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Pre algebra solving for x

Introduction :

Pre algebra solving for x means to find the value of x from the given expression. The given expression may have an integer values or any other variables. To get the value of x we have to do some addition, subtraction, multiplication or division methods. Here x is an unknown variable. Even though there are many known variables, we have to solve only for x.

Example problems – Pre algebra solving for x:
  • Example 1 - Pre algebra solving for x:

Solve the given expression for x: x + 23 = 120

Solution:

The given expression is x + 23 = 120

Subtract 23 on both sides,

x + 23 = 120

x + 23 - 23 = 120 - 23

`x = 97`

  • Example 2 - Pre algebra solving for x:

Solve the given expression for x: 2x + 8 = 44.

Solution:

The given expression is 2x + 8 = 44.

Subtract 8 on both sides,

2x + 8 - 8 = 44 - 8.

2x = 36

Divide by 2 on both sides,

`(2x) / 2 = 36 / 2`

`x = 18`

  • Example 3 - Pre algebra solving for x:

Solve the given expression for x: x + 3a = 54

Solution:

The given expression is x + 3a = 54

Subtract 3a on both sides,

x + 3a - 3a = 54 - 3a

`x =54-3a`

The answer is in terms of a.

  • Example 4 - Pre algebra solving for x:

Solve the given expression for x: `(x + 9) / p = 2 + q `

Solution:

The given expression is `(x+9)/p` = 2 + q

Multiply by 'p' on both sides,

`((x+9)/p)xxp= (2+q)xxp`

x + 9 = 2p + pq

Subtract 9 on both sides

x + 9 - 9= 2p + pq - 9

`x = pq + 2p - 9`

  • Example 5 - Pre algebra solving for x:

Solve the given expression for x: `(x - m) / (2m + 1) = 3`

Solution:

The given expression is `(x-m)/(2m+1)= 3`

Now multiply by 2m+1 on both sides

`((x-m)/(2m+1))xx(2m+1)= 3 xx(2m+1)`

x – m = 6m + 3

Add 'm' on both sides

x – m + m = 6m + 3 + m

`x = 7m + 3`

Practice problems - Pre algebra solving for x:
  • Problem 1:

Solve the given expression for x: x + 8c = 1/3   

Answer: `x = (1-24c)/3`

  • Problem 2:

Simplify the Expression for x: 2x+4 =8          

Answer: `x = 2`

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