Area of a Rectangle

Problem: The length of a rectangular garden is 5 meters more than its width. If the area of the garden is 36 square meters, find the dimensions.

Solution Steps:

  1. Variables: Let width = x meters. Then length = x + 5 meters.
  2. Equation: Area = Length × Width => x(x + 5) = 36.
  3. Standard Form: x² + 5x - 36 = 0.
  4. Solve (using solver): A=1, B=5, C=-36 => x = 4 or x = -9.
  5. Interpret: Width cannot be negative, so x = 4 meters.
  6. Answer: Width = 4m, Length = 4+5 = 9m.

Consecutive Integer Problems (Quadratic)

These involve integers that follow each other and their products or sums of squares often lead to quadratic equations.

Example: Product of Consecutive Integers

Problem: The product of two consecutive positive integers is 156. Find the integers.

Solution Steps:

  1. Variables: Let the first integer = x. The next is x + 1.
  2. Equation: Product = 156 => x(x + 1) = 156.
  3. Standard Form: x² + x - 156 = 0.
  4. Solve: A=1, B=1, C=-156 => x = 12 or x = -13.
  5. Interpret: Integers must be positive, so x = 12.
  6. Answer: Integers are 12 and 13.

Work-Rate Problems (Quadratic)

When combining rates leads to equations with variables in denominators, cross-multiplication can result in quadratic terms.

Example: Two Pipes Filling a Tank

Problem: Pipe A can fill a tank in 'x' hours. Pipe B can fill the same tank in 'x + 2' hours. If both pipes working together can fill the tank in 2 hours and 24 minutes (2.4 hours), find how long each pipe takes alone.

Solution Steps:

  1. Variables: Pipe A time = x hrs. Pipe B time = x + 2 hrs. Combined time T = 2.4 hrs.
  2. Rates: Rate A = 1/x. Rate B = 1/(x+2).
  3. Equation: 2.4/x + 2.4/(x+2) = 1.
  4. Standard Form: 5x² - 14x - 24 = 0 (after clearing denominators and simplifying).
  5. Solve (using solver): A=5, B=-14, C=-24 => x = 4 or x = -1.2.
  6. Interpret: Time 'x' cannot be negative, so x = 4 hours.
  7. Answer: Pipe A takes 4 hours. Pipe B takes x+2 = 6 hours.

Motion (Distance/Speed/Time) Problems (Quadratic)

These use D=ST. Quadratics often arise when speeds or times are related, or when dealing with relative speeds, especially if it involves solving for time which might then appear in a denominator.

Example: Round Trip with Different Speeds

Problem: A car travels 240 km. On the return trip over the same distance, the speed was 20 km/h greater. If the total time for the round trip was 7 hours, what was the speed on the first leg of the journey?

Solution Steps:

  1. Variables: Speed on first leg = x km/h. Speed on return = x + 20 km/h.
  2. Times: T1 = 240/x. T2 = 240/(x+20).
  3. Equation: T1 + T2 = 7 => 240/x + 240/(x+20) = 7.
  4. Standard Form: 7x² - 340x - 4800 = 0.
  5. Solve (using solver): A=7, B=-340, C=-4800. Approx. x ≈ 60 or x ≈ -11.43.
  6. Interpret: Speed 'x' cannot be negative, so x ≈ 60 km/h.
  7. Answer: The speed on the first leg was approximately 60 km/h.

Age Problems (Quadratic)

When the problem involves products of ages at different times, or squares of ages.

Example: Product of Ages

Problem: The product of Reena's age 4 years ago and her age 2 years hence is 40. Find Reena's current age.

Solution Steps:

  1. Variables: Current age = x. Age 4 yrs ago = x - 4. Age 2 yrs hence = x + 2.
  2. Equation: Product = 40 => (x - 4)(x + 2) = 40.
  3. Standard Form: x² - 2x - 48 = 0.
  4. Solve: A=1, B=-2, C=-48 => x = 8 or x = -6.
  5. Interpret: Age 'x' cannot be negative, so x = 8 years.
  6. Answer: Reena's current age is 8 years.

Tips for Quadratic Problems

  • Look for phrases indicating products (e.g., "product of," "area") or squares.
  • Be careful when expanding products like (x+a)(x+b).
  • Always simplify to the Ax² + Bx + C = 0 form before using the solver.
  • Remember that quadratic equations can have 0, 1, or 2 real solutions. The context of the problem will determine which are valid.