Algebra House
  • Home
  • Ask a Question
  • Examples
  • State Test Practice
    • Algebra 1 State Test Practice
    • Geometry State Test Practice
  • ACT and SAT
  • Calculators
    • Problem Solver
    • Graphing Calculator
    • 3D Graphing Calculator
    • Scientific Calculator
  • Class
    • Algebra 2 Notes, Assignments, and Videos
    • Algebra 3 Notes, Assignments, and Videos
  • Contact
    • Contact Algebra House

Coin Problem

12/9/2014

 
Harold has $4.35 of change in quarters and dimes, with 5 more dimes than quarters. How many of each coin does he have?  
x = number of quarters
x + 5 = number of dimes   {there are 5 more dimes than quarters} 

25x + 10(x + 5) = 435   {number of coins times value of coin equals total value}
25x + 10x + 50 = 435   {used distributive property}
35x + 50 = 435   {combined like terms}
35x = 385   {subtracted 50 from each side}
x = 11   {divided each side by 35}
x + 5 = 16   {substituted 11, in for x, into x + 5} 

11 quarters and 16 dimes

- Algebra House


Comments are closed.

    Ask a Question





    Examples

    All
    All Word Problems
    Basic Math
    Combining Like Terms
    Distributive Property
    Equations Of Lines
    Exponents
    Factoring
    Foil Method
    Functions
    Geometry
    Graphing
    Imaginary Numbers
    Inequalities
    Linear Equations
    Literal Equations
    Miscellaneous
    Percents
    Pythagorean Theorem
    Quadratic Equations
    Radicals
    Simplifying
    Slopes And Intercepts
    Solving Equations
    System Of Equations
    Word Problems Age
    Word Problems Distance
    Word Problems Geometry
    Word Problems Integers
    Word Problems Misc.
    Word Problems Mixture
    Word Problems Money
    Word Problems Number



Algebra House