Question

How do I solve the following equations for the value of x: 3x+=4x-10, 4x+10=6x-8, 3x+10=5x, 3x-11=8x-21, 3(x+8)=12

Answer 1

look like there is a number missing form the first one
but lets look at the second:-
4x + 10 = 6x - 8
we need to bring all the terms in x ( that is 4x and 6x) to one side of the '=' and the numbers on the other side

Answer 2

the equation for the first one is not missing any numbers

Answer 3

so first we subtract 4x from LHS (left hand side):-
4x - 4x + 10 = 6x - 4x - 9
4x - 4x = 0 and 6x - 4x = 2x so:-
10 = 2x - 9
now we need to get the 9 on LHS
so since -9 + 9 = 0 we add 9 to both sides:
10 + 9 = 2x - 9 + 9
19 = 2x
now if we divide both sides by 2 we get the value of x
19/2 = 2x / 2

Answer 4

ok - with the first one I thought it might be because you have a + with nothing after it

Answer 5

oops sorry - I made a typo in the second equation the last number is 8 not 9
so the last line will be
18/2 = 2x

so you only have to divide 18 by 2 and you have found the value of x and the equation is solved

Answer 6

do you follow the method ok?

Answer 7

the first 4 equation can be solved in the same way
eqaution 5
3(x + 8) = 12
you first have to distribute the 3 over the parentheses
so its
3*x + 3*8 = 12
3x + 24 = 12

Answer 8

Are you having difficulties solving this?

Answer 9

can you help me solve the rest please I'm having little difficulties