7x-1 = 2x+5
First get the unknowns on one side of the equation.
7x - 2x = 6
5x = 6
x = 1.2
You can check that by substituting.
7(1.2) - 1 = 2(1.2) + 5
8.4 - 1 = 2.4 + 5
7.4 = 7.4
Now -- please try a couple on your own. We'll be glad to check them for you.
I have to do 10 algebra problems by tomorrow........and i don't know how..... :(
1. 7x-1 = 2x+5
2. 3y+1 = 4y-6
3. 10-y = 2y+3
4. 6x+2 = 1-x
5. 3a-10 = 4+a
6. 5y+4 = 4y+5
7. b+3 = 2b-7
8. 2x-1 = x-3
9. 2w+3 = 4w-5
10. 6n+4 = n-11
please help in any way you can...thnx
7 answers
ok.
3y+1=4y-6 is Y= 1
umm.... i think that i need more help with 10-y=2y+3.....
6x+2=1-x is X= 0.42
3a-10=4+a is A= 7
5y+4=4y+5 is Y= 1
b+3=2b-7 is B= 3.3
2x-1=x-3 is X= 2
2w+3=4w-5 is W= 1.3
6n+4=n-11 is N= 2.14
3y+1=4y-6 is Y= 1
umm.... i think that i need more help with 10-y=2y+3.....
6x+2=1-x is X= 0.42
3a-10=4+a is A= 7
5y+4=4y+5 is Y= 1
b+3=2b-7 is B= 3.3
2x-1=x-3 is X= 2
2w+3=4w-5 is W= 1.3
6n+4=n-11 is N= 2.14
umm....does any1 kno if this is rite?
3y+1=4y-6 is Y= 1
1 + 6 = y
7 = y
10-y=2y+3
10 - 3 = 3y
1 + 6 = y
7 = y
10-y=2y+3
10 - 3 = 3y
I have this word problem that I can't figure out. The problem is:
A 15 lb weight is positioned 10 in. form a fulcrum. At what distance from the fulcrum must a 12 lb weigt be positioned to keep the scale balanced?
The substitute that I had today said that the 12 lb. weight would have to be further away from the fulcrum than the 15 lb., but my mom says the opposite.
A 15 lb weight is positioned 10 in. form a fulcrum. At what distance from the fulcrum must a 12 lb weigt be positioned to keep the scale balanced?
The substitute that I had today said that the 12 lb. weight would have to be further away from the fulcrum than the 15 lb., but my mom says the opposite.
5y+4=4y+5
I have a trapezoid, and the bases are lengths 18 inches and 12 inches. Dividing the trapezoid into unequal parts are triangles. In other words, the trapezoid is divided into 7 unequal triangles. 4 triangles are shaded. How do I find the area of the shaded region of the trapezoid?