Use the distributive property to rewrite the algebraic expression. 3(x + 7)
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Answer:
a.
b. x = -8 and y = 4
Step-by-step explanation:
This question is incomplete. I will type the complete question below before giving my solution.
For integers a, b, c, consider the linear Diophantine equation
Suppose integers x0 and yo satisfy the equation; that is,
what other values
and
also satisfy ax + by = c? Formulate a conjecture that answers this question.
Devise some numerical examples to ground your exploration. For example, 6(-3) + 15*2 = 12.
Can you find other integers x and y such that 6x + 15y = 12?
How many other pairs of integers x and y can you find ?
Can you find infinitely many other solutions?
From the Extended Euclidean Algorithm, given any integers a and b, integers s and t can be found such that
the numbers s and t are not unique, but you only need one pair. Once s and t are found, since we are assuming that gcd(a,b) divides c, there exists an integer k such that gcd(a,b)k = c.
Multiplying as + bt = gcd(a,b) through by k you get
So this gives one solution, with x = sk and y = tk.
Now assuming that ax1 + by1 = c is a solution, and ax + by = c is some other solution. Taking the difference between the two, we get
Therefore,
This means that a divides b(y−y1), and therefore a/gcd(a,b) divides y−y1. Hence,
for some integer r. Substituting into the equation
or
Thus if ax1 + by1 = c is any solution, then all solutions are of the form
In order to find all integer solutions to 6x + 15y = 12
we first use the Euclidean algorithm to find gcd(15,6); the parenthetical equation is how we will use this equality after we complete the computation.
Therefore gcd(6,15) = 3. Since 3|12, the equation has integral solutions.
We then find a way of representing 3 as a linear combination of 6 and 15, using the Euclidean algorithm computation and the equalities, we have,
Because 4 multiplies 3 to give 12, we multiply by 4
So one solution is
&
All other solutions will have the form
where ∈ Ζ
Hence by putting r values, we get many (x, y)
the scale factor or ratio of two similar figures, will be the same as the ratio of their perimeter, or even the ratio of two corresponding sides.
for example, we have say, an irregular octagon with a perimeter of say 13, and a similar octagon with a perimeter of 39, then
which is another way to say, the smaller is 1/3 of the larger one, or the larger one is 3 times as the smaller one.