You can write three equations in the numbers of nickels (n), dime (d), and quarters (q).
n + d + q = 23 . . . . . . . there are 23 coins total
0n +d -q = 2 . . . . . . . . .there are 2 more dimes than quarters
5n +10d +25q = 250 . .the total value is $2.50
The collection includes 11 nickels, 7 dimes, and 5 quarters.
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I used the matrix function of my calculator to solve these equations. You can find q by subtracting from the last equation five times the sum of the first two equations.
(5n +10d +25q) -5((n +d +q) +(d -q)) = (250) -5(23 +2)
25q = 125 . . . . . . . simplify
q = 5
From the second equation,
d = q +2 = 7
And from the first,
n = 23 -5 -7 = 11
Those are 185, 186 and 187
What you do is create this system of equations:
x+y+z=558 (three integers add up to 558)
x+1=y (they are consecutive, so that is translated as a number+1)
y+1=z (same logic as before)
And then you just resolve that system.
Answer:
See Explanation
Step-by-step explanation:
<em>The question is incomplete as what is required of the question is not stated.</em>
<em>However, since the question is only limited to distance, a likely question could be to calculate the distance from Bayville to Colleyville.</em>
Represent the distance from Atlanta to Colleyville with AC
Represent the distance from Atlanta to Bayville with AB
Represent the distance from Bayville to Colleyville with BC
So, we have that:
The relationship between AB, AC and BC is:
Make BC the subject of formula:
Convert fraction to decimal
<em>Hence, the distance from Bayville to Colleyville is 14.8 miles</em>
7x+y-9=0
move constant to the left by adding its opposite to both sides. the sum of the two =0
Combine the like terms by adding the coefficients,
->5x^4 can't add anything so leave it there,
->-3x^3 can add -3x^3 which gives you -6x^3,
->We can't do anything with -11x^2 so leave it there as well,
->6x add to -(-8x) gives you 14x,
You will get for your final answer is 5x^4-6x^3-11x^2+14x.
