SOLUTION: victor has 16.25 in coin if there were twice as many 5 centavo coins as 25 centavo coin and as many 10 centavo coin as 5 centavo coin and 25 centavo coin combine how many coin of e

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Question 662421: victor has 16.25 in coin if there were twice as many 5 centavo coins as 25 centavo coin and as many 10 centavo coin as 5 centavo coin and 25 centavo coin combine how many coin of each type were there?
Answer by raymond123456(4) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: EXPLAIN simplification of an expression
Your Result:


YOUR ANSWER


  • This is an equation! Solutions: x=2.
  • Graphical form: Equation x%2Bx-x%2B2x=6 was fully solved.
  • Text form: x+x-x+2x=6 simplifies to 0=0
  • Cartoon (animation) form: simplify_cartoon%28+x%2Bx-x%2B2x=6+%29
    For tutors: simplify_cartoon( x+x-x+2x=6 )
  • If you have a website, here's a link to this solution.

DETAILED EXPLANATION

Look at .
Eliminated similar terms highlight_red%28+x+%29,highlight_red%28+x+%29,highlight_red%28+-x+%29,highlight_red%28+2%2Ax+%29 replacing them with highlight_green%28+%281%2B1-1%2B2%29%2Ax+%29
It becomes highlight_green%28+%281%2B1-1%2B2%29%2Ax+%29=6.
Look at .
Added fractions or integers together
It becomes %28highlight_green%28+3+%29%29%2Ax=6.
Look at highlight_red%28+%28highlight_red%28+3+%29%29%2Ax+%29=6.
Remove unneeded parentheses around factor highlight_red%28+3+%29
It becomes highlight_green%28+3+%29%2Ax=6.
Look at 3%2Ax=highlight_red%28+6+%29.
Moved these terms to the left highlight_green%28+-6+%29
It becomes 3%2Ax-highlight_green%28+6+%29=0.
Look at highlight_red%28+3%2Ax-6+%29=0.
Solved linear equation highlight_red%28+3%2Ax-6=0+%29 equivalent to 3*x-6 =0
It becomes highlight_green%28+0+%29=0.
Result: 0=0
This is an equation! Solutions: x=2.

Universal Simplifier and Solver


Done!
Solved by pluggable solver: Linear System solver (using determinant)
Solve:
+system%28+%0D%0A++++1%5Cx+%2B+1%5Cy+=+2%2C%0D%0A++++1%5Cx+%2B+-1%5Cy+=+0+%29%0D%0A++

Any system of equations:


has solution

or



(x=1, y=1}
Solved by pluggable solver: SOLVE a linear equation
Solve 2%2Ax%2B1=5. Move 1 to the right: 2%2Ax=5-1. Divide by 2: x=%285-1%29%2F2=4%2F2+=+2
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-2x%2B-3+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-2%29%5E2-4%2A1%2A-3=16.

Discriminant d=16 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--2%2B-sqrt%28+16+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-2%29%2Bsqrt%28+16+%29%29%2F2%5C1+=+3
x%5B2%5D+=+%28-%28-2%29-sqrt%28+16+%29%29%2F2%5C1+=+-1

Quadratic expression 1x%5E2%2B-2x%2B-3 can be factored:
1x%5E2%2B-2x%2B-3+=+1%28x-3%29%2A%28x--1%29
Again, the answer is: 3, -1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B-3+%29