SOLUTION: Suppose you have a coin colllection of dimes and quaters containing 46 coins. if you have $6.70, how mant of each type of coin do you have?

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Question 221479: Suppose you have a coin colllection of dimes and quaters containing 46 coins. if you have $6.70, how mant of each type of coin do you have?
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose you have a coin collection of dimes and quarters containing 46 coins. if you have $6.70, how many of each type of coin do you have?

Step 1. Let x be the number of dimes.

Step 2. Let 0.10x be the dollar value of dimes

Step 3. Let 46-x be the number of quarters

Step 4. Let 0.25(46-x) be the dollar value of quarters.

Step 5. Then 0.10x+0.25(46-x)=6.70 be the total dollar value of dimes and quarters.

Step 6. The following steps will solve the equation in Step 5.

Solved by pluggable solver: EXPLAIN simplification of an expression
Your Result:


YOUR ANSWER


  • This is an equation! Solutions: x=32.
  • Graphical form: Equation 0.10x%2B0.25%2A%2846-x%29=6.70 was fully solved.
  • Text form: 0.10x+0.25*(46-x)=6.70 simplifies to 0=0
  • Cartoon (animation) form: simplify_cartoon%28+0.10x%2B0.25%2A%2846-x%29=6.70+%29
    For tutors: simplify_cartoon( 0.10x+0.25*(46-x)=6.70 )
  • If you have a website, here's a link to this solution.

DETAILED EXPLANATION

Look at 0.10%2Ax%2B0.25%2A%28highlight_red%28+46+%29-x%29=6.70.
Moved 46 to the right of expression
It becomes 0.10%2Ax%2B0.25%2A%28-x%2Bhighlight_green%28+46+%29%29=6.70.
Look at 0.10%2Ax%2B0.25%2A%28-x%2B46%29=highlight_red%28+6.70+%29.
Moved these terms to the left highlight_green%28+-6.70+%29
It becomes 0.10%2Ax%2B0.25%2A%28-x%2B46%29-highlight_green%28+6.70+%29=0.
Look at 0.10%2Ax%2Bhighlight_red%28+0.25%2A%28-x%2B46%29+%29-6.70=0.
Expanded term 0.25 by using associative property on %28-x%2B46%29
It becomes 0.10%2Ax-highlight_green%28+0.25%2Ax+%29%2Bhighlight_green%28+0.25%2A46+%29-6.70=0.
Look at 0.10%2Ax-0.25%2Ax%2Bhighlight_red%28+0.25%2A46+%29-highlight_red%28+6.70+%29=0.
Replace constants in expression: highlight_red%28+0.25%2A46+%29,highlight_red%28+-6.70+%29 with highlight_green%28+4.8+%29
It becomes highlight_green%28+4.8+%29%2B0.10%2Ax-0.25%2Ax=0.
Look at highlight_red%28+4.8+%29%2B0.10%2Ax-0.25%2Ax=0.
Moved 4.8 to the right of expression
It becomes 0.10%2Ax-0.25%2Ax%2Bhighlight_green%28+4.8+%29=0.
Look at highlight_red%28+0.10%2Ax+%29-highlight_red%28+0.25%2Ax+%29%2B4.8=0.
Eliminated similar terms highlight_red%28+0.10%2Ax+%29,highlight_red%28+-0.25%2Ax+%29 replacing them with highlight_green%28+%280.10-0.25%29%2Ax+%29
It becomes highlight_green%28+%280.10-0.25%29%2Ax+%29%2B4.8=0.
Look at %28highlight_red%28+0.10+%29-highlight_red%28+0.25+%29%29%2Ax%2B4.8=0.
Added fractions or integers together
It becomes %28-highlight_green%28+15%2F100+%29%29%2Ax%2B4.8=0.
Look at highlight_red%28+%28-highlight_red%28+15%2F100+%29%29%2Ax+%29%2B4.8=0.
Remove unneeded parentheses around factor highlight_red%28+15+%29,highlight_red%28+1%2F100+%29
It becomes -highlight_green%28+15+%29%2Fhighlight_green%28+100+%29%2Ax%2B4.8=0.
Look at -highlight_red%28+15+%29%2Fhighlight_red%28+100+%29%2Ax%2B4.8=0.
Factors 15 and 100 have greatest common factor (GCF) of 5. Reducing fraction.
It becomes -highlight_green%28+3+%29%2Fhighlight_green%28+20+%29%2Ax%2B4.8=0.
Look at highlight_red%28+-3%2F20%2Ax%2B4.8+%29=0.
Solved linear equation highlight_red%28+-3%2F20%2Ax%2B4.8=0+%29 equivalent to -0.15*x+4.8 =0
It becomes highlight_green%28+0+%29=0.
Result: 0=0
This is an equation! Solutions: x=32.

Universal Simplifier and Solver


Done!



With x=32, then 46-x=14. Check dollar value in Step 5.

0.10*32+0.25*14=3.20+3.50=6.70 which is a true statement.

Step 7. ANSWER: The number of dimes is 32 and the number of quarters is 14.

I hope the above steps were helpful.

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Good luck in your studies!

Respectfully,
Dr J