Asked by Ruthie
perform indicated operations.
(a+3)/(5a+25)-(a-1)/(3a+15)
(a+3)/5(a+5)-(a-1)/3(a+5)=(a+3)5(a+5)/5(a+5)3(a+5)-(a-1)5(a+5)/5(a+5)3(at5)= 5a^2+25a+15a+75+5a^2+25a-5a-25/5(a+5)3(a+5)
i got (10a^2+60a-50)/5(a+5)3(a+5)and that's wrong. where did i go wrong?
(a+3)/(5a+25)-(a-1)/(3a+15)
(a+3)/5(a+5)-(a-1)/3(a+5)=(a+3)5(a+5)/5(a+5)3(a+5)-(a-1)5(a+5)/5(a+5)3(at5)= 5a^2+25a+15a+75+5a^2+25a-5a-25/5(a+5)3(a+5)
i got (10a^2+60a-50)/5(a+5)3(a+5)and that's wrong. where did i go wrong?
Answers
Answered by
Steve
(a+3)/(5a+25)-(a-1)/(3a+15)
(a+3)/5(a+5)-(a-1)/3(a+5)
=(a+3)<b>3</b>(a+5)/5(a+5)3(a+5)-(a-1)5(a+5)/5(a+5)3(a+5)
Aside from that, I'd not include (a+5) twice, making it
=3(a+3)/15(a+5) - 5(a-1)/15(a+5)
= (3a+9-5a+5)/15(a+5)
= 2(7-a)/15(a+5)
(a+3)/5(a+5)-(a-1)/3(a+5)
=(a+3)<b>3</b>(a+5)/5(a+5)3(a+5)-(a-1)5(a+5)/5(a+5)3(a+5)
Aside from that, I'd not include (a+5) twice, making it
=3(a+3)/15(a+5) - 5(a-1)/15(a+5)
= (3a+9-5a+5)/15(a+5)
= 2(7-a)/15(a+5)
Answered by
Reiny
Since the a+5 is already common, you don't need it twice
LCD = 15(x+5)
so your second line
= 3(a+3)/(15(a+5)) - 5(a-1)/(15(a+5))
= (3a+9 - 5a + 5)/(15(a+5))
= (-2a + 14)/(15(a+5))
=-2(a-7)/(15(a+5))
LCD = 15(x+5)
so your second line
= 3(a+3)/(15(a+5)) - 5(a-1)/(15(a+5))
= (3a+9 - 5a + 5)/(15(a+5))
= (-2a + 14)/(15(a+5))
=-2(a-7)/(15(a+5))
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