Solve each equation. Then check your solution.
Please help me on this! Thanks in advance. :)
For both of these math questions, it seems like there is no solution. Am I doing this correctly?
5y - 2y = 3y + 2
My current answer is 0=2, no solution. Is this correct?
And another question. Please and thank you.
3(a+1) - 5 = 3a-2
2006-08-27 07:27:41 · 10 answers · asked by Shaikoh 2 in Science & Mathematics ➔ Mathematics
For the first, you are correct. There is no solution.
For the second:
3a + 3 - 5 = 3a - 2
3a - 2 = 3a - 2
This is true for all values of a.
2006-08-27 07:35:46 · answer #1 · answered by TychaBrahe 7 · 0⤊ 0⤋
There is no solution to the first equation.
5y - 2y = 3y + 2
3y = 3y + 2
3y - 3y = 2
0 = 2
For the second equation:
3(a + 1) - 5 = 3a - 2
3a + 3 - 5 = 3a - 2
3a - 2 = 3a - 2
3a - 3a = - 2 + 2
0 = 0
which means all possible values for a is correct.
Ü hope it helped u.
2006-08-27 07:37:01 · answer #2 · answered by ←deadstar→ 3 · 0⤊ 0⤋
I highly recommend doing your own homework, but since I am compassionate, I will answer your question.
Consider both problems geometrically. If we make the variables both the independent (x) variables, then we have 4 lines (2 per problem). In the first, both lines have the same slope (coefficient of the variable, = 3) but different intercepts (0 and 2), therefore, no intersection and so no solution.
In the second problem, again, both slopes are 3, but the intercepts are also equal, at -2. This means that they are the same line, and any value of the variable (a) satisfies the equation/is a solution.
2006-08-27 07:39:25 · answer #3 · answered by aristotle2600 3 · 0⤊ 0⤋
You are correct - there is no solution. After you combine terms, the equation is: 3y = 3y - 2. There's no way 3y can equal two less than itself.
The second equation is an identity: 3a - 2 = 3a - 2. Therefore, the solution is the set of all real numbers.
2006-08-27 07:41:23 · answer #4 · answered by pgd_malaka 6 · 0⤊ 0⤋
The idea is to combine like terms. So you would first work on the left side. This gives you 3y. So your equation now becomes 3y = 3y + 2. Then you need to combine like terms from one side of the = to the other. This is done by subtracting 3y on both sides. So you get 3y - 3y = 3y + 2 - 3y which simplifies to 0 = 2. Since this statement is not true, there is NO SOLUTION. Which means there is no finite value of y that will satisfy the equation.
2006-08-27 07:34:36 · answer #5 · answered by pecosbill2000 3 · 0⤊ 0⤋
5y - 2y = 3y + 2
3y = 3y + 2
0 = 2
No Solution
3(a + 1) - 5 = 3a - 2
3a + 3 - 5 = 3a - 2
3a - 2 = 3a - 2
No Solution
2006-08-27 07:34:49 · answer #6 · answered by Anonymous · 0⤊ 0⤋
Where did you get this equation: 5y-2y=3y+2???
You better check them out.
2006-08-27 07:33:31 · answer #7 · answered by Beauty 3 · 0⤊ 0⤋
No your answer is not correct.
In first case y is infinity. U add anything to infinity, its infinity. Math do have solution to each equations, isn't it?
For second case all the values of "a" are correct.
2006-08-27 07:32:10 · answer #8 · answered by Anonymous · 0⤊ 1⤋
Yep.....when nothing else to do I start around page 6 and work backwards.
2016-03-26 21:58:51 · answer #9 · answered by Anonymous · 0⤊ 0⤋
i would say you are correct on both results
2006-08-27 07:31:09 · answer #10 · answered by Soledad 2 · 0⤊ 0⤋