Connor has 14 quarters and 28 dimes.
Step-by-step explanation:
Given,
Worth of coins = $6.30 = 6.30*100 = 630 cents
We know that,
1 quarter = 25 cents
1 dime = 10 cents
Let,
Quarter = x
Dime = y
According to given statement;
25x+10y=630 Eqn 1
The number of dimes is 14 more than the number of quarters
y = x+14 Eqn 2
Putting value of y from Eqn 2 in Eqn 1
![25x+10(x+14)=630\\25x+10x+140=630\\35x=630-140\\35x=490](https://tex.z-dn.net/?f=25x%2B10%28x%2B14%29%3D630%5C%5C25x%2B10x%2B140%3D630%5C%5C35x%3D630-140%5C%5C35x%3D490)
Dividing both sides by 35
![\frac{35x}{35}=\frac{490}{35}\\x=14](https://tex.z-dn.net/?f=%5Cfrac%7B35x%7D%7B35%7D%3D%5Cfrac%7B490%7D%7B35%7D%5C%5Cx%3D14)
Putting x=14 in Eqn 2
![y=14+14\\y=28](https://tex.z-dn.net/?f=y%3D14%2B14%5C%5Cy%3D28)
Connor has 14 quarters and 28 dimes.
Keywords: linear equation, substitution method
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