i just have to simplify. what do you do with the variables?
1.) 12xy - 3y^2 over (divided by) 3xy
does it equal 4 - y^2 ???
2.) 14a^2b + 21ab^2 over 28ab
i just have to simplify. what do you do with the variables?
1.) 12xy - 3y^2 over (divided by) 3xy
does it equal 4 - y^2 ???
2.) 14a^2b + 21ab^2 over 28ab
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Verified answer
1. 4-y/x
2. a/2 + 3b/4
1) (12xy-3y^2)/3xy
3 divided by 3 is one and 3 divided by 12 is 4. So on your paper, cross out the 3's and put 3 times 4 next to the division part. Then the y. Since y^2=y X y, you will divide the y^2 by y. That gives you y. The y in the denominator also cancels out the y in the 12xy binomial. So your answer is 4x-y.
2) (14a^2b + 21ab^2)/28ab
Here we can use the GCF, which is the Greatest Common Factor. Since 7 fits into 14a^2b, 21av^2 and 28ab, we can divide all the NUMBERS by 7. Now it will read (2a^2b + 3ab^2)/4ab. Divide all by ab and it will read (2a + 3b)/4. Simplified.
not clear where the parenthesis are, so I have to guess...
1.) (12xy - 3y²)/3xy
12xy/3xy - 3y²/3xy
4 -y/x
2.) (14a²b + 21ab²)/28ab
14a²b/28ab + 21ab²/28ab
a/2 + 3b/4
1.) 12xy - 3y^2 / 3xy
4 - y^2
or
y^2 - 4
2.) 14a^2b + 21ab^2 / 28ab <---factor each equation
7(2a^2b + 3ab^2) / 7(4ab)
2a^2b + 3ab^2 / 4ab
can you factor out something? like in #1, you can factor and cancel 3y, then you have 3y(4x-y)/3xy
i simplified that and got 4-y, so yes
#2 i factored out 7ab and had 2a+3b/3 i think thats as far as you can go.
your answer in 1 is correct...you can't simplify 2