how would you solve 2/3x = 24?

Question

and 3(x-5) +7 = 16? could you please explain?

Answer 1

To solve 2/3x = 24, multiply both sides by 3 (to get 2x = 72), and then divide both sides by 2 (to get x = 36).

As for the other

3(x - 5) + 7 = 16
3(x - 5) = 16 - 7 = 9 (subtract the 7 from both sides)
3x -15 = 9 (do the multiplication)
3x = 9 + 15 = 24 (add 15 to both sides)
x = 8 (divide both sides by 3)

Alternatively, after subtracting the 7 from both sides, you could divide both sides by 3:

x - 5 = 3
x = 5 + 3 = 8 (add 5 to both sides)

Hope this helps.

Answer 2

2/3x = 24
2x = 24 * 3
x = 72 : 2
x = 36
Solution: {x belongs to R| x = 36}.
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Evaluate:
3(x-5) +7 = 16
3x - 15 + 7 = 16
3x = 16 + 15 - 7
3x = 24
x = 24 -:- 3
x = 8
Solution: {x belongs to R| x = 8}
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Answer 3

When your solving equations just think about going in reverse order of operations.

If your adding, then subtract. If you dividing, then multiply.
When solving equations, you'll need to use the inverse operation.

2/3x = 24
since it's divided by 3 , then multiply both sides by 3

2x = 72
since it's times 2, then divide by 2
x = 36.

Next equation:

get rid of the parathesis by using the distributive property

3x-15 + 7 = 16.
since its add 7, then subtract 7

3x-15 = 9
since is subtract 15, then add 15

3x = 24
since is multiplied by 3, then divide by 3
x = 8

Answer 4

(2/3)x = 24
x = 24(3/2) = 36

3(x-5) + 7 = 16
3(x-5) = 9
x - 5 = 3
x = 8

Answer 5

yet otherwise of questioning of this - i became into taught via my instructor as quickly as of imagining the = sign as a bridge, and you're able to desire to get all the a number of 'gadgets' (to that end comparable variables mutually with all 'x' mutually) onto the two edge of the bridge... yet every time they bypass THE BRIDGE, they 'replace factors' - and so replace into ITS opposite (or if its a function replace into its 'anti'-function). So making use of this concept, at the start you're able to be able to confirm to get all the numbers (constants) on the main suitable hand edge and all the 'x' words on the left. that's finished in whichever order you like, as long as you you ought to prepare the typical rules... 5x + 2 = 3x + 24 pass +2 to the RHS 5x = 3x + 24 - 2 pass 3x to the LHS 5x - 3x = 24 - 2 we now have comparable words amassed on the two edge, we are in a position to simplify them via common addition and subtraction: (5 - 3)x = 24 - 2 2x = 22 Now there is largely one term of each sort on the two edge of the =, we are in a position to confirm what x is (that's 1x) via shifting the *2 over, as 2x = 2*x = x*2. So changing 2x with 2*x we get... x*2 = 22 shifting *2 over we get /2 (common Multiplications continuously replace to Divisions) x = 22 / 2 x = eleven end : you have solved the equation!

Answer 6

Question 1
(2/3).x = 24
2x = 72
x = 36

Question 2
3x - 15 + 7 = 16
3x = 24
x = 8

Answer 7

note

2/3 x = 24
multiply both sides by the reciprocal 3/2

(3/2)(2/3)x = (24/1)(3/2)
x = 12(3) = 36