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:
Solution:
The given expression is x + 23 = 120
Subtract 23 on both sides,
x + 23 = 120
x + 23 - 23 = 120 - 23
`x = 97`
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`
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.
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`
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:
Answer: `x = (1-24c)/3`
Answer: `x = 2`
I like to share this Algebra Word Problems with you all through my blog.
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:
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:
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:
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:
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:
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:
Answer: `x = (1-24c)/3`
- Problem 2:
Answer: `x = 2`
I like to share this Algebra Word Problems with you all through my blog.
