You know that there are 46 dimes and quarters in a jar. When you count up the money you find you have a total of $9.40. How many dimes are there? How many quarters?
setup a system of equations d + q = 46 .10d + .25q = 9.40
So do you solve for D or Q? and how
i would solve for d.imes and q.uarters yes. and your material should give you at least 3 ways to approach a solution. use the one you are most comfortable with.
elimination seems to be the simplest approach to me since the top is practically bare of coeffs ... but many like subsitution better
let d = (23+n) and q=(23-n) d + q = 46 23 + n + 23 - n = 46 ... by default substitute them into the second equation and solve for n.
Ummm ok how do you do substitution? I know what substitution is and I know the equation, I just don't know how to use the two together.
Where does 23 come from?
one type of substitution i just mentioned the other type is to solve one equation for a term, and replace it in the other equation if ax + by = c, we can solve for a: a = (c-by)/x this is one type of relationship that must be satisfied so we simply let all a's be (c-by)/x
23 is a convenient number since 23 + 23 = 46 and n-n = 0 its just useful to define them that way
d = 23 + n + q = 23 - n ------------ 46 = 46 + 0
.10d + .25q = 9.40 substitute for d and q .10(23+n) + .25(23-n) = 9.40 now we only have one variable to solve
n= -9
very good now we know d and q d = 23 + n = 23 - 9 q = 23 - n = 23 + 9
So what do we do now?
^^^^ look at the post above ..
d = 23 + n = 23 - 9 q = 23 - n = 23 + 9 doesn't answer the question
im pretty sure it does ..... what question are we trying to answer?
How many dimes? How many quarters?
so how many d.imes do we have?
d = 23 + n, and n=-9
14?
yep how many q.uarters so we have? q = 23 - n ; and n=-9
or if you want to use our fundamental equation: 14 + q = 46
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