solve 3x-7=2x+8 use rules of algebra to solve?

Answer 1

3x - 7 = 2x + 8
x = 8 + 7
x = 15

Answer 2

Using algebra to isolate x, you want to move all of the X terms to one side and all of the constant terms to the other.

3x-7 = 2x + 8 subtract 7 from both sides

3x = 2x + 15 subtract 2x from both sides

x = 15

Answer 3

get x on one side and add like terms

3x - 7 = 2x + 8
subtract 2x from each side
3x - 7 - 2x = 2x + 8 - 2x
x - 7 = 8
add 7 to both sides
x - 7 + 7 = 8 + 7
x = 15

Answer 4

14√(x) is the same as x^(1/14) when dealing with long radicals or it best to turn them into powers first that why i did it the way i did last time but this time i will do it in steps for you {7√[a^(-3x+10)b^(x-5)c^(2x+7)]}/{14√[a^... first we got the a part {7√[a^(-3x+10)]}/{14√[a^(20x-8)]} first steps is to make it all one radical if double the exponent of the top part then we can change that radical go from a 7 to a 14 so we have same on top and bottom {14√[a^(-6x+20)]}/{14√[a^(20x-8)]} now we have same radical on top and bottom so we can pull it all the way out 14√{[a^(-6x+20)]/[a^(20x-8)]} ok now we want to pull those numbers off the bottom to do this we just need subtract our exponents like this 14√[a^((-6x+20)-(20x-8)) 14√[a^(-26x+28)] ok part a done {7√[b^(x-5)]}/{14√[b^(-6x+2)]} {14√[b^(2x-10)]}/{14√[b^(-6x+2)]} 14√{[b^(2x-10)]/[b^(-6x+2)]} 14√[b^(8x-12)] ok part b done {7√[c^(2x+7)]}/{14√[c^(-2x-10)]} {14√[c^(4x+14)]}/14√[c^(-2x-10)]} 14√{[c^(4x+14)]/[c^(-2x-10)]} 14√[c^(6x+24)] ok part c done now bring all three parts back together 14√{[a^(-26x+28)][b^(8x-12)][c^(6x+24)]... hope this time i was easier to understand EDIT ok i see someone found a way to simply sorry missed that should be 7√{[a^(-13x+14)][b^(4x-6)][c^(3x+12)]}

Answer 5

3x-7=2x+8
3x - 2x = 8 + 7
1x =15
x = 15/1
x = 15

Answer 6

3x-7=2x+8
x=15

Get the x to one side and the numbers to the other and solve.

Answer 7

3x - 2x = 8 + 7
x = 15

Answer 8

3x-7=2x+8

subtract 2x from each side:
x-7=8

add 7 to each side:
x=15

15 is your answer

Answer 9

x=15.
i can do those in my head now.
basically in algebra you can do the same thing to both sides of an equation. like...if i said 25 cents = 1 quarter. then added 5 pennies to both sides, i would have the same amount on both sides now since i added the same thing to both sides. same thing when you are solving equations. the goal is to get the variable alone on one side of the equation. to do this you add/subtract and later multiply and divide to get this. it takes practice. u should ask your teacher if you don't understand a concept.