What is the solution for 2x + y =12 x = 9 - 2y
1 answer:
You might be interested in
Answer: y =
Step-by-step explanation:
x+6y=12
6y=-x+12
y=- + 2
If its parallel it means that the slope will be the same
y = mx + b
-4 = -1/6 * (-6) + b
b = -5
Thus,
y = -1/6 x - 5
<h3>Distributive Property</h3>
The distributive property lets you group or ungroup terms using parentheses. It lets you multiply an external factor by every term in parentheses, expressing the result as a sum:
... a(b + c) = ab + ac . . . . . . factor <em>a</em> multiplies the terms <em>b</em> and <em>c</em>
and it lets you remove a common factor from different terms, putting that factor outside parentheses:
... ab + ac = a(b + c)
The letters <em>a</em>, <em>b</em>, <em>c</em> here can stand for any number or expression.
<h3>Homework</h3>20. To do this problem, you need to eliminate the parentheses using the distributive property. Then, "collect terms," which is another application of the distributive property. The external factor outside the parentheses is (-2/3). Multiply that by each term in parentheses, and add the results.
After that, recognize that c is a factor of two of the terms. Add their coefficients to simplify that sum to one term.
22. You are to evaluate the expression with x=2 two ways: as is, and after you simplify it by combining like terms.
<u>As is:</u>
<u>Simplified:</u>
I prefer simplifying the expression first. The number of calculations and chances for error are reduced.
23. It is convenient to check for equivalence after simplifying both expressions.
<u>First Expression:</u>
<u>Second Expression:</u>
The simplified forms of the expressions are identical, so we conclude the expressions are equivalent.
25. The area of a rectangle is the product of its length and width. Here, you are asked to simplify the product of (3+x) ft and 3 ft.
In the last form of this expression, we have used "standard form" which has the degree of the variable decreasing in terms left to right. The units are factored out to make the expression a bit less cumbersome.