2x + 3y = 6
What is the question? You cannot solve for both x and y without a second independent equation, but you can express one variable in terms of the other. For example,
2x = 6 - 3y, so
x = -3y/2 + 3
solve each literal equation for the indicated varible
2x + 3y = 6 (for x)
next question
A=1/2 h (B +b ) ( for b)
Ok, if you multiply both sides by 2 that clears the fraction. If you then divide by h (with the requirement that h is non-zero), that clears h. Finally, subtracting B from both sides leaves only b on the right hand. I'll let you write the steps out.
Do you know what this equation is the formula for?
No rodger i have no clue what I am doing I am so confused I need someone to go over it step by step
No problem Tammy, I'll try to help whenever I'm on line and have some time to spare.
I only wanted to see if you recognized the formula for the area of a trapezoid. You expressed the smaller base in terms of the other variables here.
1--2x + 3y = 6
2--Dividing by 2 yields x + y + y/2 = 3
3--y/2 = 3 - x - y
4--y/2 must be an integer k making y = 2k
5--Substituting into (1) yields 2x + 6k = 6 making x = 3 - 3k
6--
k...0...1...2...3
x...3...0..-3..-6
y...0...2...4...6
Clearly, no pair of positive solutions exixt for both x and y.
Just a couple observations:
(1)In the second equation,why did you split y as y + y/2?
(2)In the third equation, why do you have y on both sides?
(3)In equation 4 why don't you simply state, y must be and even integer?
(4)Ok, so you now substitute back into equation 1. You've taken us a long way to state the following:
-2x+3y=6 and y must be an even integer, thus y = 2k. Substituting this gives, -2x+3(2k)=-2x+6k=6 which is the same as -x+3k=6, or 3k=x+6
Now we can state our "clearly" statement with a 'little' more clearness. Yes, if x is positive then x+6 is positive and so is 3k = x+3, hence k is positive. However, after we simplify a statement the word "clearly" should be unnecessary. The steps themselves should show the reader how "clear" the observation is implicitly.
I don't want to dissuade you from using "clearly", but if you submit some work using that expression and it comes back in error, be prepared for sarcasm.
Finally, it is clear to me that you have a good handle on algebraic operations. You should do well.
hmmm...maybe I'm doing too much math today. After reviewing my own work I see I took the long way too.
-2x + 3y = 6
1--2x + 3y = 6 (Do you men to use 2 minus signs, or this that a single one?)
If you mean a single minus sign then we have -2x + 3y = 6, thus 3y = 2x + 6.
Now use any positive value for x and you see y is positive. (1,7/3), (3,4) so forth. A typical response back would be "You're statement is 'clearly' false".
If you indeed meant -(-2)x + 3y = 6, then we have 2x +3y = 6, and (1, 4/3) works. Still false. However, if you meant this and had required only positive integer solutions then your statement would be true (check for yourself); how clear this is would depend on the reader.
bro its 2018
I apologize for the confusion. Let's go through the steps again to solve for x in the equation 2x + 3y = 6.
Step 1: Start with the equation 2x + 3y = 6.
Step 2: Subtract 3y from both sides of the equation to isolate the term with x. This gives us 2x = 6 - 3y.
Step 3: Simplify the right side of the equation by combining like terms. This gives us 2x = -3y + 6.
Step 4: Divide both sides of equation by 2 to solve for x. This gives us x = (-3y + 6) / 2.
So, the solution for x is x = (-3y + 6) / 2.
Now, let's move on to the next question:
A = 1/2 h(B + b), solve for b.
Step 1: Start with the equation A = 1/2 h(B + b).
Step 2: Clear the fraction by multiplying both sides of the equation by 2. This gives us 2A = h(B + b).
Step 3: Distribute the h to the terms inside the parentheses. This gives us 2A = hB + hb.
Step 4: Subtract hB from both sides of the equation to isolate the term with b. This gives us 2A - hB = hb.
Step 5: Divide both sides of the equation by h to solve for b. This gives us b = (2A - hB) / h.
So, the solution for b is b = (2A - hB) / h.
I hope that helps! Let me know if you have any further questions.