You have 14 coins that are nickles and quarters that equals 2 dollars how do you solve using substitution?

The question is incorrect but if half nickels and quarters exist then you could say the answer is 7.5 nickels and 6.5 quarters.

The problem:

14 coins that are either nickels are quarters

The total dollar amount of these coins is 2 dollars

Step-by-step explanation:

Let n be the number of nickels.

Let q be the number of quarters.

We are given that n+q=14 and that .05n+.25q=2.

I'm going to choose to solve the system by elimination.

Multiplying the first equation by .25 gives:

.25n+.25q=.25(14)

.25n+.25q=3.5

So let's line the equations up now:

.25n+.25q=3.5

.05n+.25q=2

If we subtract the equations, the variable q will be eliminated.                Thus, we will able to solve for the variable n.

.20n+0q=1.5

.20n=1.5

Dividing both sides by .2 gives:

n=7.5

If n=7.5 and n+q=14, then we have 7.5+q=14 by substitution property.

We can subtract 7.5 on both sides giving us:

q=14-7.5=6.5

We cannot have 7.5 nickels and 6.5 quarters but this satisfies the given information.

Check:

7.5+6.5=(7+6)+(.5+.5)=13+1=14

7.5(.05)+6.5(.25)=2

Now, the requested way which is by substitution:

Let n be the number of nickels.

Let q be the number of quarters.

We are given that n+q=14 and that .05n+.25q=2.

We can solve the first equation for either n or q and then substitute that into the other equation allowing us to solve for the other variable.

Let's solve for q by subtraction n on both sides:

q=14-n

We are going to replace q in the second equation with (14-n) giving us:

.05n+.25(14-n)=2

Distribute:

.05n+3.5-.25n=2

Combine like terms:

-0.2n+3.5=2

Subtract 3.5 on both sides:

-0.2n=-1.5

Divide both sides by -0.2:

n=7.5

Since q=14-n and n=7.5, then q=14-7.5=6.5 .

We are already check the solution above.

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.